From: samuel.thomas.blake@gmail.com   
      
   Hi All,   
      
   For fun I have been implementing the rational integral routines from   
   Bronstein's Symbolic Integration I in Mathematica. As part of my testing I   
   stumbled across the following example integral   
      
   \int x/(2 + 4 x + 5 x^2 + 2 x^3 + x^4) dx   
      
   Rubi does well on this example   
      
   In[66]:= Int[x/(2 + 4 x + 5 x^2 + 2 x^3 + x^4), x]   
      
   Out[66]= -(ArcTan[(1 + 2 x)/Sqrt[7 - 4 Sqrt[2]]]/(   
     2 Sqrt[2 (7 - 4 Sqrt[2])])) +   
    ArcTan[(1 + 2 x)/Sqrt[7 + 4 Sqrt[2]]]/(2 Sqrt[2 (7 + 4 Sqrt[2])]) -   
    ArcTanh[(7 + 4 (1/2 + x)^2)/(4 Sqrt[2])]/(2 Sqrt[2])   
      
   While my rational function integrator returns   
      
   In[67]:= IntegrateRational[x/(2 + 4 x + 5 x^2 + 2 x^3 + x^4), x]   
      
   Out[67]= 1/2 (((-1 + I Sqrt[7 - 4 Sqrt[2]]) Log[   
        1/2 (1 - I Sqrt[7 - 4 Sqrt[2]]) + x])/(   
      2 (2 + 5/2 (-1 + I Sqrt[7 - 4 Sqrt[2]]) +   
         3/4 (-1 + I Sqrt[7 - 4 Sqrt[2]])^2 +   
         1/4 (-1 + I Sqrt[7 - 4 Sqrt[2]])^3)) + ((-1 -   
         I Sqrt[7 - 4 Sqrt[2]]) Log[   
        1/2 (1 + I Sqrt[7 - 4 Sqrt[2]]) + x])/(   
      2 (2 + 5/2 (-1 - I Sqrt[7 - 4 Sqrt[2]]) +   
         3/4 (-1 - I Sqrt[7 - 4 Sqrt[2]])^2 +   
         1/4 (-1 - I Sqrt[7 - 4 Sqrt[2]])^3)) + ((-1 +   
         I Sqrt[7 + 4 Sqrt[2]]) Log[   
        1/2 (1 - I Sqrt[7 + 4 Sqrt[2]]) + x])/(   
      2 (2 + 5/2 (-1 + I Sqrt[7 + 4 Sqrt[2]]) +   
         3/4 (-1 + I Sqrt[7 + 4 Sqrt[2]])^2 +   
         1/4 (-1 + I Sqrt[7 + 4 Sqrt[2]])^3)) + ((-1 -   
         I Sqrt[7 + 4 Sqrt[2]]) Log[   
        1/2 (1 + I Sqrt[7 + 4 Sqrt[2]]) + x])/(   
      2 (2 + 5/2 (-1 - I Sqrt[7 + 4 Sqrt[2]]) +   
         3/4 (-1 - I Sqrt[7 + 4 Sqrt[2]])^2 +   
         1/4 (-1 - I Sqrt[7 + 4 Sqrt[2]])^3)))   
      
   I find similar results from AXIOM and FriCAS. Is this a limitation of Rioboo's   
   algorithm?   
      
   Cheers,   
      
   Sam   
      
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