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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              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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