"clicliclic@freenet.de" schrieb:
>
> Sam Blake schrieb:
> >
> > Here are some nice cube root-type integrals which may or may not be
> > integrable using your Goursat adaption. I am guessing that as Rubi
> > cannot integrate these that they may not be Goursat type integrals?
> >
> > (1 + x)/(x*(-1 + x^3)^(1/3))
> > (-1 + x^2)/(x*(-1 + x^3)^(2/3))
> > ((1 + x)*(-1 + x^3)^(1/3))/((-1 + x)^2*x)
> > ((-1 + x)^2*(1 + x))/(x*(1 + x + x^2)*(-1 + x^3)^(1/3))
> > ((1 + x)*(-1 + x^3)^(2/3))/((-1 + x)^3*x)
> > (x*(-2 + x^3))/((-1 + x^3)^(1/3)*(-1 + x^3 + x^6))
> > (x^3*(2 + x^3))/((1 + x^3)^(2/3)*(1 + x^3 + x^6))
> > ((1 - x^3)^(1/3)*(-2 + x^3))/(x^3*(-1 + x^3 + x^6))
> > ((2 + x + x^2)*(-1 + x^3)^(1/3))/(x*(-1 + x^2)^2*(-3 - 2*x + x^2 + x^3))
> > (1 + 3*x)/((-1 + 3*x)*(-x + x^3)^(1/3))
> > ((-1 + x)*(1 + 3*x))/((-1 + 3*x)*(-x + x^3)^(2/3))
> > ((-1 + x)^2*(1 + 3*x))/(x*(1 + x)*(-1 + 3*x)*(-x + x^3)^(1/3))
> > (x*(1 + x)*(1 + 3*x))/((-1 + x)*(-1 + 3*x)*(-x + x^3)^(2/3))
> > (x*(1 + x)*(1 + 3*x))/((-1 + x)^2*(-1 + 3*x)*(-x + x^3)^(1/3))
> >
>
> There are 14 integrands here. The first two are readily integrated by
> Derive 6.10 and should therefore be doable by standard recipes from
> books like G&R and Timofeev:
>
> INT((x + 1)/(x*(x^3 - 1)^(1/3)), x)
> INT((x^2 - 1)/(x*(x^3 - 1)^(2/3)), x)
>
> So Rubi could and should be tought to do these as well.
>
> The subsequent twelve integrands, however, fail in Derive 6.10:
>
> INT((x + 1)*(x^3 - 1)^(1/3)/(x*(x - 1)^2), x)
> INT((x + 1)*(x - 1)^2/(x*(x^2 + x + 1)*(x^3 - 1)^(1/3)), x)
> INT((x + 1)*(x^3 - 1)^(2/3)/(x*(x - 1)^3), x)
> INT(x*(x^3 - 2)/((x^3 - 1)^(1/3)*(x^6 + x^3 - 1)), x)
> INT(x^3*(x^3 + 2)/((x^3 + 1)^(2/3)*(x^6 + x^3 + 1)), x)
> INT((1 - x^3)^(1/3)*(x^3 - 2)/(x^3*(x^6 + x^3 - 1)), x)
> INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),
> x)
> INT((3*x + 1)/((3*x - 1)*(x^3 - x)^(1/3)), x)
> INT((x - 1)*(3*x + 1)/((3*x - 1)*(x^3 - x)^(2/3)), x)
> INT((x - 1)^2*(3*x + 1)/(x*(x + 1)*(3*x - 1)*(x^3 - x)^(1/3)), x)
> INT(x*(x + 1)*(3*x + 1)/((x - 1)*(3*x - 1)*(x^3 - x)^(2/3)), x)
> INT(x*(x + 1)*(3*x + 1)/((x - 1)^2*(3*x - 1)*(x^3 - x)^(1/3)), x)
>
> My Goursat tests for cube-root integrands determine all but the four
> integrals from #6 to #9 to be directly of Goursat type, which means
> that the integrands become rational under one of the characteristic
> substitutions. Of the other four, the first three become Goursat
> integrable under the fairly obvious substitution t = x^3.
>
> A more involved substitution must have been used to scramble integrand
> #9; even FriCAS fails on this one, but succeeds once t = x^3 has been
> applied. The substitution doesn't make the integrand look nicer
> though, and while the change cannot be considered momentous either,
> for FriCAS it evidently is. The integrand simplifies a bit when an
> algebraic part of the antiderivative is removed:
>
> INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/
> (x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
> (x^3 - 1)^(1/3)/(x^2 - 1) +
> INT((x^2 - 1)*(x^2 + x + 2)/
> (x*(x^3 - 1)^(2/3)*(x^3 + x^2 - 2*x - 3)), x)
>
> There is no reason to assume that Rubi can handle Goursat type
> integrals in general.
>

For the record, and as an "optimal" reference solution for Nasser's
integrator tests, a compact antiderivative of the non-Goursat integral
#9:

INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),
x)

is given by a sum of the following five components:

- INT(x*(x^2 + x + 2)/((x + 1)*(x^2 - 1)*(x^3 - 1)^(2/3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1)

INT(2/(3*x*(x^3 - 1)^(2/3)), x) =
- 2*SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(x^3 - 1)^(1/3)))
+ 1/3*LN((x^3 - 1)^(1/3) + 1) - 1/9*LN(x^3)

- INT(x^2*(2*x^6 - 19*x^3 - 10)
/(3*(x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(2*(x^3 - 1)^(2/3) + x^3 - 4)
/((x^3 - 1)^(1/3)*(2*(x^3 - 1)^(1/3) + 3))))
+ 1/6*LN(- 2*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) + x^3 - 4)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

INT(x*(x^3 - 1)^(1/3)*(3*x^3 - 1)/(x^9 - 2*x^6 + x^3 - 27), x) =
SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x*(x^3 - 1)^(2/3))/3))
+ 1/6*LN(x*(x^3 - 1)^(2/3) - 3) - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

INT((x^9 - 3*x^6 + 8*x^3 + 3)
/((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3)
/(3*((x^3 - 1)^(1/3) + x^2))))
+ 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

All but the first and last of these turn into classical cases of
elementary integrals under the substitutions t = x^3 or t = 1/x^3.

Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should be
merged into one. Is the FriCAS antiderivative suboptimal because its
leaf count exceeds twice that of this optimal result?

Martin.

On 10/30/2023 2:02 PM, clicliclic@freenet.de wrote:

> Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should be
> merged into one. Is the FriCAS antiderivative suboptimal because its
> leaf count exceeds twice that of this optimal result?
>
> Martin.

fyi, I've added your 14 integrals as separate new test file
to CAS integration tests. Maybe this will make it easier to reference things.

Created a new test file (#213) of them, and converted them to all
the other CAS formats and added them to
CAS integration tests summer 2023 version.

So now you can see the full report for all 8 CAS systems. Link at end.

Result % solved is
-------------------
1. Maple 100.00% 14/14
2. Fricas 92.86% 13/14
3. Mathematica 78.57% 11/14
4. Rubi 71.43% 10/14
5. Maxima 14.29% 2/14
6. Sympy 14.29% 2/14
7. Mupad 7.14% 1/14
8. Giac 0.00% 0/14

In terms of grading:

A B C F
Fricas 85.714 7.143 0.000 7.143
Mathematica 78.571 0.000 0.000 21.429
Rubi 35.714 0.000 35.714 28.571
Maxima 14.286 0.000 0.000 85.714
Maple 7.143 0.000 92.857 0.000
Mupad 0.000 7.143 0.000 92.857
Giac 0.000 0.000 0.000 100.000
Sympy 0.000 0.000 14.286 85.714

For optimal anti, I picked the smallest one by trying each integral
on different CAS's. So it is possible a better optimal exists that
I missed.

Only Maple was able to solve them all but the grading was not the best.
FriCAS got the best A grading and was in second place in terms of number
solved. Maxima, mupad, giac and sympy did not do well on these hard
integrals.

Link to the new test report is

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>

Which you can also access from the main web page

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/index.htm>

by going to the "links to individual test reports" page and scroll all
the way down.

--Nasser

"Nasser M. Abbasi" schrieb:
>
> On 10/30/2023 2:02 PM, clicliclic@freenet.de wrote:
>
> > Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should
> > be merged into one. Is the FriCAS antiderivative suboptimal because
> > its leaf count exceeds twice that of this optimal result?
> >
>
> fyi, I've added your 14 integrals as separate new test file
> to CAS integration tests. Maybe this will make it easier to reference
> things.
>

Just to avoid misunderstandings: these 14 integrals are by Sam Blake
who posted them on 11 December 2022; they may even be duplicated from
his test suite of algebraic integrals - I simply don't know.

> Created a new test file (#213) of them, and converted them to all
> the other CAS formats and added them to
> CAS integration tests summer 2023 version.
>
> So now you can see the full report for all 8 CAS systems. Link at end.
>
> Result % solved is
> -------------------
> 1. Maple 100.00% 14/14
> 2. Fricas 92.86% 13/14
> 3. Mathematica 78.57% 11/14
> 4. Rubi 71.43% 10/14
> 5. Maxima 14.29% 2/14
> 6. Sympy 14.29% 2/14
> 7. Mupad 7.14% 1/14
> 8. Giac 0.00% 0/14
>

Rubi's success rate on these 14 cube root integrals is impressive even
though many solutions appear to involve functions higher than
elementary. And I notice that my antiderivative for integral #9 was
very far from optimal! The optimal solution is just something like:

INT((x^2 + x + 2)*(x^3 - 1)^(1/3)
/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1)
+ SQRT(3)/3*ATAN(SQRT(3)/3*(1 + 2*(x^2 - 1)/(x^3 - 1)^(1/3)))
+ 1/2*LN((x^3 - 1)^(1/3) - x^2 + 1)
- 1/6*LN(x^2*(x^4 - 3*x^2 - x + 3))

Wow,

Martin.

> In terms of grading:
>
> A B C F
> Fricas 85.714 7.143 0.000 7.143
> Mathematica 78.571 0.000 0.000 21.429
> Rubi 35.714 0.000 35.714 28.571
> Maxima 14.286 0.000 0.000 85.714
> Maple 7.143 0.000 92.857 0.000
> Mupad 0.000 7.143 0.000 92.857
> Giac 0.000 0.000 0.000 100.000
> Sympy 0.000 0.000 14.286 85.714
>
> For optimal anti, I picked the smallest one by trying each integral
> on different CAS's. So it is possible a better optimal exists that
> I missed.
>
> Only Maple was able to solve them all but the grading was not the
> best. FriCAS got the best A grading and was in second place in terms
> of number solved. Maxima, mupad, giac and sympy did not do well on
> these hard integrals.
>
> Link to the new test report is
>
> <https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>
>
> Which you can also access from the main web page
>
> <https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/index.htm>
>
> by going to the "links to individual test reports" page and scroll all
> the way down.
>

PS: Can any of the systems solve the fifth component integral:

INT((x^9 - 3*x^6 + 8*x^3 + 3)
/((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3)
/(3*((x^3 - 1)^(1/3) + x^2))))
+ 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

in elementary terms?

On Tuesday, October 31, 2023 at 5:59:01 AM UTC+11, nob...@nowhere.invalid wrote:
> "clicl...@freenet.de" schrieb:
> >
> > Sam Blake schrieb:
> > >
> > > Here are some nice cube root-type integrals which may or may not be
> > > integrable using your Goursat adaption. I am guessing that as Rubi
> > > cannot integrate these that they may not be Goursat type integrals?
> > >
> > > (1 + x)/(x*(-1 + x^3)^(1/3))
> > > (-1 + x^2)/(x*(-1 + x^3)^(2/3))
> > > ((1 + x)*(-1 + x^3)^(1/3))/((-1 + x)^2*x)
> > > ((-1 + x)^2*(1 + x))/(x*(1 + x + x^2)*(-1 + x^3)^(1/3))
> > > ((1 + x)*(-1 + x^3)^(2/3))/((-1 + x)^3*x)
> > > (x*(-2 + x^3))/((-1 + x^3)^(1/3)*(-1 + x^3 + x^6))
> > > (x^3*(2 + x^3))/((1 + x^3)^(2/3)*(1 + x^3 + x^6))
> > > ((1 - x^3)^(1/3)*(-2 + x^3))/(x^3*(-1 + x^3 + x^6))
> > > ((2 + x + x^2)*(-1 + x^3)^(1/3))/(x*(-1 + x^2)^2*(-3 - 2*x + x^2 + x^3))
> > > (1 + 3*x)/((-1 + 3*x)*(-x + x^3)^(1/3))
> > > ((-1 + x)*(1 + 3*x))/((-1 + 3*x)*(-x + x^3)^(2/3))
> > > ((-1 + x)^2*(1 + 3*x))/(x*(1 + x)*(-1 + 3*x)*(-x + x^3)^(1/3))
> > > (x*(1 + x)*(1 + 3*x))/((-1 + x)*(-1 + 3*x)*(-x + x^3)^(2/3))
> > > (x*(1 + x)*(1 + 3*x))/((-1 + x)^2*(-1 + 3*x)*(-x + x^3)^(1/3))
> > >
> >
> > There are 14 integrands here. The first two are readily integrated by
> > Derive 6.10 and should therefore be doable by standard recipes from
> > books like G&R and Timofeev:
> >
> > INT((x + 1)/(x*(x^3 - 1)^(1/3)), x)
> > INT((x^2 - 1)/(x*(x^3 - 1)^(2/3)), x)
> >
> > So Rubi could and should be tought to do these as well.
> >
> > The subsequent twelve integrands, however, fail in Derive 6.10:
> >
> > INT((x + 1)*(x^3 - 1)^(1/3)/(x*(x - 1)^2), x)
> > INT((x + 1)*(x - 1)^2/(x*(x^2 + x + 1)*(x^3 - 1)^(1/3)), x)
> > INT((x + 1)*(x^3 - 1)^(2/3)/(x*(x - 1)^3), x)
> > INT(x*(x^3 - 2)/((x^3 - 1)^(1/3)*(x^6 + x^3 - 1)), x)
> > INT(x^3*(x^3 + 2)/((x^3 + 1)^(2/3)*(x^6 + x^3 + 1)), x)
> > INT((1 - x^3)^(1/3)*(x^3 - 2)/(x^3*(x^6 + x^3 - 1)), x)
> > INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),
> > x)
> > INT((3*x + 1)/((3*x - 1)*(x^3 - x)^(1/3)), x)
> > INT((x - 1)*(3*x + 1)/((3*x - 1)*(x^3 - x)^(2/3)), x)
> > INT((x - 1)^2*(3*x + 1)/(x*(x + 1)*(3*x - 1)*(x^3 - x)^(1/3)), x)
> > INT(x*(x + 1)*(3*x + 1)/((x - 1)*(3*x - 1)*(x^3 - x)^(2/3)), x)
> > INT(x*(x + 1)*(3*x + 1)/((x - 1)^2*(3*x - 1)*(x^3 - x)^(1/3)), x)
> >
> > My Goursat tests for cube-root integrands determine all but the four
> > integrals from #6 to #9 to be directly of Goursat type, which means
> > that the integrands become rational under one of the characteristic
> > substitutions. Of the other four, the first three become Goursat
> > integrable under the fairly obvious substitution t = x^3.
> >
> > A more involved substitution must have been used to scramble integrand
> > #9; even FriCAS fails on this one, but succeeds once t = x^3 has been
> > applied. The substitution doesn't make the integrand look nicer
> > though, and while the change cannot be considered momentous either,
> > for FriCAS it evidently is. The integrand simplifies a bit when an
> > algebraic part of the antiderivative is removed:
> >
> > INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/
> > (x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
> > (x^3 - 1)^(1/3)/(x^2 - 1) +
> > INT((x^2 - 1)*(x^2 + x + 2)/
> > (x*(x^3 - 1)^(2/3)*(x^3 + x^2 - 2*x - 3)), x)
> >
> > There is no reason to assume that Rubi can handle Goursat type
> > integrals in general.
> >
> For the record, and as an "optimal" reference solution for Nasser's
> integrator tests, a compact antiderivative of the non-Goursat integral
> #9:
> INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),
> x)
> is given by a sum of the following five components:
>
> - INT(x*(x^2 + x + 2)/((x + 1)*(x^2 - 1)*(x^3 - 1)^(2/3)), x) > (x^3 - 1)^(1/3)/(x^2 - 1)
> INT(2/(3*x*(x^3 - 1)^(2/3)), x) =
> - 2*SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(x^3 - 1)^(1/3)))
> + 1/3*LN((x^3 - 1)^(1/3) + 1) - 1/9*LN(x^3)
>
> - INT(x^2*(2*x^6 - 19*x^3 - 10)
> /(3*(x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
> SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(2*(x^3 - 1)^(2/3) + x^3 - 4)
> /((x^3 - 1)^(1/3)*(2*(x^3 - 1)^(1/3) + 3))))
> + 1/6*LN(- 2*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) + x^3 - 4)
> - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)
>
> INT(x*(x^3 - 1)^(1/3)*(3*x^3 - 1)/(x^9 - 2*x^6 + x^3 - 27), x) =
> SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x*(x^3 - 1)^(2/3))/3))
> + 1/6*LN(x*(x^3 - 1)^(2/3) - 3) - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)
>
> INT((x^9 - 3*x^6 + 8*x^3 + 3)
> /((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
> SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3)
> /(3*((x^3 - 1)^(1/3) + x^2))))
> + 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
> - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)
>
> All but the first and last of these turn into classical cases of
> elementary integrals under the substitutions t = x^3 or t = 1/x^3.
>
> Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should be
> merged into one. Is the FriCAS antiderivative suboptimal because its
> leaf count exceeds twice that of this optimal result?
>
> Martin.

Hey Martin,

Here's my solution for this integral:

In[431]:= IntegrateAlgebraic[((x^2 + x + 2) (x^3 - 1)^(1/3))/(x (x^2 - 1)^2 (x^3 + x^2 - 2 x - 3)), x]

Out[431]= (-1 + x^3)^(1/3)/(-1 + x^2) - ArcTan[(Sqrt[3] (-1 + x^3)^(1/3))/(-2 + 2 x^2 + (-1 + x^3)^(1/3))]/Sqrt[3] +
1/3 Log[1 - x^2 + (-1 + x^3)^(1/3)] - 1/6 Log[1 - 2 x^2 + x^4 + (-1 + x^2) (-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]

Where the substitution u == (1-x^2)/(-1+x^3)^(1/3) reduces the integral to 1/(u^2 (1+u^3)).

Cheers,

Sam

On 10/31/2023 12:46 PM, clicliclic@freenet.de wrote:

> PS: Can any of the systems solve the fifth component integral:
>
> INT((x^9 - 3*x^6 + 8*x^3 + 3)
> /((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
> SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3)
> /(3*((x^3 - 1)^(1/3) + x^2))))
> + 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
> - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)
>
> in elementary terms?

I've added this as integral 15 and updated the report at

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>

it shows that only Fricas and Maple can solve it. But Fricas only one that
gives solution in terms of elementary functions:

1/9*3^(1/2)*arctan((258*3^(1/2)*(2*x^5-59*x^2)*(x^3-1)^(2/3)-258*3^(1/2)*(13*x^7+56*x^4-24*x)*(x^3-1)^(1/3)-3^
(1/2)*(169*x^9+7789*x^6+7135*x^3-10368))/(2197*x^9+13021*x^6-25667*x^3+13824))+1/18*ln((x^9-83*x^6+82*x^3+18*(
2*x^5-5*x^2)*(x^3-1)^(2/3)-9*(x^7-13*x^4+3*x)*(x^3-1)^(1/3)-27)/(x^9-2*x^6+x^3-27))

I see now that Sam's integrator also solved this giving

(-1 + x^3)^(1/3)/(-1 + x^2) - ArcTan[(Sqrt[3] (-1 + x^3)^(1/3))/(-2 + 2 x^2 + (-1 + x^3)^(1/3))]/Sqrt[3] +
1/3 Log[1 - x^2 + (-1 + x^3)^(1/3)] - 1/6 Log[1 - 2 x^2 + x^4 + (-1 + x^2) (-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]

Which is smaller than Fricas result. Next time I build the pages will
use Sam's result as the optimal for this problem as I only saw it now.

So file #213 now has 15 integrals not 14. I also updated the optimal for 3 integrals
to better ones I found later. Currently A grade result is

Fricas 86.667
Mathematica 66.667
Rubi 20.000
Maxima 13.333
Maple 0.000
Giac 0.000
Mupad 0.000
Sympy 0.000

--Nasser

On 10/31/2023 6:19 PM, Nasser M. Abbasi wrote:
> On 10/31/2023 12:46 PM, clicliclic@freenet.de wrote:
>
>> PS: Can any of the systems solve the fifth component integral:
>>
>> INT((x^9 - 3*x^6 + 8*x^3 + 3)
>> /((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
>> SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3)
>> /(3*((x^3 - 1)^(1/3) + x^2))))
>> + 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
>> - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)
>>
>> in elementary terms?
>
> I've added this as integral 15 and updated the report at
>
> <https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>
>
> it shows that only Fricas and Maple can solve it. But Fricas only one that
> gives solution in terms of elementary functions:
>
> 1/9*3^(1/2)*arctan((258*3^(1/2)*(2*x^5-59*x^2)*(x^3-1)^(2/3)-258*3^(1/2)*(13*x^7+56*x^4-24*x)*(x^3-1)^(1/3)-3^
> (1/2)*(169*x^9+7789*x^6+7135*x^3-10368))/(2197*x^9+13021*x^6-25667*x^3+13824))+1/18*ln((x^9-83*x^6+82*x^3+18*(
> 2*x^5-5*x^2)*(x^3-1)^(2/3)-9*(x^7-13*x^4+3*x)*(x^3-1)^(1/3)-27)/(x^9-2*x^6+x^3-27))
>
> I see now that Sam's integrator also solved this giving
>

Opps, sorry, I see now Sam was answering about #9 and not #15.
So one can ignore this below. There is nothing I need to change or update.

> (-1 + x^3)^(1/3)/(-1 + x^2) - ArcTan[(Sqrt[3] (-1 + x^3)^(1/3))/(-2 + 2 x^2 + (-1 + x^3)^(1/3))]/Sqrt[3] +
> 1/3 Log[1 - x^2 + (-1 + x^3)^(1/3)] - 1/6 Log[1 - 2 x^2 + x^4 + (-1 + x^2) (-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]
>
> Which is smaller than Fricas result. Next time I build the pages will
> use Sam's result as the optimal for this problem as I only saw it now.
>

--Nasser

On Wednesday, November 1, 2023 at 4:42:47 AM UTC+11, nob...@nowhere..invalid wrote:
> "Nasser M. Abbasi" schrieb:
> >
> > On 10/30/2023 2:02 PM, clicl...@freenet.de wrote:
> >
> > > Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should
> > > be merged into one. Is the FriCAS antiderivative suboptimal because
> > > its leaf count exceeds twice that of this optimal result?
> > >
> >
> > fyi, I've added your 14 integrals as separate new test file
> > to CAS integration tests. Maybe this will make it easier to reference
> > things.
> >
> Just to avoid misunderstandings: these 14 integrals are by Sam Blake
> who posted them on 11 December 2022; they may even be duplicated from
> his test suite of algebraic integrals - I simply don't know.
> > Created a new test file (#213) of them, and converted them to all
> > the other CAS formats and added them to
> > CAS integration tests summer 2023 version.
> >
> > So now you can see the full report for all 8 CAS systems. Link at end.
> >
> > Result % solved is
> > -------------------
> > 1. Maple 100.00% 14/14
> > 2. Fricas 92.86% 13/14
> > 3. Mathematica 78.57% 11/14
> > 4. Rubi 71.43% 10/14
> > 5. Maxima 14.29% 2/14
> > 6. Sympy 14.29% 2/14
> > 7. Mupad 7.14% 1/14
> > 8. Giac 0.00% 0/14
> >
> Rubi's success rate on these 14 cube root integrals is impressive even
> though many solutions appear to involve functions higher than
> elementary. And I notice that my antiderivative for integral #9 was
> very far from optimal! The optimal solution is just something like:
> INT((x^2 + x + 2)*(x^3 - 1)^(1/3)
> /(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) > (x^3 - 1)^(1/3)/(x^2 - 1)
> + SQRT(3)/3*ATAN(SQRT(3)/3*(1 + 2*(x^2 - 1)/(x^3 - 1)^(1/3)))
> + 1/2*LN((x^3 - 1)^(1/3) - x^2 + 1)
> - 1/6*LN(x^2*(x^4 - 3*x^2 - x + 3))
>
> Wow,
>
> Martin.
> > In terms of grading:
> >
> > A B C F
> > Fricas 85.714 7.143 0.000 7.143
> > Mathematica 78.571 0.000 0.000 21.429
> > Rubi 35.714 0.000 35.714 28.571
> > Maxima 14.286 0.000 0.000 85.714
> > Maple 7.143 0.000 92.857 0.000
> > Mupad 0.000 7.143 0.000 92.857
> > Giac 0.000 0.000 0.000 100.000
> > Sympy 0.000 0.000 14.286 85.714
> >
> > For optimal anti, I picked the smallest one by trying each integral
> > on different CAS's. So it is possible a better optimal exists that
> > I missed.
> >
> > Only Maple was able to solve them all but the grading was not the
> > best. FriCAS got the best A grading and was in second place in terms
> > of number solved. Maxima, mupad, giac and sympy did not do well on
> > these hard integrals.
> >
> > Link to the new test report is
> >
> > <https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>
> >
> > Which you can also access from the main web page
> >
> > <https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/index.htm>
> >
> > by going to the "links to individual test reports" page and scroll all
> > the way down.
> >
> PS: Can any of the systems solve the fifth component integral:
> INT((x^9 - 3*x^6 + 8*x^3 + 3)
> /((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
> SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3)
> /(3*((x^3 - 1)^(1/3) + x^2))))
> + 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
> - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)
>
> in elementary terms?

Hi Martin,

Regarding the integral of

(x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3))

Seeing the term in your answer

- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

makes me think that this integrand must be split into a rational term + an algebraic term(or terms). In general, IntegrateAlgebraic will not compute such an integral unless the splitting of the rational part and the algebraic part can be achieved with Apart. Which in this case it cannot.

Cheers,

Sam

Sam Blake schrieb:
>
> On Wednesday, November 1, 2023 at 4:42:47 AM UTC+11, nob...@nowhere.invalid wrote:
> >
> > [...]
> >
> > Rubi's success rate on these 14 cube root integrals is impressive
> > even though many solutions appear to involve functions higher than
> > elementary. And I notice that my antiderivative for integral #9 was
> > very far from optimal! The optimal solution is just something like:
> >
> > INT((x^2 + x + 2)*(x^3 - 1)^(1/3)
> > /(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
> > (x^3 - 1)^(1/3)/(x^2 - 1)
> > + SQRT(3)/3*ATAN(SQRT(3)/3*(1 + 2*(x^2 - 1)/(x^3 - 1)^(1/3)))
> > + 1/2*LN((x^3 - 1)^(1/3) - x^2 + 1)
> > - 1/6*LN(x^2*(x^4 - 3*x^2 - x + 3))
> >
> > [...]
> >
> > PS: Can any of the systems solve the fifth component integral:
> > INT((x^9 - 3*x^6 + 8*x^3 + 3)
> > /((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
> > SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3)
> > /(3*((x^3 - 1)^(1/3) + x^2))))
> > + 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
> > - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)
> >
> > in elementary terms?
>
> Regarding the integral of
>
> (x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3))
>
> Seeing the term in your answer
>
> - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)
>
> makes me think that this integrand must be split into a rational term
> + an algebraic term(or terms). In general, IntegrateAlgebraic will
> not compute such an integral unless the splitting of the rational
> part and the algebraic part can be achieved with Apart. Which in this
> case it cannot.
>

In the antiderivatives of cube-root integrands, one regularly finds
paired logarithms that correspond with algebraic factorizations of the
integrand's denominator in terms of the radical.

For the denominator of your integral one has:

x^2*(x - 1)*(x^3 + x^2 - 2*x - 3) = (x^2 - 1 - (x^3 - 1)^(1/3))
*((x^2 - 1)^2 + (x^2 - 1)*(x^3 - 1)^(1/3) + (x^3 - 1)^(2/3))

Similarly for the denominator of the third and fifth components of my
multi-component antiderivative:

x^9 - 2*x^6 + x^3 - 27 =
(- 2*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) + x^3 - 4)
*((x^3 - 1)^(1/3)*((x^3 - 1)^(1/3)*(2*x^3 + 1) + x^3 + 8)
+ x^6 - 2*x^3 + 10)

x^9 - 2*x^6 + x^3 - 27 =
(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
*((x^3 - 1)^(1/3)*(3*(x^3 + 3)*(x^3 - 1)^(1/3) + x^2*(x^3 + 8))
+ 3*x*(2*x^3 + 1))

which incidently shows that algebraic factorization is not unique. More
compact antiderivatives result when they are expressed in terms of the
smaller factor and the full denominator polynominal.

For the component integrals, I just simplified the antiderivative
produced by FriCAS for t = x^3 substituted in your integral - this was
the only solution I had access to. Unfortunately, I didn't try to lump
its logarithms and its arc tangents - I should have been surprised.

I have no idea how to recocognize that the fifth component integral is
elementary. FriCAS can solve this one "as is".

Martin.