From:
here is a Lagrange inversion expansion with $r=\frac23$ using inverse beta regularized $\text I^{-1}_x(a,b)$, factorial power $u^{(v)}$, the Gauss hypergeometric function $_2\text F_1(a,b;c;z)$, and the principal root: $$\begin{align}f(x)=\int_0^x \frac{dx}{\sqrt[3]x(1-x)}\implies f^{-1}(x)=\lim_{c\to0}\text I^{-1}_{cx}\left(\frac23,c\right)=\left(1+\sum_{n=1}^\infty\frac{\left(-e^{-x}\right)^n}{n!}\lim_{x\to1}\frac{d^{n-1}(\sqrt[3]{-1}x+1)^{\sqrt[3]{-1}n}\left(1-(-1)^\frac23x\right)^{-(-1)^\frac23n}}{dx^{n-1}}\right)^3\end{align}$$
to get:
$$\begin{align}\left(1+\sqrt3\sum_{2\ne n=1}^\infty\frac{(-1)^\frac{(5+\sqrt3i)n+1}6e^{-nx}}{n!}(\sqrt[-3]{-1}n)^{(n-1)}\,_2\text F_1\left(1-n,-\sqrt[3]{-1}n;2-\sqrt[3]{-1}n;\sqrt[3]{-1}\right)\right)^3=\left(1-\sqrt3\sum_{n=1}^\infty\sum_{m=0}^b\frac{(-1)^\frac{7n-2m+5}6e^{-\left(\frac{\sqrt3}6\pi+x\right)n}\left(\sqrt[3]{-1}n\right)!\left(\sqrt[-3]{-1}\right)!}{\left((-1)^\frac23n+m+1\right)!\left(\sqrt[-3]{-1}n-m\right)!\Gamma(n-m)m!n}\right)^3\end{align}\tag1$$ The possibly hypergometric double sum has an empty $n=2$ term, likely $b\ge n-1$, and interchangeable sums. This sum is an analytic continuation of $\lim\limits_{c\to0}\text I^{-1}_{cx}\left(\frac23,0\right), x\ge0$ since it inverts $f(x)$ for complex inputs too. It implies $y\ge0$ and a period of $2\pi$ for $y$.
Unfortunately, standard double hypergeometric functions fail since they require coefficients of $m,n$, as gamma function arguments, to be integers. Additionally, the Gauss hypergeometric function integral representations converge for $n<0$. Therefore, none of these probably help get an integral representation or closed form of $\lim\limits_{c\to0}\text I^{-1}_{cx}\left(\frac23,0\right)$
Also, the hypergeometric part fits $_2\text F_1(a,b;b+2,z)$, but there is no special case formula for it.
An equivalent definition is:
$$e^{\sqrt3y}\cos(y)=x\implies y=\tan^{-1}\left(\frac{2\sqrt[3]{\lim\limits_{c\to0}\text I^{-1}_{c\left(\frac\pi{\sqrt3}-\ln(2x)\right)}\left(\frac23,c\right)}+1}{\sqrt3}\right)+\frac\pi6, 0\le x\le\frac{e^\frac\pi{\sqrt3}}2$$ Shown here, but maybe it is extendible outside the restriction.
Is there any closed form or integral representation for $(1)$?
