clicliclic@freenet.de schrieb:   
   >   
   > 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.   
   >   
      
   The sign problem may reflect that there is just a single numerical root   
   of degree four in this integrand. Its rational part:   
      
   (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)   
      
   multiplied by 6*(- sqrt(sqrt(2) + 1) + sqrt(2) + 2)/7 equals:   
      
   (3*(t^2 + 8*t + 22)*x - t^3 + 12*t^2 + 90*t + 392)   
   /((x - t)*((2*t + 8)*x + t^2 + 4*t + 18))   
      
   for t = - 6*sqrt(sqrt(2) + 1) + 3*sqrt(2) + 2, one among four roots of   
   the quartic t^4 - 8*t^3 - 84*t^2 - 464*t - 956 = 0.   
      
   Martin.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)