I'm trying to solve the following integral $$ I(x)= \int_0^1 \frac{[1-(1-w)^2][1-w^2]}{[1-w(1-w)]^2}\exp\left( - \frac{1}{xw(1-w)} \right)dw. $$ for $x>0$. The aim is to obtain some closed form solution, or a solution in terms of some special function depending on $x$.
I've attempted a solution through the change of variable $w=\frac{1}{2}(1+\tanh(t))$ and the properties of hyperbolic functions. This has led me to \begin{equation} \begin{split} I(x)&= 2\int_{-\infty}^{\infty}\frac{5+2(e^{2t}+e^{-2t})}{(e^{2t}+e^{-2t}+1)(e^{2t}+e^{-2t}+2)^2} \exp\left( -\frac{1}{x} (e^{2t}+e^{-2t}) \right)dt\\ &=\int_0^\infty \frac{1}{s}\frac{5+2(s+s^{-1})}{(s+s^{-1}+1)(s+s^{-1}+2)^2}\exp\left( -\frac{1}{x}(s+s^{-1}) \right)ds\\ &=:\int_0^\infty f(s) ds \end{split} \end{equation} Afterwards, I've tried to split the integral as $\int_0^1f(s)ds$ and $\int_1^\infty f(s)ds$ and perform the change of variable $y=s+1/s$, but this hasn't produced anything useful (apparently). Any suggestion would be much appreciated.
With $w=\frac{1+t}2$, the integral simplifies to
$$I(x) =\int_0^1 \left( 1- \left(\frac{4t}{t^2+3}\right)^2\right)e^{-\frac4{x(1-t^2)}}dt $$
which is unlikely to yield a close-form result.