Showing integral $\int_0^k \text{sinh}^{-2/3}(x)\mathrm{d}x$, has a hypergeometric solution. For my undergraduate physics research I encountered the integral,
\begin{align}
I=\int_0^k\frac{1}{\text{sinh}^{2/3}x}\mathrm dx, \ \ \ k=\text{arsinh}(\alpha),\ \ \  \alpha\in(0,\infty)\tag{1}
\end{align}
This has the solution, as is given by WA,
\begin{align}
I=a^{1/3}\ _2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\alpha^2\right).\tag{2}
\end{align}
I've convinced myself it is correct, as I've plotted the result and evaluated it numerically via SymPy, but I'm stuck trying to derive the answer myself.
My plan of attack has been to transform Eqn. [1] into the standard integral representations of $_2F_1$ I can find online, those being
\begin{align}
_2F_1(a,b;c;z)=&\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1\frac{t^{b-1}(1-t)^{c-b+1}}{(1-zt)^a}\mathrm dt,\tag{3}\\
\\
_2F_1(a,b;c;z)=&\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^\infty\frac{t^{c-b+1}(1+t)^{a-c}}{(1-z+t)^a}\mathrm dt.\tag{4}\\ \\ 
_2F_1(a,b;c;z)=&\frac{1}{\Gamma(b)}\int_0^\infty e^{-t}t^{b-1}\ _1F_1(a;c;zt)\mathrm dt\tag{5}
\end{align}
Substituting my values for $a, b, c,$ and $z$ into Eqns. [3] & [4] I get
\begin{align}
_2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\alpha^2\right)&=\frac{\Gamma(7/6)}{\pi^{1/2}}\int_0^1\frac{t^{1/2}}{(1-t)^{1/3}(1+\alpha^2t)^{1/6}}\mathrm dt,\ \ \ \ \ \ \text{or}\\ \\
_2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\alpha^2\right)&=\frac{\Gamma(7/6)}{\pi^{1/2}}\int_0^\infty\frac{t^{5/4}}{(1+t)(1+\alpha^2+t)^{1/6}}\mathrm dt.
\end{align}
For $I$, Using the definition of sinh$(x)$, substituting $e^x=t$, and factoring I arrive at
\begin{align}
I=2^{2/3}\int_1^{e^k}\frac{\mathrm dt}{t^{1/3}(t-1)^{2/3}(t+1)^{2/3}}.
\end{align}
I've really just got a hundred questions on what exactly to do.
 A: 
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.

A: 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.
