From: Albert_Rich@msn.com   
      
   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 Appel   
   F1[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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