Showing integral $\int_0^k \text{sinh}^{-2/3}(x)\mathrm{d}x$, has a hypergeometric solution.

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Define the function $\mathcal{I}:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ via the definite integral

$$\mathcal{I}{\left(a\right)}:=\int_{0}^{\operatorname{arsinh}{\left(a\right)}}\mathrm{d}\eta\,\frac{1}{\left[\sinh{\left(\eta\right)}\right]^{2/3}}.$$

You're probably going to kick yourself when realize what the first substitution should be, but sometimes the most obvious paths are most overlooked.

Given $a\in\mathbb{R}_{>0}$, we obtain

$$\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{\operatorname{arsinh}{\left(a\right)}}\mathrm{d}\eta\,\frac{1}{\left[\sinh{\left(\eta\right)}\right]^{2/3}}\\ &=\int_{0}^{a}\mathrm{d}x\,\frac{1}{x^{2/3}\sqrt{1+x^{2}}};~~~\small{\left[\eta=\operatorname{arsinh}{\left(x\right)}\right]}\\ &=\int_{0}^{a^{2}}\mathrm{d}y\,\frac{1}{2y^{5/6}\sqrt{1+y}};~~~\small{\left[x=\sqrt{y}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{a^{2}}{2a^{5/3}t^{5/6}\sqrt{1+a^{2}t}};~~~\small{\left[y=a^{2}t\right]}\\ &=\frac{a^{1/3}}{2}\int_{0}^{1}\mathrm{d}t\,\frac{1}{t^{5/6}\sqrt{1+a^{2}t}}.\\ \end{align}$$

You should now have no problem translating this into the hypergeometric form using the first integral representation you listed, so I leave you to take it from here.

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My version of Mathematica gives a slightly different result. I got $$I=\frac{(-1)^{1/6}\Gamma(1/6)\sqrt{\pi}}{\Gamma(2/3)}+3\alpha^{1/3}{}_{2}F_1\left(\left[\frac{1}{6},\frac{1}{2}\right];\frac{7}{6},-\alpha^2\right)$$ With a whole bunch of conditions for convergence after that.