clicliclic@freenet.de wrote:   
   >   
   > clicliclic@freenet.de schrieb:   
   > >   
   > > 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.   
   > >   
   >   
   > Still simpler, I suppose one may also put sqrt(2) = t^2 - 1, s2 = (t^2   
   >  - 1)*t^5, s3 = (t^2 + 2)*t, where t = sqrt(sqrt(2) + 1) is one root of   
   > the quartic t^4 - 2*t^2 - 1 = 0. This shows that there is no unique   
   > solution to the independentization of a given set of roots. Brrrrrr.   
   >   
   > If a reasonably fast algorithm is available that tests for the   
   > independence of algebraic numbers, users could be informed that their   
   > input must be revised. In the present case, relations among the roots   
   > sqrt(2), sqrt(sqrt(2) + 1), s2, and s3 of the polynomials t^2 - 2,   
   > t^4 - 2*t^2  - 1, t^4 - 164*t^2 - 4, and t^4 - 46*t^2 - 49 would have   
   > to be explored. But the integer-relation problem expressing the first   
   > numeric value in terms of the three others would involve 4*4*4 = 64   
   > coefficients. Brrrrrr.   
      
   There is general algorithm.  One "just" needs to check that defining   
   equation of the new root is irreducible with respect to previously   
   introduced roots.  In this case doing computation step by set we get:   
      
   (1) -> factor(x^2 - 2)   
      
            2   
      (1)  x  - 2   
                                             Type: Factored(Polynomial(Integer))   
   (2) -> r1 := sqrt(2)   
      
            +-+   
      (2)  \|2   
                                                           Type: AlgebraicNumber   
   (3) -> factor(x^2 - (sqrt(2) + 1), [r1])   
      
            2    +-+   
      (3)  x  - \|2  - 1   
                                     Type: Factored(Polynomial(AlgebraicNumber))   
   (4) -> r2 := sqrt(sqrt(2) + 1)   
      
            +--------+   
            | +-+   
      (4)  \|\|2  + 1   
                                                           Type: AlgebraicNumber   
   (5) -> factor(x^2 - (58*sqrt(2) + 82), [r1, r2])   
      
                               +--------+                   +--------+   
                      +-+      | +-+               +-+      | +-+   
      (5)  (x + (- 3 \|2  - 4)\|\|2  + 1 )(x + (3 \|2  + 4)\|\|2  + 1 )   
                                     Type: Factored(Polynomial(AlgebraicNumber))   
   (6) -> factor(x^2 - (17*sqrt(2) + 23), [r1, r2])   
      
                             +--------+                 +--------+   
      
   [continued in next message]   
      
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