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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 9,798 of 10,432    |
|    Albert Rich to clicl...@freenet.de    |
|    Re: More teething help    |
|    28 Jan 18 22:49:07    |
      From: Albert_Rich@msn.com              On Sunday, January 28, 2018 at 7:37:33 PM UTC-10, clicl...@freenet.de wrote:       > Albert Rich schrieb:       > >       > > I would like to be able to assure users that Rubi will find at least       > > near optimal antiderivatives for all integrands of the form       > >       > > x^m*(a+b*x^3)^p*(c+d*x^3)^q       > >       > > where m and p are integers, and q is a half-integer. For that I need       > > to know:       > >       > > In light of the ongoing theoretical discussion on this thread, is the       > > antiderivative of x/((b+x^3)*sqrt(1+x^3)) elementary for any values of       > > b other than 4 and -8. If so, can someone give me an example and its       > > elementary antiderivative.       > >       >       > Testing for standard Goursat cases (Eq. 5 on p. 111 of Goursat's 1887       > paper) leads to (b + 8)*(b^2 - 20*b - 8) = 0, any such b suffices for       > an elementary antiderivative, which FriCAS should be able to determine.       > I have not implemented the material in paragraphs 4 and 5 of Goursat's       > paper, which is special to the radical SQRT(x^3-1), but nothing new       > seems to follow from it (Goursat may not have been aware of the fact       > that his eq. 5 alone suffices to detect b = -8).       >       > Testing for Goursat cases of the equivalent cube-root integrand leads       > to (b + 8)*(b - 4) = 0, as mentioned in a post of Sun, 21 August 2016       > 18:12:49 +0200 already (the thread was entitled "Rubi 4.9.2 do not       > integrate Timofeev #319 while 4.9 does"). Such tests of cube-root       > integrands do not appear in Goursat's paper (or elsewhere to my       > knowledge), but the case of b = 4 apparently goes undetected otherwise.       >       > Needless to say, the cases of b = 10 +- 6*SQRT(3) result in new cube-       > root pseudo-elliptics as well.       >       > For any possible further (i.e. non-Goursat) pseudo-elliptic instances       > of your integrand, somebody more familiar with elliptic curves must be       > consulted.       >       > Martin.              Knowing that x/((10+6*sqrt(3)+x^3)*sqrt(1+x^3)) has an elementary       antiderivative, how does one goes about finding it? I was not able to get       FriCAS to help out...              Albert              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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