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sci / sci.math.symbolic / A rational integral in Rubi

SubjectAuthor
* A rational integral in RubiSam Blake
+- Re: A rational integral in RubiNasser M. Abbasi
`- Re: A rational integral in Rubiclicliclic@freenet.de

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Subject: A rational integral in Rubi
From: Sam Blake
Newsgroups: sci.math.symbolic
Date: Wed, 11 Oct 2023 04:06 UTC
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Subject: A rational integral in Rubi
From: samuel.thomas.blake@gmail.com (Sam Blake)
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I was pleased to see the following concise result from Rubi

In[8203]:= Int[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8203]= ArcTan[u^2/(Sqrt[2] (1 + u))]/Sqrt[2]

My implementation of Bronstein's Risch algorithm (from Symbolic Integration I) returns two arctangents

In[8206]:= Risch[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8206]= (ArcTan[(-1 + u)/Sqrt[2]] - ArcTan[(2 + 2 u + (-1 + u) u^2)/Sqrt[2]])/Sqrt[2]

and Mathematica returns the naive form

In[8207]:= Integrate[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8207]= 1/4 RootSum[
2 + 4 #1 + 2 #1^2 + #1^4 &, (2 Log[u - #1] #1 + Log[u - #1] #1^2)/(
1 + #1 + #1^3) &]

FriCAS also returns two arctangents

(8) -> integrate((2*u + u^2)/(2 + 4*u + 2*u^2 + u^4), u)

+-+ 3 2
(u - 1)\|2 u - u + 2 u + 2
atan(-----------) - atan(-----------------)
2 +-+
\|2
(8) -------------------------------------------
+-+
\|2
Type: Union(Expression(Integer),...)

Perhaps there is still some room for improvement in the log to arctan conversions in symbolic integrators...

Sam

Subject: Re: A rational integral in Rubi
From: Nasser M. Abbasi
Newsgroups: sci.math.symbolic
Organization: A noiseless patient Spider
Date: Wed, 11 Oct 2023 04:29 UTC
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From: nma@12000.org (Nasser M. Abbasi)
Newsgroups: sci.math.symbolic
Subject: Re: A rational integral in Rubi
Date: Tue, 10 Oct 2023 23:29:51 -0500
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On 10/10/2023 11:06 PM, Sam Blake wrote:
> I was pleased to see the following concise result from Rubi
>
> In[8203]:= Int[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
> Out[8203]= ArcTan[u^2/(Sqrt[2] (1 + u))]/Sqrt[2]
>

Fyi, the rules used are (3 rules)

TraceScan[Print[#/. Rubi`Private`ShowStep[i_,___]:>i]&,Int[(2 u+u^2)/(2+4 u+2 u^2+u^4),u],Rubi`Private`ShowStep[i_,___]]
1607
2119
209

Rule 1607 is

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol]
:> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; FreeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

2119 is

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(n_.)))/((a_) + (b_.)*(x_)^(k_.) + (c_.)*(x_)^(n_.) + (d_.)*(x_)^(n2_)), x_Symbol] :>
Dist[A^2*((m - n + 1)/(m + 1)), Subst[Int[1/(a + A^2*b*(m - n + 1)^2*x^2), x],
x, x^(m + 1)/(A*(m - n + 1) + B*(m + 1)*x^n)], x] /; FreeQ[{a, b, c, d, A, B, m, n}, x]
&& EqQ[n2, 2*n] && EqQ[k, 2*(m + 1)] && EqQ[a*B^2*(m + 1)^2 -
A^2*d*(m - n + 1)^2, 0] && EqQ[B*c*(m + 1) - 2*A*d*(m - n + 1), 0]

Rule 209 is

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :>
Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

> My implementation of Bronstein's Risch algorithm (from Symbolic Integration I) returns two arctangents
>
> In[8206]:= Risch[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
> Out[8206]= (ArcTan[(-1 + u)/Sqrt[2]] - ArcTan[(2 + 2 u + (-1 + u) u^2)/Sqrt[2]])/Sqrt[2]
>
> and Mathematica returns the naive form
>
> In[8207]:= Integrate[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
> Out[8207]= 1/4 RootSum[
> 2 + 4 #1 + 2 #1^2 + #1^4 &, (2 Log[u - #1] #1 + Log[u - #1] #1^2)/(
> 1 + #1 + #1^3) &]
>
> FriCAS also returns two arctangents
>
> (8) -> integrate((2*u + u^2)/(2 + 4*u + 2*u^2 + u^4), u)
>
> +-+ 3 2
> (u - 1)\|2 u - u + 2 u + 2
> atan(-----------) - atan(-----------------)
> 2 +-+
> \|2
> (8) -------------------------------------------
> +-+
> \|2
> Type: Union(Expression(Integer),...)
>
> Perhaps there is still some room for improvement in the log to arctan conversions in symbolic integrators...
>
> Sam

--Nasser

Subject: Re: A rational integral in Rubi
From: clicliclic@freenet.d
Newsgroups: sci.math.symbolic
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Date: Thu, 12 Oct 2023 09:38 UTC
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Newsgroups: sci.math.symbolic
Subject: Re: A rational integral in Rubi
Date: Thu, 12 Oct 2023 11:38:59 +0200
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Sam Blake schrieb:
>
> I was pleased to see the following concise result from Rubi
>
> In[8203]:= Int[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
> Out[8203]= ArcTan[u^2/(Sqrt[2] (1 + u))]/Sqrt[2]
>
> My implementation of Bronstein's Risch algorithm (from Symbolic
> Integration I) returns two arctangents
>
> In[8206]:= Risch[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
> Out[8206]= (ArcTan[(-1 + u)/Sqrt[2]] - ArcTan[(2 + 2 u + (-1 + u) u^2)/Sqrt[2]])/Sqrt[2]
>
> and Mathematica returns the naive form
>
> In[8207]:= Integrate[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
> Out[8207]= 1/4 RootSum[
> 2 + 4 #1 + 2 #1^2 + #1^4 &, (2 Log[u - #1] #1 + Log[u - #1] #1^2)/(
> 1 + #1 + #1^3) &]
>
> FriCAS also returns two arctangents
>
> (8) -> integrate((2*u + u^2)/(2 + 4*u + 2*u^2 + u^4), u)
>
> +-+ 3 2
> (u - 1)\|2 u - u + 2 u + 2
> atan(-----------) - atan(-----------------)
> 2 +-+
> \|2
> (8) -------------------------------------------
> +-+
> \|2
> Type: Union(Expression(Integer),...)
>
> Perhaps there is still some room for improvement in the log to arctan
> conversions in symbolic integrators...
>

Bronstein's Risch algorithm presumably involves Rioboo's decomposition
of arc tangents to ensure continuity on the real axis (see chapter 2.8
in his Symbolic Integration I). Indeed, ATAN(u^2/(SQRT(2)*(1 + u)))/
SQRT(2) is discontinuous at x = -1. However, a user can easily change
this antiderivative to -ATAN(SQRT(2)*(1 + u)/u^2)/SQRT(2), but he then
needs a limit to evaluate it at x = 0.

Martin.

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