Inverse relation including Hypergeometric function For the following function:
\begin{equation}
y\left( _{2}F_{1}\left( 1-\frac{1}{p}, \frac{1}{p}, 1+\frac{1}{p}; t^p \right)t \right)=t
\end{equation}
I am trying to find an explicit formula. Therefore I consider the following relation:
\begin{equation}
x=_{2}F_{1}\left( 1-\frac{1}{p}, \frac{1}{p}, 1+\frac{1}{p}; t^p \right)t
\end{equation}
where $_{2}F_{1}\left(a,b,c;t\right)$ the ordinary hypergeometric function. Moreover, if my math up until now is correct, it also holds:
\begin{equation}
_{2}F_{1}\left( 1-\frac{1}{p}, \frac{1}{p}, 1+\frac{1}{p}; t^p \right)=t \left( 1-t^p\right)^{1/p} \quad _{2}F_{1}\left(\frac{2}{p}, 1, 1+\frac{1}{p}; t^p \right)
\end{equation}
resulting in the following:
\begin{equation}
x=t \left( 1-t^p\right)^{1/p} \quad _{2}F_{1}\left(\frac{2}{p}, 1, 1+\frac{1}{p}; t^p \right)
\end{equation}
I would like to solve for $t$ but I have reached the case where:
\begin{equation}
t^p(1-t^p)=\left( \frac{x}{_{2}F_{1}\left(\frac{2}{p}, 1, 1+\frac{1}{p}; t^p \right)} \right)^p
\end{equation}
and it leads to an implicit relation.
I would like to ask whether it is possible in this case to solve for $t$?
Many thanks!

Here is the closed form for an inverse function by rewriting with the incomplete beta function $\text B_z(a,b)$
$$x=\,_2\text F_1\left(1-\frac1p,\frac 1p;1+\frac1p;t^p\right)t=\frac{\text B_{t^p}\left(\frac1p,\frac1p\right)}p$$
Therefore, $t(x)$ is an inverse beta regularized $\text I^{-1}_s(a,b)$ expression:
$$t=\sqrt[p]{\text I^{-1}_\frac{px}{\text B\left(\frac1p,\frac1p\right)}\left(\frac1p,\frac1p\right)}= \sqrt[p]{\text I^{-1}_\frac{2^{\frac2p-1}\Gamma\left(\frac1p+\frac12\right)x}{\sqrt\pi\Gamma\left(\frac1p+1\right)}\left(\frac1p,\frac1p\right)};0\le \frac{px}{\text B\left(\frac1p,\frac1p\right)} \le1,p>0$$
which works
You can “generalize” the function by transforming it if it is meant to be periodic. Also, this function can be put into a form with $\text I^{-1}_y\left(\frac1p,\frac12\right)$
