How to determine $ \int \sec^{\frac{1}{3}}(x) dx $? How to determine $$ \int \sec^{\frac{1}{3}}(x) dx ?$$

An attempt:
$$ \int \frac{1}{\cos^{1/3}(x)}dx = \int \frac{\sin^{2}(x) + \cos^{2}(x)}{\cos^{1/3}(x)}dx$$
$$ = \int \frac{\sin^{2}(x)}{\cos^{1/3}(x)}dx + \int \cos^{2/3}(x) \cos(x) dx $$
Let $ A=\int \frac{\sin^{2}(x)}{\cos^{1/3}(x)}dx $, we focus on $\int \cos^{2/3}(x) \cos(x) dx$.
$$\int \underbrace{\cos^{2/3}(x)}_{u} \underbrace{\cos(x) dx}_{dv} = \sin(x) \cos^{2/3}(x) + (2/3) \underbrace{\int \frac{\sin^{2}(x)}{\cos^{1/3}(x)}dx }_{A}  $$
We encounter $A$ once more. 

How to continue? Or is there a better approach? Thanks.
The hint says use Trig. substitution two times
 A: Your question is equivalent to finding a primitive for $\sin(x)^{-1/3}$, or to finding a primitive for $\frac{x^{-1/3}}{\sqrt{1-x^2}}$. The latter is of course a hypergeometric function related to elliptic integrals:
$$ \int_{0}^{z}\frac{x^{-1/3}}{\sqrt{1-x^2}}\,dx = z^{2/3}\sum_{n\geq 0}\frac{3\binom{2n}{n}}{(6n+2)4^n}z^{2n}=\frac{3}{2}z^{2/3}\cdot{}_2 F_1\left(\tfrac{1}{3},\tfrac{1}{2};\tfrac{4}{3};z^2\right).$$
Some special values can be found through Euler's Beta function. For instance:
$$\begin{eqnarray*} \int_{0}^{\pi/2}\cos(x)^{-1/3}\,dx &=& \int_{0}^{\pi/2}\sin(x)^{-1/3}\,dx = \int_{0}^{1}\frac{dz}{z^{1/3}\sqrt{1-z^2}}=\frac{1}{2}\int_{0}^{1}u^{-2/3}(1-u)^{-1/2}\,du\\&=&\frac{1}{2}B\left(\tfrac{1}{3},\tfrac{1}{2}\right)=\frac{\Gamma\left(\tfrac{1}{3}\right)\,\Gamma\left(\tfrac{1}{2}\right)}{2\,\Gamma\left(\tfrac{5}{6}\right)}=\sqrt{\frac{\Gamma\left(\tfrac{1}{6}\right)^3}{2^{11/3}\sqrt{3\pi}}}=\frac{2^{1/2}3^{1/4}\pi}{\text{AGM}(1+\sqrt{3},\sqrt{8})}.\end{eqnarray*} $$
The representation through the AGM mean gives this tight inequality, for instance:
$$ \int_{0}^{\pi/2}\cos(x)^{-1/3}\,dx \leq \pi\sqrt{\frac{\sqrt{3}}{\sqrt{2}+\sqrt{6}}}.$$
