
Re: A bug in Reduce package 'algint'?
Posted:
Jan 22, 2013 2:08 PM


On 22.01.2013 18:42, clicliclic@freenet.de wrote: > > acer schrieb: >> >> Le lundi 21 janvier 2013 16:16:30 UTC5, Axel Vogt a écrit : >>> On 21.01.2013 19:55, clicliclic wrote: >>>> >>>> Wanting to refresh my knowledge of the capabilties of the Reduce algebra >>>> system, I have recently browsed the website. The system comes with the >>>> 'algint' package by J. Davenport which boosts the integrator >>>> capabilities for algebraic functions. The package documentation >>>> >>>> <http://www.reducealgebra.com/docs/algint.pdf> >>>> >>>> introduces the example integrand sqrt(sqrt(a^2 + x^2) + x)/x. A correct >>>> antiderivative for this is >>>> >>>> 2*(sqrt(sqrt(a^2 + x^2) + x) >>>>  sqrt(a)*atanh(sqrt(sqrt(a^2 + x^2) + x)/sqrt(a)) >>>>  sqrt(a)*atan(sqrt(sqrt(a^2 + x^2) + x)/sqrt(a))) >>>> >>>> The antiderivative printed in the documentation, however, is either very >>>> wrong or garbled beyond recognition. >>>> >>> >>> Maple 16 returns 2*sqrt(2*x)*hypergeom([1/4, 1/4, 1/4],[1/2, 3/4],a^2/x^2) >>> >>> May be right, but not that 'usefull' w.r.t. your result >>> >> >> Just because Axel's brought up Maple (and not wishing to sidetrack Martin), in Maple 16.02, >> >> expr := sqrt(sqrt(a^2 + x^2) + x)/x: >> p := u = sqrt(a^2 + x^2) + x: >> new := student[changevar](p, Int(expr,x), u): >> sol := eval(value(new),p): >> lprint(sol); >> >> ((2*a^2*((a^2+x^2)^(1/2)+x)^2+a^4+((a^2+x^2)^(1/2)+x)^4) > /((a^2+x^2)^(1/2)+x)^2)^(1/2) > *((a^2+x^2)^(1/2)+x)*4^(1/2)*(((a^2+x^2)^(1/2)+x)^(1/2) > +a^(1/2)*arctan(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2)) > +a^(1/2)*arctanh(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2))) > /(a^2+((a^2+x^2)^(1/2)+x)^2) >> >> other := expand(radnormal(sol),power): >> lprint(other): >> >> 2*((a^2+x^2)^(1/2)+x)^(1/2)2*a^(1/2)*arctan(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2))2*a^(1/2)*arctanh(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2)) >> >> What should we expect from the system, automatically? >> > > My outside viewpoint: if the integrand is given in terms of elementary > functions, an elementary antiderivative whenever one exists. Even if it > can be expressed more compactly in terms of higher functions. But then > an infinite 3F2 series is not really simple. > > The Wolfram Integrator produces an elementary answer, but includes an > unnecessary extra factor. Perhaps this is arrived at by automatic > simplification of 3F2(1/4, 1/4, 1/4; 1/2, 3/4; a^2/x^2) ? > > How does Maple arrive at 3F2 in the first place? Solution of an ODE? > Series expansion of the integrand? > > Sidetracking won't harm my ramblings, > > Martin. >
I have not checked, how Maple 'finds' the solution, but it does not have implemented the 'catalog' of rules you have worked out and at hand now. Unfortunately.

