From: hebisch@math.uni.wroc.pl   
      
   clicliclic@freenet.de wrote:   
   >   
   > clicliclic@freenet.de schrieb:   
   > >   
   > > integrate((2 - (k + 1)*x)/((1 - (k + 1)*x)   
   > >   *(x*(1 - x)*(1 - k*x))^(1/3)), x)   
   > >   
   > > >> Error detected within library code:   
   > >    impossible   
   > >   
   > > It is good to be cognizant of one's limitations ;).   
   > >   
   >   
   > ... and another quick FriCAS 1.2.3 on-line experiment:   
   >   
   > integrate((a + b*x + c*x^2)/((1 - x + x^2)*(1 - x^3)^(1/3)), x)   
   >   
   > >> Error detected within library code:   
   >    integrate: implementation incomplete   
   >    (residue poly has multiple non-linear factors)   
   >   
   > Is there no workable treatment of "multiple non-linear factors" other   
   > than Kauer's Groebner-basis heuristics?   
      
   Beware that Kauer's heuristics is incomplete.  AFAICS the   
   your first example is one when this heurisctic fail.  Namely,   
   rationalizing substitution gives integral of the form   
      
   \sum w*log(x - w*(x*(1 - x)*(1 - k*x))^(1/3))   
      
   where w runs over cube roots of 1/k.  Derivative of the   
   log has pole at 0 and at point over 1/(k+1).  Trager   
   method depends on rearanging integral in such a form that   
   logands are regular at 0.  So Trager method produces   
   3 divisors, each one corresponding to point with   
   x = 1/(k+1) and y beeing one ot the roots of (x*(1 - x)*(1 - k*x)).   
   The same happens in Kaures method.  Next Kaures tries   
   to find function having zero as specified by the divisor   
   or its power and arbitrary behaviour at infinity.  For this   
   to succeed the divisor have to be torsion divisor.  But   
   this is unlikely to be the case.  At least my partial   
   implementation of Kauers method gives up on this example.   
      
   The example is amenable to Trager method.  If I replace k   
   by k^3, then FriCAS can compute the integral.  Such replacement   
   is equivalent to extending field of constants by cube root   
   of k and FriCAS could do this automatically.  It is not   
   done ATM because this can lead to excessive execution time   
   (in this case it is 287.49 sec on fast machine).  I did   
   not analyze the second example, but it seem to be similar.   
      
   Let me add that that trying Kauers method I substituted   
   5 for k.  Without such substitution Groebner basis   
   computations were quite slow.   
      
   --   
                                 Waldek Hebisch   
   hebisch@math.uni.wroc.pl   
      
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