antispam@math.uni.wroc.pl schrieb:   
   >   
   > clicliclic@freenet.de wrote:   
   > >   
   > > clicliclic@freenet.de schrieb:   
   > > >   
   > > > Here's a much more deadly failure:   
   > > >   
   > > > setSimplifyDenomsFlag(true)   
   > > >   
   > > > integrate((7*x + 6*sqrt(17*sqrt(2) + 23) + 9*sqrt(2) + 34)   
   > > > /((x^2 + 2*x*(3*sqrt(sqrt(2) + 1) + 1) + 6*sqrt(58*sqrt(2) + 82)   
   > > >  - 18*sqrt(2) - 26)*sqrt(x^3 - 30*x - 56)), x)   
   > > >   
   > > > is returned unevaluated on the web interface. But my oracle says that   
   > > > the integand is pseudo-elliptic according to Goursat:   
   > > >   
   > > > goursat3((7*x + 6*SQRT(17*SQRT(2) + 23) + 9*SQRT(2) + 34)   
   > > > /(x^2 + 2*x*(3*SQRT(SQRT(2) + 1) + 1) + 6*SQRT(58*SQRT(2) + 82)   
   > > >  - 18*SQRT(2) - 26), x, -56, -30, 0, 1)   
   > > >   
   > > > [false, false, false, true]   
   > > >   
   > > > The last answer counts.   
   > > >   
   > >   
   > > Since the oracle was characteristically cryptic, I should supply some   
   > > details. Putting x = (6 - 3*SQRT(2))*t - 4 one finds:   
   > >   
   > > SQRT(x^3 - 30*x - 56) =   
   > > 3*SQRT(12 - 6*SQRT(2))*SQRT(t*(1 - t)*(1 - (SQRT(2) - 1)^2*t))   
   > >   
   > > So Legendre's modulus k of the elliptic radical takes the nice special   
   > > value SQRT(2) - 1. The quadratic in the denominator of the integrand   
   > > divides the division polynomial of order four for the elliptic curve   
   > > y^2 = x^3 - 30*x - 56. The antiderivative consists of two terms:   
   > >   
   > > 6*(2*SQRT(SQRT(2) + 1) - SQRT(2))/7   
   > >  * INT((7*x + 6*SQRT(17*SQRT(2) + 23) + 9*SQRT(2) + 34)   
   > >  / ((x^2 + 2*x*(3*SQRT(SQRT(2) + 1) + 1) + 6*SQRT(58*SQRT(2) + 82)   
   > >  - 18*SQRT(2) - 26)*SQRT(x^3 - 30*x - 56)), x)   
   > >  = 2/(SQRT(3*SQRT(2) + 6) - SQRT(3)*2^(3/4))   
   > >  * ATANH((SQRT(3*SQRT(2) + 6) - SQRT(3)*2^(3/4))*(x - 3*SQRT(2) - 2)   
   > >  / SQRT(x^3 - 30*x - 56)) - SQRT(6 - 3*SQRT(2))/3   
   > >  * ATAN(SQRT(6*SQRT(2) + 12)*(x + 4)/SQRT(x^3 - 30*x - 56))   
   > >   
   > > Having this integral declared non-elementary constitutes a bug that   
   > > seriously undermines trust in the FriCAS integrator.   
   > >   
   >   
   > You hit known limitation of current implementation.  Your integral   
   > contains dependent roots.  If I eliminate dependent roots as below:   
   >   
   > setSimplifyDenomsFlag(true)   
   > ff := (7*x + 6*sqrt(17*sqrt(2) + 23) + 9*sqrt(2) + 34) _   
   > /((x^2 + 2*x*(3*sqrt(sqrt(2) + 1) + 1) + 6*sqrt(58*sqrt(2) + 82) _   
   >  - 18*sqrt(2) - 26)*sqrt(x^3 - 30*x - 56))   
   >   
   > s2 := sqrt(2)*sqrt(sqrt(2) + 1)^5   
   > s3 := (sqrt(2) + 3)*sqrt(sqrt(2) + 1)   
   > ff2 := eval(ff, [sqrt(58*sqrt(2) + 82) = s2, sqrt(17*sqrt(2) + 23) = s3])   
   > integrate(ff2, x)   
   >   
   > I get:   
   >   
   > [solution snipped]   
   >   
   > which is large but seem to correspond to your desired value.  OTOH   
   > there are 3 other combinations of signs of roots and FriCAS returns   
   > unevaluated for those combinations.   
   >   
   > Note: FriCAS should find out that roots are dependent, but this part   
   > currently remains unimplemented.   
      
   Yes, the only numerical roots appearing here are sqrt(2) and   
   sqrt(sqrt(2) + 1), corresponding to sqrt(w) and sqrt(w - 1) in the PS   
   to my post of 04 Apr 2018, but I didn't connect this hazy background   
   fact with my rather clear knowledge that FriCAS may give wrong results   
   in the case of dependent roots. My interest was simply in checking the   
   integrability of certain algebraics; this question has been resolved   
   without using FriCAS once its results were found unreliable, by the   
   way.   
      
   Does your statement about the signs of roots mean that the correct   
   method of reducing dependent roots must be found by trying different   
   sign combinations? Your s2, s3 look very natural to me.   
      
   Martin.   
      
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