From: samuel.thomas.blake@gmail.com   
      
   On Sunday, October 27, 2019 at 5:16:01 AM UTC+10, nob...@nowhere.invalid wrote:   
   > Albert Rich schrieb:   
   > >    
   > > On Tuesday, October 22, 2019 at 10:22:57 AM UTC-10, clicl...@freenet.de   
   wrote:    
   > > >    
   > > > FriCAS 1.3.5 determines the algebraic integral:    
   > > >    
   > > > integrate((3*x + 2)/((9*x^2 + 52*x - 12)*(3*x^2 + 4)^(1/3)), x)    
   > > >    
   > > > to be non-elementary, even though it just equals:    
   > > >    
   > > > SUBST(INT(3/(t^3 + 196), t), t, (3*x - 10)/(3*x^2 + 4)^(1/3))    
   > > >    
   > >   
   > > I added this problem to Rubi's integration test-suite.    
   > >    
   > > Short of Risch, is there an algorithm a rule-based integrator could    
   > > use to determine this substitution?    
   > >   
   > That's the idea: nudging FriCAS to learn more about cube roots!    
   >    
   > I simply analyzed which integrands of the form (a1 + b1*x)/((c1 + d1*x    
   > + e1*x^2)*croc(x)) could be rationalized as 1/(alpha + beta*t^3) for    
   > t = (p + q*x)/croc(x), where croc(x) represents the cube root of any    
   > cubic whose roots are not too expensive to handle. This integrand form    
   > is invariant under Möbius transformations, whence the analysis can be    
   > simplified by specializing the radicand to a + c*x^2, say, and    
   > generalizing to (a + b*x)*(c + d*x + e*x^2) only afterwards. Quite    
   > obviously, such an ansatz leads to a system of polynomial equations    
   > among the coefficients.    
   >    
   > The solutions comprise some families of Goursat cases, which name I    
   > apply when -p/q is a radicand root, and one non-Goursat family, of    
   > which the integral mistreated by FriCAS is a member (the FriCAS failure    
   > appears to be systematic). Any Goursat or non-Goursat solution can be    
   > easily implemented as a Rubi rule, but the former cases are better    
   > handled through tests and substitutions equivalent to those in    
   > Goursat's 1887 paper for square roots of quartics, since that approach    
   > applies to arbitrary rational factors in the integrand and encompasses    
   > rules for the appropriate integrand splitting.    
   >    
   > Just in case, here's my explicit non-Goursat antiderivative:    
   >    
   > INT((3*x + 2)/((x + 6)*(9*x - 2)*(3*x^2 + 4)^(1/3)), x) =    
   > - 1/(28*14^(1/3))*(LN(14^2*(3*x^2 + 4) + (3*x - 10)^3)    
   > - 3*LN(14^(2/3)*(3*x^2 + 4)^(1/3) + 3*x - 10)    
   > + 2*SQRT(3)*ATAN(1/SQRT(3)*(1 + 2*(10 - 3*x)    
   > /(14^(2/3)*(3*x^2 + 4)^(1/3)))))    
   >    
   > Martin.   
      
   Better late than never - Mathematica 13.2 gets this one out in under half a   
   second:   
      
   In[7146]:= $Version   
   Out[7146]= "13.2.0 for Mac OS X x86 (64-bit) (November 18, 2022)"   
      
   In[7147]:= Integrate[(2 + 3 x)/((4 + 3 x^2)^(1/3) (-12 + 52 x + 9 x^2)), x] //   
   Timing   
      
   Out[7147]= {0.495113, -(1/(28 14^(1/3)))(2 Sqrt[3]   
         ArcTan[(10 14^(1/3) - 3 14^(1/3) x + 7 (4 + 3 x^2)^(1/3))/(   
         7 Sqrt[3] (4 + 3 x^2)^(1/3))] -    
       2 Log[-10 14^(1/3) + 3 14^(1/3) x + 14 (4 + 3 x^2)^(1/3)] +    
       Log[100 14^(2/3) - 60 14^(2/3) x + 9 14^(2/3) x^2 +    
         196 (4 + 3 x^2)^(2/3) + 14 (10 - 3 x) (56 + 42 x^2)^(1/3)])}   
      
   and the general form of this integral in around 1 second:    
      
   In[7156]:= Integrate[(3 a d - 2 b c x + a c x^2)/((d + c x^2)^(1/3) (b^3 + d e   
   + 3 a b^2 x + 3 a^2 b x^2 + c e x^2 + a^3 x^3)), x] // Timing   
      
   Out[7156]= {0.903716, (1/(2 e^(2/3)))(2 Sqrt[3] ArcTan[(Sqrt[3] e^(1/3) (d + c   
   x^2)^(1/3))/(-2 b - 2 a x + e^(1/3) (d + c x^2)^(1/3))] +    
      2 Log[b + a x + e^(1/3) (d + c x^2)^(1/3)] -    
      Log[b^2 + 2 a b x + a^2 x^2 - e^(1/3) (b + a x) (d + c x^2)^(1/3) +e^(2/3)   
   (d + c x^2)^(2/3)])}   
      
   Sam   
      
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