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   sci.math.symbolic      Symbolic algebra discussion      10,432 messages   

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   Message 10,035 of 10,432   
   samuel.thomas.blake@gmail.com to clicl...@freenet.de   
   Re: An algebraic integral in FriCAS   
   03 May 20 06:21:22   
   
   On Sunday, May 3, 2020 at 11:07:50 PM UTC+10, clicl...@freenet.de wrote:   
   > samuel.thomas.blake@gmail.com schrieb:   
   > >   
   > > Can you please double check those solutions?   
   >   
   > All four?   
   >   
   > Martin.   
      
   My bad (misreading ArcTan and ArcTanh in your solutions). In mathematica they   
   are   
      
   In[1081]:=   
   D[ArcTanh[((x + I)/(1 - 6*x^2 + x^4)^(1/4))] +   
      ArcTan[((x + I)/(1 - 6*x^2 + x^4)^(1/4))] +   
      ArcTanh[((x - I)/(1 - 6*x^2 + x^4)^(1/4))] +   
      ArcTan[((x - I)/(1 - 6*x^2 + x^4)^(1/4))], x] // Simplify   
      
   Out[1081]= (-3 + x^2)/((-1 + x^2) (1 - 6 x^2 + x^4)^(1/4))   
      
   In[1082]:=   
   D[-ArcTanh[((x + I)/(1 - 6*x^2 + x^4)^(1/4))] +   
      ArcTan[((x + I)/(1 - 6*x^2 + x^4)^(1/4))] -   
      ArcTanh[((x - I)/(1 - 6*x^2 + x^4)^(1/4))] +   
      ArcTan[((x - I)/(1 - 6*x^2 + x^4)^(1/4))], x] // Simplify   
      
   Out[1082]= -((1 + x^2)^2/((-1 + x^2) (1 - 6 x^2 + x^4)^(3/4)))   
      
   Sam   
      
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