Waldek Hebisch schrieb:   
   >   
   > clicliclic@freenet.de wrote:   
   > >   
   > > Hello!   
   > >   
   > > How long does it take the popular integrators to evaluate the following   
   > > elementary indefinite integrals? Imagine them to appear in terms of some   
   > > series expansion:   
   > >   
   > > INT(1/((1+x+2*x^2)^5*(3-2*x)^(11/2)),x)   
   > >   
   > > INT(1/((1+x+2*x^2)^10*(3-2*x)^(21/2)),x)   
   > >   
   > > INT(1/((1+x+2*x^2)^20*(3-2*x)^(41/2)),x)   
   > >   
   > > and:   
   > >   
   > > INT(1/((1+x+2*x^2)^5*(3-2*x+x^2)^(11/2)),x)   
   > >   
   > > INT(1/((1+x+2*x^2)^10*(3-2*x+x^2)^(21/2)),x)   
   > >   
   > > INT(1/((1+x+2*x^2)^20*(3-2*x+x^2)^(41/2)),x)   
   > >   
   > > Does Maple beat FriCAS? Does Rubi beat Mathematica?   
   > >   
   > > Martin.   
   >   
   > For FriCAS results for 'integrate' are somewhat embarrassing:   
   > 0.20s, 0.49s, 21.85s, 0.53s, 1.89s and the sixth one would   
   > take more than few hours.  Core integrator is reasonably fast,   
   > calling 'lfintegrate' I get: 0.02s, 0.04s, 0.11s, 0.07s,   
   > 0.20s, 1.02s.  'integrate' tries to make result nicer and   
   > in the process attempts to factor largish integers which   
   > takes a lot of time...   
   >   
   > Remark: 'lfintegrate' gives correct result only if certain   
   > assumptions are satified.  'integrate' transforms integrals   
   > so that 'lfintegrate' can handle them, calls 'lfintegrate'   
   > and then tries to make the result nicer.  Your integrals   
   > are simple and well-behaved, so can be sent directly to   
   > 'lfintegrate'.   
   >   
      
   If 'integrate' on the last one could be accelerated to about 60 seconds,   
   there would be no reason for embarassment anymore ...   
      
   Martin.   
      
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