antispam@math.uni.wroc.pl schrieb:   
   >   
   > clicliclic@freenet.de wrote:   
   > >   
   > > Some results are huge and of inferior quality because they involve   
   > > unresolved (yet resolvable) algebraic quantities; searching for "%%" I   
   > > find integral numbers 2.44 - 2.47 (#136 - #139), 2.50.2 - 2.50.4 (#142   
   > >  - #144), 2.82 (#177), and 5.66 (#401), amounting to nine integrals:   
   > >   
   > >   integrate(1/(a^5+x^5), x)   
   > >   
   > >   integrate(x/(a^5+x^5), x)   
   > >   
   > >   integrate(x^2/(a^5+x^5), x)   
   > >   
   > >   integrate(x^3/(a^5+x^5), x)   
   > >   
   > >   integrate(1/(x^2*(a^5+x^5)), x)   
   > >   
   > >   integrate(1/(x^3*(a^5+x^5)), x)   
   > >   
   > >   integrate(1/(x^4*(a^5+x^5)), x)   
   > >   
   > >   integrate(x^6/(3+2*x^5)^3, x)   
   > >   
   > >   integrate(tan(x)/(-1+sqrt(tan(x)))^2, x)   
   > >   
   >   
   > This is mostly ugly output.  FriCAS knows what to do with involved   
   > algebraics and they differentiate back with no problem.  Probably   
   > it would be better to return 'rootSum' here.   
   >   
   > > More importantly, I suspect that the results for integral number 1.64   
   > > (#69):   
   > >   
   > >   integrate(asin(x/a)^(3/2)/sqrt(a^2-x^2), x)   
   >   
   > I think we discussed this one -- FriCAS chooses branch   
   > which is OK for real a > 0 and x > 0 assuming positive   
   > roots.   
   >   
   > > and numbers 5.80 - 5.82 (#416 - #418):   
   > >   
   > >   integrate((cos(2*x)-sqrt(sin(2*x)))/sqrt(cos(x)^3*sin(x)), x)   
   > >   
   > >   "integrate((-2*sin(2*x)+sqrt(cos(x)*sin(x)^3))   
   > >   /(-sqrt(cos(x)^3*sin(x))+sqrt(tan(x))), x)"   
   > >   
   > >   integrate(((sin(x)/cos(x)^7)^(1/3)-3*tan(x))   
   > >   /(cos(x)^5*sin(x))^(2/3), x)   
   > >   
   > > do not really differentiate back and consequently are mathematically   
   > > wrong - I didn't attempt to check them as your 2D output is hard to   
   > > process.   
   >   
   > AFAICS 5.82 is OK: after differentiation one have to simplify   
   > trigonometrics but algebraics are as they should be, so no   
   > branch problems.  5.81 is converted to string because FriCAS   
   > signals error -- the error is presumably due to dependent   
   > algebraics.  ATM I can not say how bad 5.80 is.  The result   
   > is clearly too complicated.  AFAICS the difference of integrand   
   > and drivative of the integral on 0..%pi/2 when evaluated   
   > using pricipal branches is zero, but when plotting one   
   > gets some discrepancies due to poor numerical stability of   
   > the result.  If I run FriCAS simplifier on the difference,   
   > the result does not simplify to 0 due to presence   
   > of 'sqrt(-tan(x/2)^3+tan(x/2))' and 'sqrt(-2*tan(x/2)^3+2*tan(x/2))'.   
   > Only the second one is present in the integrand and first   
   > one looks spurious.   
   >   
   > > There may be others - can the evaluation of integral number   
   > > 5.77 (#413):   
   > >   
   > >   integrate(sqrt(sin(x)^5/cos(x)), x)   
   > >   
   > > be verified?   
   >   
   > FriCAS 'normalize' simplifies difference of derivative of the   
   > result and the integrand to 0.  But 'normalize' treats branches   
   > the same as integrator so there is some possiblity for   
   > branch violation.  AFAICS badness of this result is due   
   > to postprocessing, I want to get rid of it, but ATM it   
   > is needed for some other integrals.   
   >   
   > BTW: The "bad" postprocessing step assumes that integrator   
   > is dealing with real functions.  It may make result   
   > more messy but as long as assumption hold the result   
   > should be correct.  Of course once you try real looking   
   > functions which are in fact complex (like square roots   
   > of negative functions) the result may be wrong.   
   >   
   > > Wrong antiderivatives should earn you penalty points; if   
   > > generated deliberately they should disqualify the FriCAS integrator   
   > > from serious consideration.   
   > >   
   > > An easy and very natural way out is to reject integrands 1.64 and 5.80   
   > >  - 5.82 as "Risch incompatible: dependent roots" (the same would apply   
   > > to "dependent logarithms").   
   >   
   > Strictly speaking roots in 5.80 are independent due to factor   
   > of 2 in one of the roots.   
   >   
   > > Thus amended, and if other doubtful results   
   > > can be confirmed, the FriCAS version you tested would score at (705 -   
   > > 17)/705 = 97.589% solved - coming third, after Rubi and Mathematica.   
   > >   
      
   Among the 14 Timofeev problems commented out, the purely algebraic   
   number 6.17 (#474):   
      
     INT(x^4/(SQRT(10) - x^2)^(9/2), x) =   
     (7*SQRT(10)*x^5 - 2*x^7)/(350*(SQRT(10) - x^2)^(7/2))   
      
   represents the most interesting FriCAS failure in my view.   
      
   I would address the unresolved algebraic quantities in your results by   
   moving the boundary up to which explicit radicals are introduced: Not   
   only quadratics should be resolved, but also quartics resolvable into   
   nested square roots, as detected from a cubic resolvent that factors   
   rationally. They are distinguished by their Galois group as well.   
      
   In discussing the cases of dependent roots I am relying on   
   antiderivatives produced by the current web interface (FriCAS 1.3.1);   
   while their agreement with your results could not be checked, they   
   corroborate most of your findings.   
      
   The result for 1.64 (#69) appears to be the same as from earlier FriCAS   
   versions and fails to differentiate back even for real a and real x. For   
   radicals interpreted as single-valued, it needs to be multiplied by   
   a/SQRT(a^2 - x^2)*SQRT(1 - (x/a)^2), or to be guarded by equivalent   
   conditions. Following Albert's practice, the result should therefore   
   enter your Timofeev score with negative weight.   
      
   I find the result for 5.77 (#413) to differentiate back unconditionally,   
   whence it should hold for multivalued as well as single-valued radicals.   
   As so often in FriCAS antiderivatives, the ATAN terms are complicated by   
   splitting (compare Albert's single "optimal" term) as well as halving,   
   the latter also introducing artificial radicals. Thus, your ATAN   
   quadruplet (as rewritten by Derive):   
      
   3*SQRT(2)*ATAN((SQRT(-(2*SQRT(2)*COS(x)^2*SQRT(SIN(x)^5/COS(x))+~   
   COS(x)*(2*SQRT(2)*SIN(x)*SQRT(SIN(x)^5/COS(x))-4*SIN(x)^3)-SIN(x~   
   )^2)/SIN(x)^2)*(2*COS(x)^4-COS(x)^2*(SQRT(2)*SQRT(SIN(x)^5/COS(x~   
   ))+3)+SQRT(2)*SIN(x)*COS(x)*SQRT(SIN(x)^5/COS(x))+1)-SQRT(2)*COS~   
   (x)*SQRT(SIN(x)^5/COS(x))*(COS(x)-SIN(x)))/(SIN(x)^2*(2*SIN(x)*C~   
   OS(x)-1)))/16+3*SQRT(2)*ATAN(SQRT(2)*(SQRT(SIN(x)^5/COS(x))*SQRT~   
   (-(2*SQRT(2)*COS(x)^2*SQRT(SIN(x)^5/COS(x))+COS(x)*(2*SQRT(2)*SI~   
   N(x)*SQRT(SIN(x)^5/COS(x))-4*SIN(x)^3)-SIN(x)^2)/SIN(x)^2)+SQRT(~   
   2)*SIN(x)^3*COS(x)-SQRT(SIN(x)^5/COS(x))+SQRT(2)*SIN(x)^4)/(2*SI~   
   N(x)^3*(COS(x)-SIN(x))))/16-3*SQRT(2)*ATAN((SQRT((2*SQRT(2)*COS(~   
   x)^2*SQRT(SIN(x)^5/COS(x))+COS(x)*(2*SQRT(2)*SIN(x)*SQRT(SIN(x)^~   
   5/COS(x))+4*SIN(x)^3)+SIN(x)^2)/SIN(x)^2)*(2*COS(x)^4+COS(x)^2*(~   
   SQRT(2)*SQRT(SIN(x)^5/COS(x))-3)-SQRT(2)*SIN(x)*COS(x)*SQRT(SIN(~   
   x)^5/COS(x))+1)+SQRT(2)*COS(x)*SQRT(SIN(x)^5/COS(x))*(COS(x)-SIN~   
   (x)))/(SIN(x)^2*(2*SIN(x)*COS(x)-1)))/16+3*SQRT(2)*ATAN(SQRT(2)*~   
   (SQRT(SIN(x)^5/COS(x))*SQRT((2*SQRT(2)*COS(x)^2*SQRT(SIN(x)^5/CO~   
   S(x))+COS(x)*(2*SQRT(2)*SIN(x)*SQRT(SIN(x)^5/COS(x))+4*SIN(x)^3)~   
   +SIN(x)^2)/SIN(x)^2)-SQRT(2)*SIN(x)^3*COS(x)-SQRT(SIN(x)^5/COS(x~   
      
   [continued in next message]   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)