Date: Sep 6, 2013 2:51 PM
Author: Peter Luschny
Subject: Re: The integration test suites for Sage.
>> Albert used Maxima 5.28 whereas I used Sage 5.10. I do not know which

>> Maxima version Sage 5.10 uses. They might be different.

> The Maxima integrator would be undergoing noticeable development then. A

> pleasant surprise.

Well, I don't know. I just switched from Sage 5.10 to 5.11 and

there are differences with regard to the Charlwood problems!

Problem 8 for example now has a monster solution; so long that

I did not care to check if it is right or wrong.

> Charlwood_problem(43)

> integrand : tan(x)/sqrt(tan(x)^4 + 1)

> antideriv : -1/4*sqrt(2)*arctanh(-1/2*(tan(x)^2-1)*sqrt(2)/sqrt(tan(x)^4+1))

> maxima : -1/4*sqrt(2)*arcsinh(2*sin(x)^2 - 1)

> After sign inversion the Maxima result appears to be correct on

> the real axis.

Yes. And what about

diff(-1/4*sqrt(2)*arctanh(-1/2*(tan(x)^2-1)*sqrt(2)/sqrt(tan(x)^4+1)),x)

= tan(x)/sqrt((tan(x))^4+1)

versus

diff(-1/4*sqrt(2)*arcsinh(cos(2*x)),x)

= sin(2*x)/sqrt(cos(4*x)+3)

tan(x)/sqrt((tan(x))^4+1) = sin(2*x)/sqrt(cos(4*x)+3) on the real axis?

> But then Maxima doesn't claim to deliver antiderivatives for the

> entire complex plane, or does it?

What are rules of the game anyway: Does the 'Charlwood test'

require antiderivatives for the entire complex plane or

only for the real line? Charlwood writes: "We consider integrals of

real elementary functions of a single real variable in the examples

that follow."

>> Charlwood_problem(49)

>> integrand : arcsin(x/sqrt(-x^2 + 1))

>> antideriv : x*arcsin(x/sqrt(-x^2 + 1)) + arctan(sqrt(-2*x^2 + 1))

>> maxima : x*arcsin(x/sqrt(-x^2 + 1)) - 1/2*(-2*I*x^2 + I)/

sqrt(2*x^2 - 1) - 1/2*I*sqrt(2*x^2 - 1) - 1/2*I*log(sqrt(2*x^2 - 1) - 1)

+ 1/2*I*log(sqrt(2*x^2 - 1) + 1)

> The Sage/Maxima result is more than just deficient: it is incorrect for

> -1/SQRT(2) < x < 1/SQRT(2) on the real axis.

Ok.

Peter