From: samuel.thomas.blake@gmail.com   
      
   Hi All,   
      
   Is there any way to a priori estimate of the time it will take the   
   Risch-Trager-Bronstein algorithm to compute the integral of   
      
   (1 - x^4)/((x^4 + x^2 + 1)*(x^5 - x^3)^(1/4))?   
      
   I left it for 30 minutes and it was still crunching away...   
      
   It takes my package 24 seconds on my old laptop   
      
   In[22299]:= IntegrateAlgebraic[(1 - x^4)/((x^4 + x^2 + 1) (x^5 - x^3)^(1/4)),   
   x] // Timing   
      
   Out[22299]= {23.9914, -(((1/3 (17 + 12 Sqrt[2]))^(1/8)   
        ArcTan[(   
        3 Sqrt[1 + Sqrt[2]] x)/(-3 Sqrt[1 + Sqrt[2]] x +   
         3 Sqrt[2 + 2 Sqrt[2]] x - 2^(3/4) 3^(7/8) (-x^3 + x^5)^(1/4))])/   
      2^(3/4)) + ((1/3 (17 + 12 Sqrt[2]))^(1/8)   
       ArcTan[(   
       3 Sqrt[1 + Sqrt[2]] x)/(-3 Sqrt[1 + Sqrt[2]] x +   
        3 Sqrt[2 + 2 Sqrt[2]] x + 2^(3/4) 3^(7/8) (-x^3 + x^5)^(1/4))])/   
     2^(3/4) -   
     ArcTan[(-2^(1/4) (3/(17 + 12 Sqrt[2]))^(1/8) x^2 + (-3 x^2 +   
        3^(3/4) (17 + 12 Sqrt[2])^(1/4) Sqrt[-x^3 + x^5])/(   
       2^(1/4) 3^(7/8) (17 + 12 Sqrt[2])^(1/8)))/(   
      x (-x^3 + x^5)^(1/4))]/(2^(3/4) (3 (17 + 12 Sqrt[2]))^(1/8)) +   
     ArcTanh[(   
      3 (17/8748 + Sqrt[2]/729)^(1/8) x^2 +   
       3^(3/4) (17/8748 + Sqrt[2]/729)^(1/8) Sqrt[-x^3 + x^5])/(   
      x (-x^3 + x^5)^(1/4))]/(   
     2^(3/4) (3 (17 + 12 Sqrt[2]))^(1/8)) + ((1/3 (17 + 12 Sqrt[2]))^(   
      1/8) ArcTanh[(((3/(17 + 12 Sqrt[2]))^(1/8) x^2)/2^(1/4) +   
        Sqrt[-x^3 + x^5]/(2^(1/4) (3 (17 + 12 Sqrt[2]))^(1/8)))/(   
       x (-x^3 + x^5)^(1/4))])/2^(3/4)}   
      
   Cheers,   
      
   Sam   
      
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