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

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   Message 9,181 of 10,432   
   Nasser M. Abbasi to Nasser M. Abbasi   
   Re: FriCAS web interface failure   
   30 Oct 16 02:36:10   
   
   From: nma@12000.org   
      
   On 10/4/2016 11:17 AM, Nasser M. Abbasi wrote:   
      
   >   
   > btw, I tried the above on the new Mathematica 11.0.1   
   > and on Maple 2016.1 and Rubi 4.9.2, Mathematica gave   
   > the simplist answer. Maple did not do it. Rubi answer   
   > contained complex numbers. Here is the result   
   >   
      
   .....   
      
   >   
   > Rubi: 4.9.2   
   > ============   
   > In[62]:= ShowSteps = False;   
   > In[63]:= Int[(a + b*x)/((3 + x^2)*(1 - x^2)^(1/3)), x]   
   >   
   > Out[63]= ((I*Sqrt[3]*a -   
   >        3*b)*((1 - x)/(I*Sqrt[3] - x))^(1/   
   >         3)*(-((1 + x)/(I*Sqrt[3] - x)))^(1/3)*   
   >      AppellF1[2/3, 1/3, 1/3, 5/3, -((1 - I*Sqrt[3])/(I*Sqrt[3] - x)),   
   >            (1 + I*Sqrt[3])/(I*Sqrt[3] - x)])/(4*(1 - x^2)^(1/3)) - ((I*   
   >         Sqrt[3]*a +   
   >        3*b)*(-((1 - x)/(I*Sqrt[3] + x)))^(1/   
   >         3)*((1 + x)/(I*Sqrt[3] + x))^(1/3)*   
   >          AppellF1[2/3, 1/3, 1/3,   
   >       5/3, (1 + I*Sqrt[3])/(I*Sqrt[3] +   
   >          x), -((1 - I*Sqrt[3])/(I*Sqrt[3] + x))])/(4*(1 - x^2)^(1/3))   
   >   
   > In[64]:= Int[(a + b*x)/((3 - x^2)*(1 + x^2)^(1/3)), x]   
   >   
   > Out[64]= ((Sqrt[3]*a +   
   >        3*b)*((I - x)/(Sqrt[3] - x))^(1/   
   >         3)*(-((I + x)/(Sqrt[3] - x)))^(1/3)*   
   >      AppellF1[2/3, 1/3, 1/3, 5/3, (I + Sqrt[3])/(Sqrt[3] - x),   
   >            -((I - Sqrt[3])/(Sqrt[3] - x))])/(4*(1 + x^2)^(1/   
   >         3)) - ((Sqrt[3]*a -   
   >        3*b)*(-((I - x)/(Sqrt[3] + x)))^(1/   
   >         3)*((I + x)/(Sqrt[3] + x))^(1/3)*   
   >          AppellF1[2/3, 1/3, 1/3,   
   >       5/3, -((I - Sqrt[3])/(Sqrt[3] + x)), (I + Sqrt[3])/(Sqrt[3] +   
   >          x)])/(4*(1 + x^2)^(1/3))   
   >   
      
   fyi, Rubi 4.10.1 now gives better answer in terms of only   
   elementary functions and no complex numbers to the above   
   two problems:   
      
   In[6]:= ShowSteps = False;   
   In[7]:= Int[(a + b*x)/((3 + x^2)*(1 - x^2)^(1/3)), x]   
      
   Out[7]= (a*   
        ArcTan[1/   
           Sqrt[3] - (2^(2/3)*(1 - x)^(2/3))/(Sqrt[   
              3]*(1 + x)^(1/3))])/(2*2^(2/3)*Sqrt[3]) - (a*   
        ArcTan[1/   
           Sqrt[3] - (2^(2/3)*(1 + x)^(2/3))/(Sqrt[   
              3]*(1 - x)^(1/3))])/(2*2^(2/3)*Sqrt[3]) +   
       (Sqrt[3]*b*   
        ArcTan[(1 + 2^(1/3)*(1 - x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3)) - (b*   
        Log[3 + x^2])/(4*2^(2/3)) + (a*   
        Log[(2 - 2*x)^(1/3) + (1 + x)^(2/3)])/(4*2^(2/3)) - (a*   
        Log[(1 - x)^(2/3) + (2 + 2*x)^(1/3)])/(4*2^(2/3)) +   
       (3*b*Log[2^(2/3) - (1 - x^2)^(1/3)])/(4*2^(2/3))   
      
   In[8]:= Int[(a + b*x)/((3 - x^2)*(1 + x^2)^(1/3)), x]   
      
   Out[8]= -((a*   
          ArcTan[1/   
             Sqrt[3] + (2^(2/3)*(Sqrt[3] -   
                 x))/(3*(1 + x^2)^(1/3))])/(6*2^(2/3))) + (a*   
        ArcTan[1/   
           Sqrt[3] + (2^(2/3)*(Sqrt[3] +   
               x))/(3*(1 + x^2)^(1/3))])/(6*2^(2/3)) -   
       (Sqrt[3]*b*   
        ArcTan[(1 + 2^(1/3)*(1 + x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3)) + (b*   
        Log[3 - x^2])/(4*2^(2/3)) - (3*b*   
        Log[2^(2/3) - (1 + x^2)^(1/3)])/(4*2^(2/3)) -   
       (a*Log[-1 - x/Sqrt[3] + 2^(1/3)*(1 + x^2)^(1/3)])/(4*2^(2/3)*   
        Sqrt[3]) + (a*   
        Log[-1 + x/Sqrt[3] + 2^(1/3)*(1 + x^2)^(1/3)])/(4*2^(2/3)*   
        Sqrt[3]) -   
       (a*Log[-((Sqrt[3] - x)^3/(3*Sqrt[3])) + 2*(1 + x^2)])/(12*2^(2/3)*   
        Sqrt[3]) + (a*   
        Log[-((Sqrt[3] + x)^3/(3*Sqrt[3])) + 2*(1 + x^2)])/(12*2^(2/3)*   
        Sqrt[3])   
      
      
   --Nasser   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

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