Date: Jun 8, 2013 1:17 PM
Author: clicliclic@freenet.de
Subject: Re: The Charlwood Fifty

Albert Rich schrieb:
>
> I just posted a revised pdf file of the Charlwood Fifty integration
> test-suite at
>
> http://www.apmaths.uwo.ca/~arich/CharlwoodIntegrationProblems.pdf
>
> [...] Also posted is a Mathematica package file of the test-suite in
>
> http://www.apmaths.uwo.ca/~arich/CharlwoodProblems.m
>
> [...] Hopefully all the antiderivatives in Charlwood Fifty test-suite
> are now optimal...
>

I haven't checked the evaluations in the file systematically, but a new
look has revealed further possibilities for improvement.

The present solutions of problems #21 and #22 from Charlwood's appendix,
INT(x^3*ASIN(x)/SQRT(1-x^4), x) and INT(x^3*ASEC(x)/SQRT(x^4-1), x), can
be written as:

1/4*(x*SQRT(1-x^4)/SQRT(1-x^2)
+ LN(1-x^2) - LN(-x + x^3 + SQRT(1-x^2)*SQRT(1-x^4)))
- 1/2*SQRT(1-x^4)*ASIN(x)

1/2*(SQRT(x^4-1)*ASEC(x) - SQRT(x^4-1)/(x*SQRT(1 - 1/x^2))
- LN(x - x^3) + LN(1 - x^2 - x*SQRT(x^4-1)*SQRT(1 - 1/x^2)))

In my eyes, ATANH constitutes a more natural option here than LN:

1/4*(x*SQRT(1-x^4)/SQRT(1-x^2) + ATANH(x*SQRT(1-x^2)/SQRT(1-x^4)))
- 1/2*SQRT(1-x^4)*ASIN(x)

1/2*(SQRT(x^4-1)*ASEC(x) - SQRT(x^4-1)/(x*SQRT(1 - 1/x^2))
+ ATANH(x*SQRT(1 - 1/x^2)/SQRT(x^4-1)))

Or, using piecewise-constant prefactors:

SQRT(1-x^4)/(4*SQRT(1-x^2)*SQRT(1+x^2))
*(x*SQRT(1+x^2) + ATANH(x/SQRT(1+x^2))) - 1/2*SQRT(1-x^4)*ASIN(x)

1/2*(SQRT(x^4-1)*ASEC(x) - x*SQRT(x^2+1)/SQRT(x^4-1)*SQRT(1 - 1/x^2)
*(SQRT(x^2+1) - ATANH(1/SQRT(x^2+1))))

A much simpler evaluation of integral #48 from the appendix is:

INT(ATAN(SQRT(x) - SQRT(x+1)), x) =
= (x+1)*ATAN(SQRT(x) - SQRT(x+1)) - SQRT(x)/2

Finally, in the antiderivative #50, the rationalization of radicand
denominators in the numerical prefactors may be considered:

INT(ATAN(x*SQRT(1-x^2)), x) =
= x*ATAN(x*SQRT(1-x^2))
- SQRT(2*SQRT(5) + 2)/2*ATAN(SQRT(2*SQRT(5) + 2)/2*SQRT(1-x^2))
+ SQRT(2*SQRT(5) - 2)/2*ATANH(SQRT(2*SQRT(5) - 2)/2*SQRT(1-x^2))

Martin.