From: samuel.thomas.blake@gmail.com   
      
   On Tuesday, October 31, 2023 at 5:59:01 AM UTC+11, nob...@nowhere.invalid   
   wrote:   
   > "clicl...@freenet.de" schrieb:    
   > >    
   > > Sam Blake schrieb:   
   > > >    
   > > > Here are some nice cube root-type integrals which may or may not be    
   > > > integrable using your Goursat adaption. I am guessing that as Rubi    
   > > > cannot integrate these that they may not be Goursat type integrals?    
   > > >    
   > > > (1 + x)/(x*(-1 + x^3)^(1/3))    
   > > > (-1 + x^2)/(x*(-1 + x^3)^(2/3))    
   > > > ((1 + x)*(-1 + x^3)^(1/3))/((-1 + x)^2*x)    
   > > > ((-1 + x)^2*(1 + x))/(x*(1 + x + x^2)*(-1 + x^3)^(1/3))    
   > > > ((1 + x)*(-1 + x^3)^(2/3))/((-1 + x)^3*x)    
   > > > (x*(-2 + x^3))/((-1 + x^3)^(1/3)*(-1 + x^3 + x^6))    
   > > > (x^3*(2 + x^3))/((1 + x^3)^(2/3)*(1 + x^3 + x^6))    
   > > > ((1 - x^3)^(1/3)*(-2 + x^3))/(x^3*(-1 + x^3 + x^6))    
   > > > ((2 + x + x^2)*(-1 + x^3)^(1/3))/(x*(-1 + x^2)^2*(-3 - 2*x + x^2 + x^3))    
   > > > (1 + 3*x)/((-1 + 3*x)*(-x + x^3)^(1/3))    
   > > > ((-1 + x)*(1 + 3*x))/((-1 + 3*x)*(-x + x^3)^(2/3))    
   > > > ((-1 + x)^2*(1 + 3*x))/(x*(1 + x)*(-1 + 3*x)*(-x + x^3)^(1/3))    
   > > > (x*(1 + x)*(1 + 3*x))/((-1 + x)*(-1 + 3*x)*(-x + x^3)^(2/3))    
   > > > (x*(1 + x)*(1 + 3*x))/((-1 + x)^2*(-1 + 3*x)*(-x + x^3)^(1/3))    
   > > >    
   > >    
   > > There are 14 integrands here. The first two are readily integrated by    
   > > Derive 6.10 and should therefore be doable by standard recipes from    
   > > books like G&R and Timofeev:    
   > >    
   > > INT((x + 1)/(x*(x^3 - 1)^(1/3)), x)    
   > > INT((x^2 - 1)/(x*(x^3 - 1)^(2/3)), x)    
   > >    
   > > So Rubi could and should be tought to do these as well.    
   > >    
   > > The subsequent twelve integrands, however, fail in Derive 6.10:    
   > >    
   > > INT((x + 1)*(x^3 - 1)^(1/3)/(x*(x - 1)^2), x)    
   > > INT((x + 1)*(x - 1)^2/(x*(x^2 + x + 1)*(x^3 - 1)^(1/3)), x)    
   > > INT((x + 1)*(x^3 - 1)^(2/3)/(x*(x - 1)^3), x)    
   > > INT(x*(x^3 - 2)/((x^3 - 1)^(1/3)*(x^6 + x^3 - 1)), x)    
   > > INT(x^3*(x^3 + 2)/((x^3 + 1)^(2/3)*(x^6 + x^3 + 1)), x)    
   > > INT((1 - x^3)^(1/3)*(x^3 - 2)/(x^3*(x^6 + x^3 - 1)), x)    
   > > INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),    
   > > x)    
   > > INT((3*x + 1)/((3*x - 1)*(x^3 - x)^(1/3)), x)    
   > > INT((x - 1)*(3*x + 1)/((3*x - 1)*(x^3 - x)^(2/3)), x)    
   > > INT((x - 1)^2*(3*x + 1)/(x*(x + 1)*(3*x - 1)*(x^3 - x)^(1/3)), x)    
   > > INT(x*(x + 1)*(3*x + 1)/((x - 1)*(3*x - 1)*(x^3 - x)^(2/3)), x)    
   > > INT(x*(x + 1)*(3*x + 1)/((x - 1)^2*(3*x - 1)*(x^3 - x)^(1/3)), x)    
   > >    
   > > My Goursat tests for cube-root integrands determine all but the four    
   > > integrals from #6 to #9 to be directly of Goursat type, which means    
   > > that the integrands become rational under one of the characteristic    
   > > substitutions. Of the other four, the first three become Goursat    
   > > integrable under the fairly obvious substitution t = x^3.    
   > >    
   > > A more involved substitution must have been used to scramble integrand    
   > > #9; even FriCAS fails on this one, but succeeds once t = x^3 has been    
   > > applied. The substitution doesn't make the integrand look nicer    
   > > though, and while the change cannot be considered momentous either,    
   > > for FriCAS it evidently is. The integrand simplifies a bit when an    
   > > algebraic part of the antiderivative is removed:    
   > >    
   > > INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/    
   > > (x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =    
   > > (x^3 - 1)^(1/3)/(x^2 - 1) +    
   > > INT((x^2 - 1)*(x^2 + x + 2)/    
   > > (x*(x^3 - 1)^(2/3)*(x^3 + x^2 - 2*x - 3)), x)    
   > >    
   > > There is no reason to assume that Rubi can handle Goursat type    
   > > integrals in general.    
   > >   
   > For the record, and as an "optimal" reference solution for Nasser's    
   > integrator tests, a compact antiderivative of the non-Goursat integral    
   > #9:   
   > INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),    
   > x)   
   > is given by a sum of the following five components:    
   >    
   > - INT(x*(x^2 + x + 2)/((x + 1)*(x^2 - 1)*(x^3 - 1)^(2/3)), x) =   
   > (x^3 - 1)^(1/3)/(x^2 - 1)   
   > INT(2/(3*x*(x^3 - 1)^(2/3)), x) =    
   > - 2*SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(x^3 - 1)^(1/3)))    
   > + 1/3*LN((x^3 - 1)^(1/3) + 1) - 1/9*LN(x^3)    
   >    
   > - INT(x^2*(2*x^6 - 19*x^3 - 10)    
   > /(3*(x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =    
   > SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(2*(x^3 - 1)^(2/3) + x^3 - 4)    
   > /((x^3 - 1)^(1/3)*(2*(x^3 - 1)^(1/3) + 3))))    
   > + 1/6*LN(- 2*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) + x^3 - 4)    
   > - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)    
   >    
   > INT(x*(x^3 - 1)^(1/3)*(3*x^3 - 1)/(x^9 - 2*x^6 + x^3 - 27), x) =    
   > SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x*(x^3 - 1)^(2/3))/3))    
   > + 1/6*LN(x*(x^3 - 1)^(2/3) - 3) - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)    
   >    
   > INT((x^9 - 3*x^6 + 8*x^3 + 3)    
   > /((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =    
   > SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3)    
   > /(3*((x^3 - 1)^(1/3) + x^2))))    
   > + 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)    
   > - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)    
   >    
   > All but the first and last of these turn into classical cases of    
   > elementary integrals under the substitutions t = x^3 or t = 1/x^3.    
   >    
   > Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should be    
   > merged into one. Is the FriCAS antiderivative suboptimal because its    
   > leaf count exceeds twice that of this optimal result?    
   >    
   > Martin.   
      
   Hey Martin,    
      
   Here's my solution for this integral:    
      
   In[431]:= IntegrateAlgebraic[((x^2 + x + 2) (x^3 - 1)^(1/3))/(x (x^2 - 1)^2   
   (x^3 + x^2 - 2 x - 3)), x]   
      
   Out[431]= (-1 + x^3)^(1/3)/(-1 + x^2) - ArcTan[(Sqrt[3] (-1 + x^3)^(1/3))/(-2   
   + 2 x^2 + (-1 + x^3)^(1/3))]/Sqrt[3] +    
   1/3 Log[1 - x^2 + (-1 + x^3)^(1/3)] - 1/6 Log[1 - 2 x^2 + x^4 + (-1 + x^2) (-1   
   + x^3)^(1/3) + (-1 + x^3)^(2/3)]   
      
   Where the substitution u == (1-x^2)/(-1+x^3)^(1/3) reduces the integral to   
   1/(u^2 (1+u^3)).    
      
   Cheers,   
      
   Sam   
      
   --- SoupGate-Win32 v1.05   
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