hypergeometric function with arguments (1/3,1/2;3/2;x^6)

Question

I am attempting to find an exact, closed, form of the integral:

$\int_0^r (1-\frac{1}{r'^6})^{-\frac{4}{3}}dr'$

Wolfram evaluates this to:

$=\frac{x\left((1-x^6)^{\frac{1}{3}} \;_2F_1(\frac{1}{3},\frac{1}{2};\frac{3}{2};x^6)-1 \right)}{2(1-x^6)^{\frac{1}{3}}}|^{x=r}_{x=0}$

I know I can simplify this using the identity:

$\;_2F_1(a,b;a+1;z)=az^{-a}B_z(a,1-b)$

And the symmetry of the hypergeometric function in the first two inputs, however, this is not giving me a closed form for this integral.

Any help simplifying this would be appreciated.

Answer 1

$$\;_2F_1(a,b;a+1;z)=\;_2F_1(b,a;a+1;z)=az^{-a}B_z(a,1-b)$$ $$\;_2F_1(\frac{1}{3},\frac{1}{2};\frac{3}{2};x^6)=\frac{1}{2}x^{-3}B_{x^6}(\frac{1}{2},\frac{2}{3})\qquad \text{with}\quad a=\frac{1}{2} \text{ and } b=\frac{1}{3}$$