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

Question

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?

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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.

Answer 1

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

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