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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 10,033 of 10,432    |
|    samuel.thomas.blake@gmail.com to clicl...@freenet.de    |
|    Re: An algebraic integral in FriCAS    |
|    03 May 20 04:05:03    |
   
   On Sunday, May 3, 2020 at 8:12:30 PM UTC+10, clicl...@freenet.de wrote:   
   > samuel.thomas.blake@gmail.com schrieb:   
   > >    
   > > On Saturday, May 2, 2020 at 9:13:51 PM UTC+10, clicl...@freenet.de wrote:   
   > > >   
   > > > [...]   
   > > >   
   > > > For example, FriCAS 1.3.6 with setSimplifyDenomsFlag(true) fails on:   
   > > >   
   > > > integrate((1 + x)/((3 - x^2)*(1 - 3*x^2)^(1/3)), x)   
   > > >   
   > > > >> Error detected within library code:   
   > > > integrate: implementation incomplete   
   > > > (residue poly has multiple non-linear factors)   
   > > >   
   > > > and on:   
   > > >   
   > > > integrate((2 - x)/((3 + x^2)*(1 - x^2)^(1/3)), x)   
   > > >   
   > > > >> Error detected within library code:   
   > > > integrate: implementation incomplete (trace 0)   
   > > >   
   > > > Hence this question to Sam: does your Gröbner-basis driven   
   > > > Kauers-code fail on these integrals too?   
   > > >   
   > >    
   > > [...]   
   > >    
   > > In regards to Kauer's integration of your integrals - they both   
   > > succeed, here are the solutions.   
   > >    
   > > In[56]:= integrate[(2 - x)/((3 + x^2) (1 - x^2)^(1/3)), x]   
   > >    
   > > Out[56]= {1/3 RootSum[343 - 5184 #1^3 + 27648 #1^6 &,   
   > > Log[(-49 (1 - x^2)^(1/3) + 120 #1^2 - 108 x #1^2 + 576 (1 - x^2)^(1/3)   
   #1^3 + 1152 x #1^5)/x] #1 &], 0}   
   > >    
   > > In[54]:= integrate[(1 + x)/((3 - x^2) (1 - 3 x^2)^(1/3)), x]   
   > >    
   > > Out[54]= {-(Log[x]/6) + (1/24)*Log[3 + 2*Sqrt[3]*x + x^2 + 2*(1 -   
   3*x^2)^(1/3) - 2*Sqrt[3]*x*(1 - 3*x^2)^(1/3) + 4*(1 - 3*x^2)^(2/3)] - Log[3 +   
   2*Sqrt[3]*x + x^2 + 2*(1 - 3*x^2)^(1/3) - 2*Sqrt[3]*x*(1 - 3*x^2)^(1/3) + 4*(1   
   - 3*x^2)^(2/3)]/(24*Sqrt[3])   
    + (1/24)*Log[3 - 2*Sqrt[3]*x + x^2 + 2*(1 - 3*x^2)^(1/3) + 2*Sqrt[3]*x*(1 -   
   3*x^2)^(1/3) + 4*(1 - 3*x^2)^(2/3)] + Log[3 - 2*Sqrt[3]*x + x^2 + 2*(1 -   
   3*x^2)^(1/3) + 2*Sqrt[3]*x*(1 - 3*x^2)^(1/3) + 4*(1 - 3*x^2)^(2/   
   )]/(24*Sqrt[3]) + (1/6)*RootSum[1 + 12*#   
   1 + 120*#1^2 + 288*#1^3 + 576*#1^4 & , Log[(1/x^2)*(3 + 6*x + x^2 - 2*(1 -   
   3*x^2)^(1/3) + 6*x*(1 - 3*x^2)^(1/3) + 36*#1 + 72*x*#1 + 12*x^2*#1 + 24*x*(1 -   
   3*x^2)^(1/3)*#1 - 48*(1 - 3*x^2)^(2/3)*#1 + 288*x*#1^2 + 48*(1 -   
   3*x^2)^(1/3)*#1^2 + 144*x*(1 - 3*x^   
   2)^(1/3)*#1^2 - 96*(1 - 3*x^2)^(2/3)*#1^2 + 576*x*#1^3)]*#1 & ], 0}   
   > >    
   >    
   > Here is a pair of Kauers candidates involving 4th roots:   
   >    
   > integrate((3 - x^2)/((1 - x^2)*(1 - 6*x^2 + x^4)^(1/4)), x)   
   >    
   > integrate((1 + x^2)^2/((1 - x^2)*(1 - 6*x^2 + x^4)^(3/4)), x)   
   >    
   > which also causes FriCAS 1.3.6 to squawk:   
   >    
   > >> Error detected within library code:   
   > integrate: implementation incomplete   
   > (residue poly has multiple non-linear factors)   
   >    
   > Martin.   
      
   These are two very nice integrals. Kauer's package (at least the version I use   
   which uses Mathematica's GroebnerBasis routine in place of calling Singular)   
   cannot integrate these.    
      
   What substitution do you propose to compute these integrals?    
      
   Sam   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   
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