On Sunday, May 3, 2020 at 11:21:24 PM UTC+10, samuel.t...@gmail.com wrote:   
   > 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   
      
   Hi Martin,   
      
   What is the method you used to compute these integrals? Did you use Derive?   
      
   I thought it may be been a substitution of Gunther(http://www.nu   
   dam.org/article/BSMF_1882__10__88_1.pdf), but I couldn't get it to fall out.   
      
   Thanks,   
      
   Sam   
      
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