Date: Jul 6, 2013 5:10 PM
Author: Albert D. Rich
Subject: Re: Charlwood Fifty test results
On Saturday, July 6, 2013 10:28:30 AM UTC-10, clicl...@freenet.de wrote:

> Since Rubi manages to return the proper T1-based result already, it

> shouldn't be hard to make it handle Charlwood's problem 41. It merely

> needs to be taught to normalize integrands like COT(x)*COS(x)/

> SQRT(1 + SIN(x)) to their sine-only form. This would be just one

> normalization among others performed regularly, and would only require

> some twisting or minor extension of existing code, I guess.

Rather than blindly normalizing such integrands to their sine-only form and hoping for the best, it is safer to drive the degree of the cosine factor toward zero using the following hot new rule: If a^2-b^2=0 and m>0 is even, then

Int[Cos[c+d*x]^m*Sin[c+d*x]^n*(a+b*Sin[c+d*x])^p, x]

goes to

1/a*Int[Cos[c+d*x]^(m-2)*Sin[c+d*x]^n*(a+b*Sin[c+d*x])^(p+1), x] -

1/b*Int[Cos[c+d*x]^(m-2)*Sin[c+d*x]^(n+1)*(a+b*Sin[c+d*x])^(p+1), x]

This rule is based on the trig identity

Cos[z]^2 = (a+b*Sin[z])/a - Sin[z]*(a+b*Sin[z])/b

which is valid if a^2-b^2=0. The new rule will make it possible for the next version of Rubi to integrate a whole new class of expressions including the integrand in question, as well as the much more difficult one

Cos[x]^4/(Sin[x]^3*(a+a*Sin[x])^(3/2))

Albert