oldk1331@gmail.com schrieb:   
   >   
   > Back to topic: yes, the form of integrand matters, at least to   
   > heuristic algorithm.  So we can grade integrator based on its output,   
   > we can also grade integrator based on its input, although this will   
   > harder.   
   >   
   > Take the famous risch algorithm wikipedia example #2:   
   >   
   >   int((x^2+2*x+1+(3*x+1)*sqrt(x+log(x)))/(x*sqrt(x+log(x))*(x+   
   qrt(x+log(x)))),x)   
   >   
   > (Is this in the CAS test suite? Can Rubi solve it?)   
   >   
   > The only winner to solve this integral so far is SymPy,   
   > f=(x**2+2*x+1+(3*x+1)*sqrt(x+log(x)))/(x*sqrt(x+log(x))*(x+sqrt(x+log(x))))   
   > integrate(f, x) can solve it, while   
   > integrate(cancel(f), x) can't, 'cancel' expands the denominator.   
   >   
   > That clearly shows the strength and limitation of heuristic algorithm.   
   >   
      
   In the test suites offered on the Rubi website, this is included as the   
   last-but-one problem of:   
      
      
      
   where the integrand reads:   
      
     ((x^2 + 2*x + 1)*SQRT(x + LOG(x)) + (3*x + 1)*LOG(x) + 3*x^2 + x)/   
     ((x*LOG(x) + x^2)*SQRT(x + LOG(x)) + x^2*LOG(x) + x^3)   
      
   which simplifies to:   
      
     (x^2 + 2*x + 1 + (3*x + 1)*SQRT(x + LOG(x)))/   
     (x*(x*SQRT(x + LOG(x)) + x + LOG(x)))   
      
   where the denominator may be factored as follows:   
      
     (x^2 + 2*x + 1 + (3*x + 1)*SQRT(x + LOG(x)))/   
     (x*SQRT(x + LOG(x))*(x + SQRT(x + LOG(x))))   
      
   However, this test problem is commented out, which may indicate that   
   Rubi cannot solve it. The FriCAS web interface currently returns:   
      
   >> Error detected within library code:   
      integrate: implementation incomplete (constant residues)   
      
   for either of the last two versions.   
      
   Martin.   
      
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