I look for some help in solving the integral
$$ \int\limits_{0}^{z_t} \mathrm{d} z\, \sqrt{a+\left[\left(\frac{1}{z}W_0(z\,e^{-z})+1\right)^2 -1 \right]}$$ involving the Lambert function $W_0$, where $ z_t$ is the first root of the integrant, that is $$ a+\left[\left(\frac{1}{z_t}W_0(z_t\,e^{-z_t})+1\right)^2 -1 \right] = 0.$$ Values of $a$ are restricted such that $z_t$ is real.
I tried the somewhat "standard" substitution $ y \equiv z\,e^{-z}$, but with no sucess, as I end up with terms involving $W_0 (-y)$ and $W_0 (+y)$ and I could not find any relation between them.
I am grateful for any ideas you have :)
Using the fact that: $$\int f^{-1}(x)dx=xf^{-1}(x)-F\left[f^{-1}(x)\right]+C$$
Try letting: $$f^{-1}(z)=\sqrt{a+\left[\left(\frac1zW_0(ze^{-z})+1\right)^2-1\right]}$$ and see if this gets you anywhere