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:

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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