```Date: Dec 14, 2012 4:47 PM
Author: clicliclic@freenet.de
Subject: Goursat pseudo-elliptics and the Wolfram Integrator

I was curious how Mathematica would handle the pseudo-elliptic integralsrecently referred to on this newsgroup. So I plugged two examples intothe Wolfram Integrator (which presumably represents Mathematica 8):  <http://integrals.wolfram.com/index.jsp>Example 1 (cubic radicand):  Integrate[(k*x^2 - 1)/((a*k*x + b)*(b*x + a)           *Sqrt[x*(1 - x)*(1 - k*x)]), x]... eeeek! The Integrator replies in terms of incomplete elliptic F,incomplete elliptic Pi, and the imaginary unit. But the antiderivativejust is:  2/(Sqrt[a*b]*Sqrt[(a + b)*(a*k + b)])    *ArcTan[Sqrt[a*b]*Sqrt[x*(1 - x)*(1 - k*x)]           /(Sqrt[(a + b)*(a*k + b)]*x)]Example 2 (quartic radicand):  Integrate[(k*x^2 - 1)/((a*k*x + b)*(b*x + a)           *Sqrt[(1 - x^2)*(1 - k^2*x^2)]), x]... "Mathematica could not find a formula for your integral. Most likelythis means that no formula exists." Waouw! Here the elementaryantiderivative is:  2/(Sqrt[(a + b)*(a*k + b)]*Sqrt[(a - b)*(a*k - b)])    *ArcTanh[Sqrt[(a + b)*(a*k + b)]*Sqrt[(1 - x^2)*(1 - k^2*x^2)]            /(Sqrt[(a - b)*(a*k - b)]*(1 - x)*(1 - k*x))]The theory behind these integrals is given in: Edouard Goursat, Note surquelques intégrales pseudo-elliptiques, Bulletin de la SociétéMathématique de France 15 (1887), 106-120, on-line at:  <http://www.numdam.org/item?id=BSMF_1887__15__106_1>This was written 125 years ago - apparently too recent for the "Risch"integrator of Mathematica 8. I expect that FriCAS can do the secondintegral too. How do Maple and Sympy behave?Martin.
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