Introduction:
Using:
Elementary closed form of $\exp\left(b\int_0^x \frac{t^{a-1}}{1-t^b}dt\right);a,b\in\Bbb N$
and inverse beta regularized, which gives quantiles meaning its parameters are nonzero, there is limit, incomplete beta function $\text B_x(u,v)$ function, and Lerch Phi $\Phi(z,s,b)$ $$f(x)=\int_0^x\frac{t^{a-1+\frac pq}}{1-t}dt=x^{a+\frac pq}\Phi\left(x,1,a+\frac pq\right)=\text B_x\left(a+\frac pq,0\right)=-\sum_{n=0}^{q-1}e^{-\frac{2\pi i n p}q}\ln\left(1-\sqrt[q]xe^{\frac{2\pi i n}q}\right)-\sum_{n=0}^{a-1}\frac{x^{n+\frac pq}}{n+\frac pq}\tag1$$$$\implies f^{-1}(x)=\lim_{c\to0}\text I^{-1}_{cx}(x,c);a,p,q\in\Bbb N,p<q,0\le x\le 1\tag2$$ The following derivations use the Lagrange inversion theorem, multinomial $\binom n{n_1,\dots,n_k}$ expansion, generalized product rule, and Kronecker delta $\delta$.
$a+\frac pq$ case?:
This method fails as shown by the $p=1,a=q=2$ case. Using Lagrange inversion: $$\sqrt{\lim_{c\to0}\text I^{-1}_{-c\ln(x)}\left(\frac52,0\right)}=1+\sum_{n=1}\frac{x^n}{n!}\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\left(\frac{x-1}{e^{-\text B_{x^2}\left(\frac52,0\right)}}\right)^n=1+\sum_{n=1}\frac{x^n}{n!}\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\left(-e^{-\frac23nx^3}e^{-2nx}(x+1)^n\right)$$ Now use the generalized product rule: $$\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\left(-e^{-\frac23nx^3}e^{-2nx}(x+1)^n\right)=\lim_{x\to1}(-1)^n\sum_{p=0}^{n-1}\sum_{q=1}^{n-1}\sum_{r=1}^{n-1}\delta_{p+q+r,n-1}\frac{(n-1)!}{p!q!r!}\frac{d^p}{dx^p}(x-1)^n\frac{d^q}{dx^q}e^{-2nx}\frac{d^r}{dx^r}e^{-\frac23 n x^3}$$ using the hypergeometric function $_p\text F_q$ the problem occurs here: $$\sum_{r=0}^{n-1}\frac{\delta_{p+q+r,n-1}}{r!}\lim_{x\to1}\frac{d^r}{dx^r}e^{-\frac23nx^3}\mathop=^?\sum_{r=0}^{n-1}\frac{\delta_{p+q+r,n-1}}{r!}\sum_{t=0}^\infty\frac{\left(-\frac23n\right)^t (3t)!}{t!(3t-r)!}=\sum_{r=0}^{n-1}\frac{\delta_{p+q+r,n-1}\,_3\text F_3(…)}{(-r)!r!}=\frac{\delta_{p+q,n-1}}{e^{\frac23n}}+0+0+\dots$$ Therefore, the series probably converges to the wrong function. What is the simplest explicit series expansion for $(2)$ or the inverse of $(1)$?
$a=0$ case:
$$\sqrt[q]{\lim_{c\to0}\text I^{-1}_{-c\ln(x)}\left(\frac pq,c\right)}=1+\sum_{n=1}^\infty\frac{x^n}{n!}\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\left(\left(\frac{x-1}{e^{-\text B_{x^q}\left(\frac pq,0\right)}}\right)^n\right)=1+\sum_{n=1}^\infty\frac{(-x)^n}{n!}\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\prod_{k=1}^{q-1}\left(1-(-1)^\frac{2k}qx\right)^{-n(-1)^\frac{2kp}q}$$ Applying the generalized product rule, $\displaystyle\sum_{n_1,\dots,n_j}=\sum_{n_1}\dots\sum_{n_j}$, and factorial power $u^{(v)}$ $$\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\prod_{k=1}^{q-1}\left(1-(-1)^\frac{2k}qx\right)^{-n(-1)^\frac{2kp}q}=\sum_{n_1,\dots,n_{q-1}=0}^{n-1}\delta_{n-1,\sum\limits_{k=1}^{q-1}n_k}\binom n{n_1,\dots,n_{q-1}}\prod_{k=1}^{q-1}\left(\frac{d^{n_k}}{dx^{n_k}} \left(1-(-1)^\frac{2k}qx\right)^{-n(-1)^\frac{2kp}q}\right)_{x=1}=\sum_{n_1,\dots n_{q-1}=0}^{n-1} \delta_{n-1,\sum\limits_{k=1}^{q-1}n_k} n!\prod_{k=1}^{q-1} \frac{(-1)^ \frac{2kn_ k}q}{n_ k!}\left(-n(-1)^{- \frac{2kp}q}\right)^{(n_k)}\left(1-(-1)^\frac{2k}q\right)^{-n(-1)^{-\frac{2kp}q}-n_k}$$ Therefore we have a multidimensional series expansion, which are hard to evaluate one by one because of Kronecker delta: $$\boxed{\lim_{c\to0}\text I^{-1}_{cx}\left(\frac pq,c\right)=\left(1+\sum_{n=1}^\infty\frac{(-e^{-x})^n}n\sum_{n_1=0}^{n-1}\dots\sum_{n_{q-1}}^{n-1}\delta_{n-1,\sum\limits_{k=1}^{q-1}n_k} \prod_{k=1}^{q-1} \frac{(-1)^ \frac{2kn_ k}q}{n_ k!}\left(-n(-1)^{- \frac{2kp}q}\right)^{(n_k)}\left(1-(-1)^\frac{2k}q\right)^{-n(-1)^{-\frac{2kp}q}-n_k}\right)^q}$$ Interestingly, the series diverges for $\frac pq=\frac14$, but the Lagrange inversion coefficients match for: $\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\left(\left(\frac{x-1}{e^{-\text B_{x^4}\left(\frac 34,0\right)}}\right)^n\right)=\left\{\vphantom{e^\frac12}\right.-2e^{-\frac\pi 2}$,$-4e^{-\pi}$, $-36e^{-\frac{3\pi}2}$, $-464e^{-2\pi}$, $\dots\left.\vphantom{e^\frac12}\right\}$
$p=0$ case:
$$\lim_{c\to0}\text I^{-1}_{-c\ln(x)}(a,c)=1+\sum_{n=1}^\infty \frac{x^n}{n!}\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\left(\left(\frac{x-1}{e^{-\text B_x(a,0)}}\right)^n\right)= 1+\sum_{n=1}^\infty \frac{(-x)^n}{n!}\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}e^{-n\sum\limits_{k=1}^{a-1}\frac{x^k}k}$$ Now expand the derivative argument as a sum with the multinomial expansion: $$\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}e^{-n\sum\limits_{k=1}^{a-1}\frac{x^k}k}=\sum_{m=0}^\infty\frac{(-n)^m}{m!}\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\left(\sum_{k=1}^{a-1}\frac{x^k}k\right)^m= \sum_{m=0}^\infty\frac{(-n)^m}{m!}\sum_{m_1,\dots,m_{a-1}=0}^m\delta_{m,\sum\limits_{k=1}^{a-1}m_k}\binom m{m_1,\dots,m_{a-1}} \lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\prod_{k=1}^{a-1}\left(\frac{x^k}k\right)^{n_k}= \sum_{m=0}^\infty(-n)^m\sum_{m_1,\dots,m_{a-1}=0}^m\delta_{m,\sum\limits_{k=1}^{a-1}m_k}\frac{\lim\limits_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\prod\limits_{k=1}^{a-1}x^{kn_k}}{\prod\limits_{k=1}^{a-1}m_k!k^{m_k}}$$ Evaluating the derivatives using factorial power, combining both steps, and algebra presents: $$\boxed{\lim_{c\to0}\text I^{-1}_{cx}(a,c)=1+\sum_{n=1}^\infty\frac{\left(-e^{-x}\right)^n}{n!}\sum_{m=0}^\infty\sum_{m_1=0}^m\dots\sum_{m_{a-1}=0}^m\frac{\delta_{m,\sum\limits_{k=1}^{a-1}m_k}\left(\sum\limits_{k=1}^{a-1}km_k\right)^{(n-1)}(-n)^m}{\prod\limits_{k=1}^{a-1}m_k!k^{m_k}}}$$ which match these Lagrange inversion coefficients: $\displaystyle\lim_{x\to1}\frac{d^{n-1}}{dx^{n-1}}\left(\left(\frac{x-1}{e^{-\text B_x(3,0)}}\right)^n\right)=\left\{\vphantom{e^\frac122}\right.$$-e^{-\frac32}$,$4e^{-3}$,$33e^{-\frac92}$,$416e^{-6}$$\left.\vphantom{e^\frac12},\dots\right\}$
Question:
The $a=0$ and $p=0$ cases are boxed and partly tested. What is the most concise explicit series expansion for $(2)$ or the inverse of $(1)$? Optionally, it should be testable in WolframAlpha or a similar software.
