> 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
> > >
> >
> > --
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>
>
> --
> Waldek Hebisch
>
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