Ok, my bad. pi is %pi in FriCAS. The first problem was my fault. But the second problem shows "potentialPole" even with %pi.
Mild Shock schrieb am Samstag, 21. Oktober 2023 um 19:55:50 UTC+2: > > Also, without ability to simplify hypergeometric answers are of > > little use, so we need first strong simplifier for hypergeometric > functions. > > I am not sure that the lack of a strong simplifier is to blame that > I don't see a slope in those systems that can solve the integral. > > Now I was trying to compute the slope, by using a definite > integral over the period instead of a indefinite integral. And I got this: > > The simpler problem with cos(t) gives me: > > (1) -> integrate(cos(t)*(2+cos(t))/100, t = 0..2*pi) > > (cos(2 pi) + 4)sin(2 pi) + 2 pi > (1) ------------------------------- > 200 > > Can I force sin(2 pi) getting computed as well? > > The more complex problem with cos(t)^(1/3) gives me: > > (2) -> integrate(cos(t)^(1/3)*(2+cos(t))/100, t = 0..2*pi) > > (2) "potentialPole" > > How can I tell the system that ^(1/3) should be real root? > > Waldek Hebisch schrieb am Samstag, 21. Oktober 2023 um 12:44:53 UTC+2: > >> On Fri, Oct 20, 2023 at 07:41:31PM -0700, 'Nasser M. Abbasi' via FriCAS - >> computer algebra system wrote: >> > Fyi, Rubi and Mathematica 13.3.1 gives answer in terms of >> Hypergeometric >> > special function. Not sure if you consider this one the "usual" special >> > functions you refer to >> >> Currently FriCAS do not use hypergeometric functions in aswers >> to integration problems. That will probably change in the future, >> but to justify the answer one needs a lot of hypergeometric >> function identities and those are scattered in the literature. >> Worse, while in many places one can find various identities, >> justifications of the identities seem to be scarce. >> >> Also, without ability to simplify hypergeometric answers are >> of little use, so we need first strong simplifier for >> hypergeometric functions. In particular we need ability >> to discover when hypergeometric function (or its derivative) >> is just a disguise and we are really dealing with something >> simpler like elementary or Liouvillian function (actually, >> theory says that all hypergeometric functions useful for >> integration are disguise, "true" hypergeometric functions >> can not appear as integrals of simpler functions). >> >> > integrand=Cos[t]^(1/3)*(2+Cos[t])/100 >> > Integrate[integrand,t] >> > >> > -(1/1400)3 Cos[t]^(4/3) Csc[t] (7 >> Hypergeometric2F1[1/2,2/3,5/3,Cos[t]^2]+2 >> > Cos[t] Hypergeometric2F1[1/2,7/6,13/6,Cos[t]^2]) Sqrt[Sin[t]^2] >> > >> > And Rubi gives >> > >> > Int[integrand,t] >> > -((3 Cos[t]^(4/3) Hypergeometric2F1[1/2,2/3,5/3,Cos[t]^2] Sin[t])/(200 >> > Sqrt[Sin[t]^2]))-(3 Cos[t]^(7/3) >> Hypergeometric2F1[1/2,7/6,13/6,Cos[t]^2] >> > Sin[t])/(700 Sqrt[Sin[t]^2]) >> > >> > Tried Maxima, Giac and Maple and these all can't solve this either. >> > --Nasser >> > >> > >> > On Friday, October 20, 2023 at 6:41:52 PM UTC-5 Waldek Hebisch wrote: >> > >> > > On Fri, Oct 20, 2023 at 02:21:21PM -0700, Mild Shock wrote: >> > > > Possible to make FriCAS solve this integral? >> > > > >> > > > /* Version: FriCAS 1.3.7, WSL2 */ >> > > > /* ^(1/3) is supposed to be the real root */ >> > > > >> > > > (2) -> integrate(cos(t)^(1/3)*(2+cos(t))/100, t) >> > > > >> > > > t 3+-------+ >> > > > ++ (cos(%A) + 2)\|cos(%A) >> > > > (2) | ----------------------- d%A >> > > > ++ 100 >> > > > >> > > >> > > Well, FriCAS claims tha answer is not elementary. Currently >> > > FriCAS can not find answer in terms of "usual" special >> > > functions and I do not know if there such an answer. >> > > >> > > Note that what is produced above can be treated as ad-hoc >> > > special function, so there is answer, the question is >> > > if this is more explicit or simpler answer. >> > > >> > > It is not clear what you mean by "Make it work"? If you >> > > know better answer you could use rewrite rule to change >> > > FriCAS result to a different one, but this is very limited in >> > > scope. Due to the way FriCAS integrator works there is no >> > > way to provide hints. >> > > >> > > If this integral is doable in terms of popular special functions, >> > > there is good chance that it will be handled in the future. >> > > But not in current version. >> > > >> > > -- >> > > Waldek Hebisch >> > > >> > >> > -- >> > You received this message because you are subscribed to the Google >> Groups "FriCAS - computer algebra system" group. >> > To unsubscribe from this group and stop receiving emails from it, send >> an email to [email protected]. >> > To view this discussion on the web visit >> https://groups.google.com/d/msgid/fricas-devel/2ebb3f63-d2e4-42b7-b3a5-9c1c6094777en%40googlegroups.com. >> >> >> >> >> -- >> Waldek Hebisch >> > -- You received this message because you are subscribed to the Google Groups "FriCAS - computer algebra system" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/fricas-devel/ff69acce-803a-4427-91d4-7bb87aea0f7bn%40googlegroups.com.
