```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> machine readable form at> > 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 newlook 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), canbe 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)/2Finally, in the antiderivative #50, the rationalization of radicanddenominators 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.
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