clicliclic@freenet.de  wrote:   
   >   
   > antispam@math.uni.wroc.pl schrieb:   
   > >   
   > > samuel.thomas.blake@gmail.com wrote:   
   > > >   
   > > > Here's a couple of other error messages I found in 1.2.6 (on OSX).   
   > > >   
   > > > [...]   
   > > >   
   > > > integrate(((-1+x^4)*(1+x^2+x^4)*(-1+x^2-x^4)^(1/2))/(1+x^4)^3,x);   
   > > >   
   > > >    >> Error detected within library code:   
   > > >    catdef: division by zero   
   > > >   
   > > >   
   > > > integrate(((4+x^6)*(-4+x^4+2*x^6)*(32-14*x^4-32*x^6-4*x^8+   
   *x^10+8*x^12)^(1/2))/(x^9*(-2+x^6)),x);   
   > > >   
   > > >    >> Error detected within library code:   
   > > >    integrate: implementation incomplete (residue poly has multiple   
   non-linear factors)   
   > >   
   > > The two above are known problems, each will take more work to fix.   
   > > Actually, the first one requires reorganization of integration   
   > > routines.  The second one is king of open problem: known methods   
   > > of solving may require prohibitivly long time and nobody knows   
   > > of really efficient method.   
   > >   
   >   
   > By the way, the radicand sqrt(-1+x^2-x^4) is negative everywhere on the   
   > real axis, and FriCAS 1.3.6 succeeds when its sign is inverted:   
   >   
   >   integrate(((-1+x^4)*(1+x^2+x^4)*(1-x^2+x^4)^(1/2))/(1+x^4)^3,x)   
   >   
   > (((-5)*x^8+(-10)*x^4+(-5))*atan((2*x*(x^4+(-1)*x^2+1)^(1/2))   
   > /(x^4+(-2)*x^2+1))+((-6)*x^5+(-4)*x^3+(-6)*x)*(x^4+(-1)*x^2+1)^(1/2))   
   > /(16*x^8+32*x^4+16)   
   >   
   > But this may be just an accident.   
      
   Well, FriCAS looks at residues of the integral.  It is easy   
   to prove that derivative of logaritm has integral residues.   
   So residues give coefficients of logaritms and decide   
   which part can combine into single logaritm and which can   
   not.  For example, if you have 1, -1, sqrt(2), -sqrt(2)   
   as resudues, then 1 and -1 form one part which can be   
   handled independenty from sqrt(2) and -sqrt(2).  FriCAS   
   represents residues using minimal polynomial, that is   
   sqrt(2) is represented as solution to r^2 - 2 = 0.   
   If there are multiple nonlinear equations for residues   
   then finding relations between them becomes computationaly   
   quite expensive and FriCAS tries a few tricks and   
   if tricks do not work it just gives up.   
      
   --   
                                 Waldek Hebisch   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)