Integral $\int_0^y \exp\left( \alpha x + \frac{1}{1-\beta e^{\gamma x}}-\frac{1/\beta }{ 1-\beta e^{-\gamma x}}\right)dx$

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.

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

$$\int \exp\left( x + \frac{1}{1-2 e^{x}}\right)dx=\frac12\int\left(\frac1{u-1}-\frac1u\right)e^udu.$$