clicliclic@freenet.de wrote:   
   >   
   > 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.   
      
   Algebraicaly a = sqrt(b) means the same as a^2 = b.  So algebraicaly   
   both signs have equal rights.  There may be extra conditions which   
   which exclude one of roots, but if they are then in a sense   
   input data is incomplete.  It is well-known that when roots   
   are independent then signs can not be distinguished by purely   
   algebraic means.  One practical consequence of this is that one   
   can do calculations as there was a single "true" root and then   
   just consitently flip signs to get result for another root.   
      
   In case of dependent roots different signs can lead to essentialy   
   different results.  If there are some extra constraints that   
   allow to choose specific root, then of course one can do calculation   
   just for a single root.  But lacking extra information _full_   
   answer must consider all sign combinations.   
      
   > 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.   
      
   Yes.  AFAICS one choice of signs corresponds to root of this quartic,   
   other choices do not correspond to roots.   
      
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