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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 9,799 of 10,432    |
|    Nasser M. Abbasi to Albert Rich    |
|    Re: More teething help    |
|    29 Jan 18 01:27:30    |
      From: nma@12000.org              On 1/29/2018 12:49 AM, Albert Rich wrote:       > 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       >              fyi,              I get on Fricas 1.3.2              (1) -> integrate(x/(((10-6*sqrt(3))+x^3)*sqrt(1+x^3)),x)               >> System error:              overflow during multiplication of large numbers                     --Nasser              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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