From: fateman@cs.berkeley.edu   
      
   Since the properties of AppellF1 are probably not too familiar   
   to most people, it may be that leaving the symbolic integral   
   alone provides a simpler description of the function. It may   
   even be quicker to evaluate a numerical quadrature   
   than one of these results  (though these are indefinite   
   integrals here...).   
      
   If you look at the DLMF, you see many formulas where   
   a special function is defined as an integral.  That is,   
   the explanation  (or one explanation) of the function   
   may be endorsing the view that the mysterious function   
   is made evident by the integral.   
   Other formulas may involve recurrences, infinite series,   
   special case simplifications in terms of better-known   
   functions, etc.   
      
   see dlmf.nist.gov   
      
   RJF   
      
      
   On 1/22/2018 11:57 AM, Albert Rich wrote:   
   > On Saturday, January 20, 2018 at 7:53:45 PM UTC-10, Nasser M. Abbasi wrote:   
   >   
   >> Mathematica 11.2 gives results using AppellF1   
   >> ================   
   >>   
   >> In[19]:= Integrate[x/((4-x^3)*Sqrt[1-x^3]),x]   
   >> Out[19]= 1/8 x^2 AppellF1[2/3,1/2,1,5/3,x^3,x^3/4]   
   >>   
   >> In[20]:= Integrate[x/((x^3+8)*Sqrt[x^3-1]),x]   
   >> Out[20]= (x^2 Sqrt[1-x^3] AppellF1[2/3,1/2,1,5/3,x^3,-(x^3/8)])/(16   
   Sqrt[-1+x^3])   
   >>   
   >> In[21]:= Integrate[(2+x^3)/((4-x^3)*Sqrt[1-x^3]),x]   
   >> Out[21]= 1/16 (8 x AppellF1[1/3,1/2,1,4/3,x^3,x^3/4]+x^4 Appe   
   lF1[4/3,1/2,1,7/3,x^3,x^3/4])   
   >>   
   >> In[22]:= Integrate[(x^3-4)/((x^3+8)*Sqrt[x^3-1]),x]   
   >> Out[22]= (Sqrt[1-x^3] (-16 x AppellF1[1/3,1/2,1,4/3,x^3,-(x^3/8)]+x^4   
   >>                AppellF1[4/3,1/2,1,7/3,x^3,-(x^3/8)]))/(32 Sqrt[-1+x^3])   
   >   
   > Interestingly, Mathematica 7 returns antiderivatives in terms of the   
   elliptic integral functions for the above indefinite integrals...   
   >   
   > Albert   
   >   
      
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