Albert Rich schrieb:   
   >   
   > On Friday, August 19, 2016 at 8:42:33 PM UTC-10, clicl...@freenet.de wrote:   
   >   
   > > Hah: There are still cube-root Goursat integrals FriCAS cannot do!   
   > > And wow: a monic quadratic in ?^3 already too complicated! My   
   > > Goursat integrability check, however, immediately certifies the   
   > > whole integral as elementary:   
   > >   
   > >   goursat5a((a + b*x)/(3 + x^2), x, 1, 0, -1, 0)   
   > >   
   > >   [false, false, false, true, true]   
   > >   
   > > the final true signifying integrability. And its evaluation does not   
   > > naturally decompose into pure a and b parts:   
   > >   
   > > INT((a + b*x)/((3 + x^2)*(1 - x^2)^(1/3)), x)   
   > >  = 2^(1/3)*(a + 3*b)/24*(LN((1 - x)^3 + 2*(1 - x^2))   
   > >  - 3*LN((1 - x) + 2^(1/3)*(1 - x^2)^(1/3))   
   > >  + 2*SQRT(3)*ATAN(1/SQRT(3)*(1 - 2^(2/3)*((1 - x)/(1 - x^2)^(1/3)))))   
   > >  - 2^(1/3)*(a - 3*b)/24*(LN((1 + x)^3 + 2*(1 - x^2))   
   > >  - 3*LN((1 + x) + 2^(1/3)*(1 - x^2)^(1/3))   
   > >  + 2*SQRT(3)*ATAN(1/SQRT(3)*(1 - 2^(2/3)*((1 + x)/(1 - x^2)^(1/3)))))   
   > >   
   > > Alternative formulations are possible, but none has one part pure a   
   > > and the other pure b.   
   > > [...]   
   >   
   > Martin's antiderivative for (a + b*x)/((3 + x^2)*(1 - x^2)^(1/3)) may   
   > be optimal.  But if either a or b is 0, there exists better ones.   
   > Using the substitution u=x^2, the following antiderivative of x/((3 +   
   > x^2)*(1 - x^2)^(1/3)) is relatively trivial:   
   >   
   >     Sqrt[3]/(2*2^(2/3)) *   
   >       ArcTan[(1 + 2^(1/3)*(1 - x^2)^(1/3))/Sqrt[3]] -   
   >     1//(4*2^(2/3)) * Log[3 + x^2] +   
   >     3/(4*2^(2/3)) * Log[2^(2/3) - (1 - x^2)^(1/3)]   
   >   
   > The following antiderivative of 1/((3 + x^2)*(1 - x^2)^(1/3)) involves   
   > two arctangents and two logs, instead of Martin's two arctangents and   
   > four logs:   
   >   
   >     1/(2*2^(2/3)*Sqrt[3]) *   
   >       ArcTan[(1 - 2^(2/3)*(1 - x)^(2/3)/(1 + x)^(1/3))/Sqrt[3]] -   
   >     1/(2*2^(2/3)*Sqrt[3]) *   
   >       ArcTan[(1 - 2^(2/3)*(1 + x)^(2/3)/(1 - x)^(1/3))/Sqrt[3]] -   
   >     1/(4*2^(2/3)) * Log[(1 - x)^(2/3) + 2^(1/3)*(1 + x)^(1/3)] +   
   >     1/(4*2^(2/3)) * Log[2^(1/3)*(1 - x)^(1/3) + (1 + x)^(2/3)]   
   >   
      
   Your antiderivative for a = 0 implies a linear relation between its   
   single LN+LN+ATAN triplet and my pair of such triplets; this relation   
   can be used to eliminate either triplet from my antiderivative. In the   
   expressions thus obtained, one triplet is multiplied simply by a, the   
   other again by a linear combination of a and b. My antiderivative and   
   those two are comparable in size and therefore, I suppose, optimality.   
      
   For reasons of transparency, in my antiderivatives I usually ignore the   
   possibility of the triplet logarithms with polynomial arguments being   
   constant or compensating mutually, either fully or in part.   
      
   > I am trying to derive a general rule for integrating expressions of   
   > the form 1/((a + b*x^2)*(c + d*x^2)^(1/3)).  Can Martin's Goursat   
   > integrability checker determine the relationship, if any, between the   
   > parameters a, b, c and d that must hold in order for an elementary   
   > antiderivative to exist?  Of course, help anyone can provide with the   
   > rule itself would be much appreciated...   
   >   
      
   By rescaling the integration variable, removing a positive factor from   
   the radical, and removing another factor from the integral, the   
   parameters in your expression can be reduced to b = 1, c = +-1, d = +-1.   
   The integrand 1/((p + x^2)*(1 + x^2)^(1/3)) is Goursat pseudo-elliptic   
   if and only if (p + 3)*(p - 9) = 0, while 1/((p + x^2)*(1 - x^2)^(1/3))   
   and 1/((p + x^2)*(x^2 - 1)^(1/3)) are Goursat pseudo-elliptic if and   
   only if (p - 3)*(p + 9) = 0.   
      
   However, I do not consider it worthwhile to add such knowledge to Rubi.   
   In a rule-based integrator, the whole lot of Goursat pseudo-elliptics   
   can and should be caught in a single check of integrands predetermined   
   to be algebraic and to possess a suitable single radicand; integrands   
   found to be pseudo-elliptic would then be decomposed as needed and the   
   components rationalized by substitution, all others should be expressed   
   in terms of Legendre's canonical integrals.   
      
   Compared to making Rubi available on free algebra systems, all of this   
   is unimportant.   
      
   Martin.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)