From: Albert_Rich@msn.com   
      
   On Wednesday, October 30, 2019 at 11:12:48 AM UTC-10, clicl...@freenet.de   
   wrote:   
   > Albert Rich schrieb:   
   > >    
   > > On Saturday, October 26, 2019 at 9:16:01 AM UTC-10, clicl...@freenet.de   
   wrote:   
   > > >   
   > > > 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)))))   
   > > >   
   > >    
   > > Thanks to Martin's post, the next release of Rubi will be able to   
   > > integrate wrt x expressions of the form   
   > >    
   > >     (g + h*x)/((a + c*x^2)^(1/3)*(d + e*x + f*x^2))   
   > >    
   > > provided   
   > >    
   > >     (f*g^2 - e*g*h + d*h^2)*(c*g^2 + 3*a*h^2 + 2*c*g*h*p) = 0   
   > > and   
   > >     a*(f*g^2 - e*g*h + 3*d*h^2) + 5*a*f*g*h*p - c*f*g*h*p^3 = 0   
   > > where   
   > >     p = (-c*d*h "+/-" sqrt(c*h*(c*d^2*h + 9*a*f*(e*g - d*h))))/(3*c*f*g)   
   > >    
   > > using the substitution rule   
   > >    
   > >     integrate((g + h*x)/((a + c*x^2)^(1/3)*(d + e*x + f*x^2)), x) =   
   > >         3*a*g*h*subst(integrate(1/(3*a*d*h - c*f*g*p^3 + a*c*f*g*x^3), x),   
   > >             x, (p + x)/(a + c*x^2)^(1/3))   
   > >    
   > > In addition, by using the shift t = b/(2*c) + x, this rule will be   
   > > generalized to integrate expressions of the form   
   > >    
   > >     (g + h*x)/((a + b*x + c*x^2)^(1/3)*(d + e*x + f*x^2))   
   > >    
   > > having the necessary application conditions.   
   > >    
   >    
   > For some reason, the quoted post did not make it to the aioe.org nntp   
   > server. Here are some comments:   
   >    
   > The factor (f*g^2 - e*g*h + d*h^2) in the first application condition   
   > seems to check for cancellation of the rational fraction, which has me   
   > wondering why Rubi allows uncancelled integrands to penetrate out to   
   > the leaves of its decision tree, thus presumably requiring such checks   
   > in countless rules. I found the non-Goursat solution to become Goursat   
   > under cancellation, with either (p/q)^2 = - a/c or (p/q)^2 = 3*a/c   
   > (antiderivatives of the second type split into a pair of the first).   
   >    
   > Instead of eliminating q, I fixed beta = 1 for the solution and arrived   
   > at application conditions not involving radicals. Additionally putting   
   > b1 = e1 = 1 in:   
   >    
   >   INT((a1 + b1*x)/((c1 + d1*x + e1*x^2)*(a + c*x^2)^(1/3)), x)   
   >    = SUBST(INT(1/(alpha + t^3), t), t, (p + q*x)/(a + c*x^2)^(1/3))   
   >    
   > thus leads to:   
   >    
   >   9*a^2 - 3*a*a1*c*d1 + a1^2*c^2*c1 = 0   
   >   3*a*(6*a1 - d1) + a1*c*(3*a1^2 - 7*c1) = 0   
   >   3*q^2 = c   
   >   9*a1*p^2 = 3*a*d1 + a1*c*c1   
   >   2*a1*c*p = - q*(3*a + a1^2*c)   
   >   3*alpha = q*(d1 - a1) - 5*p   
   >    
   > where my equation for p^2 is redundant, leaving just one square root to   
   > extract for q. Solving the application conditions for c1 and d1 gives:   
   >    
   >   8*a1^2*c^2*c1 = - 3*(12*a^2 - (3*a + a1^2*c)^2)   
   >   8*a*a1*c*d1 = 12*a^2 + (3*a + a1^2*c)^2   
   >    
   > Remarkably, the rational denominator factors symbolically for any   
   > instance of this non-Goursat solution:   
   >    
   >   3*a*a1*c*(c1 + d1*x + x^2) = (3*a + a1*c*x)*(a1*c*c1 + 3*a*x)   
   >    
   > Finally, general cubic radicands (a + b*x)*(c + d*x + e*x^2) can be   
   > tackled via:   
   >    
   >   x = - (a*u - (a*d - 2*b*c))/(b*u - (2*a*e - b*d))   
   >   u = (x*(2*a*e - b*d) + a*d - 2*b*c)/(b*x + a)   
   >    
   > implying:   
   >    
   >   (a + b*x)*(c + d*x + e*x^2) = - 2*(a^2*e - a*b*d + b^2*c)^2   
   >     *(u^2 + 4*c*e - d^2)/(b*u - (2*a*e - b*d))^3   
   >    
   > A simple example with cubic radicand and e1 = 0 has already appeared in   
   > a Juli post of mine: INT((x - 1)/((x +1)*(x^3 + 2)^(1/3)), x).   
   >    
   > Knowing Albert, I expect that integrators will now encounter a whole   
   > bunch of instances of the non-Goursat solution in the integration test   
   > suite!   
   >    
   > Martin.   
      
   Setting BOTH your beta=1 and q=1 leads to an even simpler rule not involving    
   square roots at all!  So no need to worry if c is negative.   
      
   The next release of Rubi will be able to integrate expressions of the form   
      
     (g + h*x)/((a + c*x^2)^(1/3)*(d + e*x + f*x^2))   
      
   provided   
      
     3*c^2*f*g^4 - 2*c*(4*c*d - 9*a*f)*g^2*h^2 - 9*a^2*f*h^4 = 0   
   and   
     c^2*f*g^4 + 21*a^2*f*h^4 + 2*a*c*g*h^2*(3*f*g - 4*e*h) = 0   
      
   using the straight-forward substitution rule   
      
     int((g + h*x)/((a + c*x^2)^(1/3)*(d + e*x + f*x^2)), x) -->   
      
       6*c*g*h^2*subst(int(1/(A + B*x^3), x), x, (p + x)/(a + c*x^2)^(1/3))   
      
   where   
     A = 3*c*f*g^2 + 2*c*e*g*h + 15*a*f*h^2,   
     B = 2*c^2*f*g*h, and   
     p = -(c*g^2 + 3*a*h^2)/(2*c*g*h).   
      
   Of course, this rule will be generalized to integrate expressions of the form   
      
       (g + h*x)/((a + b*x + c*x^2)^(1/3)*(d + e*x + f*x^2))   
      
   having the necessary application conditions.   
      
   Albert   
      
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