From: Albert_Rich@msn.com   
      
   On Saturday, October 29, 2016 at 9:36:13 PM UTC-10, Nasser M. Abbasi wrote:   
   >   
   > fyi, Rubi 4.10.1 now gives better answer in terms of only   
   > elementary functions and no complex numbers to the above   
   > two problems:   
   >   
   > In[6]:= ShowSteps = False;   
   > In[7]:= Int[(a + b*x)/((3 + x^2)*(1 - x^2)^(1/3)), x]   
   >   
   > Out[7]= (a*   
   >      ArcTan[1/   
   >         Sqrt[3] - (2^(2/3)*(1 - x)^(2/3))/(Sqrt[   
   >            3]*(1 + x)^(1/3))])/(2*2^(2/3)*Sqrt[3]) - (a*   
   >      ArcTan[1/   
   >         Sqrt[3] - (2^(2/3)*(1 + x)^(2/3))/(Sqrt[   
   >            3]*(1 - x)^(1/3))])/(2*2^(2/3)*Sqrt[3]) +   
   >     (Sqrt[3]*b*   
   >      ArcTan[(1 + 2^(1/3)*(1 - x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3)) - (b*   
   >      Log[3 + x^2])/(4*2^(2/3)) + (a*   
   >      Log[(2 - 2*x)^(1/3) + (1 + x)^(2/3)])/(4*2^(2/3)) - (a*   
   >      Log[(1 - x)^(2/3) + (2 + 2*x)^(1/3)])/(4*2^(2/3)) +   
   >     (3*b*Log[2^(2/3) - (1 - x^2)^(1/3)])/(4*2^(2/3))   
   >   
   > In[8]:= Int[(a + b*x)/((3 - x^2)*(1 + x^2)^(1/3)), x]   
   >   
   > Out[8]= -((a*   
   >        ArcTan[1/   
   >           Sqrt[3] + (2^(2/3)*(Sqrt[3] -   
   >               x))/(3*(1 + x^2)^(1/3))])/(6*2^(2/3))) + (a*   
   >      ArcTan[1/   
   >         Sqrt[3] + (2^(2/3)*(Sqrt[3] +   
   >             x))/(3*(1 + x^2)^(1/3))])/(6*2^(2/3)) -   
   >     (Sqrt[3]*b*   
   >      ArcTan[(1 + 2^(1/3)*(1 + x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3)) + (b*   
   >      Log[3 - x^2])/(4*2^(2/3)) - (3*b*   
   >      Log[2^(2/3) - (1 + x^2)^(1/3)])/(4*2^(2/3)) -   
   >     (a*Log[-1 - x/Sqrt[3] + 2^(1/3)*(1 + x^2)^(1/3)])/(4*2^(2/3)*   
   >      Sqrt[3]) + (a*   
   >      Log[-1 + x/Sqrt[3] + 2^(1/3)*(1 + x^2)^(1/3)])/(4*2^(2/3)*   
   >      Sqrt[3]) -   
   >     (a*Log[-((Sqrt[3] - x)^3/(3*Sqrt[3])) + 2*(1 + x^2)])/(12*2^(2/3)*   
   >      Sqrt[3]) + (a*   
   >      Log[-((Sqrt[3] + x)^3/(3*Sqrt[3])) + 2*(1 + x^2)])/(12*2^(2/3)*   
   >      Sqrt[3])   
   >   
   >   
   > --Nasser   
      
   Yes, the long awaited (at least by me) version 4.10 of Rubi is now available   
   at http://www.apmaths.uwo.ca/~arich/.  As these examples show, Rubi 4.10 does   
   a much better job at finding optimal antiderivatives of elliptic, and   
   pseudo-elliptic integrals.   
      
   For the integrand in the first example above, when a=1 and b=0,   
      
       1/ ((3 + x^2)*(1 - x^2)^(1/3))   
      
   Rubi 4.10 returns an elementary antiderivative consisting of two arctan terms   
   and two log terms; whereas Mathematica 11 returns an Appell hypergeometric   
   function.  But even more interestingly, for the integral of the equivalent   
   expression   
      
       (1 - x^2)^(2/3) / (3 - 2*x^2 - x^4)   
      
   where the common divisor has not been canceled, Rubi returns the same   
   elementary antiderivative; whereas Mathematica indicates it is not even   
   integrable in closed-form.  Maple 2016 is not able to integrate either form of   
   the integrand.   
      
   Albert   
      
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