Waldek Hebisch schrieb: > > firstname.lastname@example.org wrote: > > 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 likely > > this means that no formula exists." Waouw! Here the elementary > > antiderivative 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 sur > > quelques int?grales pseudo-elliptiques, Bulletin de la Soci?t? > > Math?matique 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 second > > integral too. > > FriCAS result: > > integrate((k*x^2 - 1)/((a*k*x + b)*(b*x + a)*sqrt((1 - x^2)*(1 - k^2*x^2))), x) >
[FriCAS result snipped as it could not be reposted on aioe.org]
> > There are two alternatives, one in terms of 'log', the other > (shorter) in terms of 'atan'. >
The large size of the logarithmic result seems to arise through a rationalization of the argument denominator; the arc tangent result is practically identical to what I gave (done manually, with Derive support).
Would the automatic rational factorization of all polynomial subexpressions in a final result be too expensive?