Integral $\int_0^y \exp\left( \alpha x + \frac{1}{1-\beta e^{\gamma x}}-\frac{1/\beta }{ 1-\beta e^{-\gamma x}}\right)dx$ I'm studying the integral
\begin{align*}
\int_0^y \exp\left(  \alpha x + \frac{1}{1-\beta e^{\gamma x}}-\frac{1/\beta }{ 1-\beta e^{-\gamma x}}\right)dx
\end{align*}
with constants $\alpha,\beta,\gamma$ (choosen such the fractions are well-defined).
I evaluated the integral numerically but Wolfram Alpha is able to solve special cases, see here or here using the exponential integral, $\text{Ei}(x)=-\int_{-x}^\infty \frac{e^{-u}}{u}du=\int_{-\infty}^x \frac{e^u}{u}du$.
Any ideas how the above integral relates to the exponential integral or other special functions?

The special cases Wolfram Alpha solves are
$$\int \exp\left(  x + \frac{1}{1-2 e^{x}}\right)dx=\frac{1}{2}\left(\text{Ei}\left(\frac{1}{1-2e^x}\right)+e^{1/(1-2e^x)}(2e^x-1)\right)$$
and
$$\int \exp\left(  x + \frac{1}{1-3 e^{-x}}\right)dx=3e\left(e^{3/(e^x-3)}-\text{Ei}\left(\frac{3}{-3+e^x}\right)+e^{1/(1-3e^{x})}(e^x-6)\right).$$
Here is a free step-by-step solution for the special cases.
 A: Assume $\beta,\gamma\neq0$ for the key case.
$$\begin{align}\int_0^ye^{\alpha x+\frac{1}{1-\beta e^{\gamma x}}-\frac{\frac{1}{\beta}}{1-\beta e^{-\gamma x}}}~dx&=\int_1^{e^{\gamma y}}u^\frac{\alpha}{\gamma}e^{\frac{1}{1-\beta u}-\frac{\frac{1}{\beta}}{1-\frac{\beta}{u}}}~d\left(\dfrac{\ln u}{\gamma}\right)\\
&=\dfrac{1}{\gamma}\int_1^{e^{\gamma y}}u^{\frac{\alpha}{\gamma}-1}e^{\frac{1}{1-\beta u}-\frac{u}{\beta(u-\beta)}}~du\\
&=\dfrac{1}{\gamma}\int_1^{e^{\gamma y}}u^{\frac{\alpha}{\gamma}-1}e^\frac{1-u^2}{(\beta u-1)(u-\beta)}~du\\
&=\dfrac{1}{\gamma}\int_1^{e^{\gamma y}}\sum\limits_{n=0}^\infty\dfrac{u^{\frac{\alpha}{\gamma}-1}(1-u)^n(u+1)^n}{n!(\beta u-1)^n(u-\beta)^n}~du\end{align}$$
The issue is far more complicated,
even the integral $\int_1^{e^{\gamma y}}\dfrac{u^{\frac{\alpha}{\gamma}-1}(1-u)^n(u+1)^n}{(\beta u-1)^n(u-\beta)^n}~du$ we decomposed should at least relates to Lauricella hypergeometric series.
A: The shape of the answer hints you that in the first case Alpha uses the substitution $u:=\dfrac1{1-2e^x}$, from which $2e^x=1-\dfrac1u$, and
$$\int \exp\left(  x + \frac{1}{1-2 e^{x}}\right)dx=\frac12\int\left(\frac1{u-1}-\frac1u\right)e^udu.$$
