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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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