I need to find an inverse function to $f(x) = (1+\frac{W(-x)}{2})\exp(-W(-x))$, where $W(x)$ is the Lambert W function.
I tried to find it using Wolfram Mathematica, but it failed to deal with it. Another approach was to consider power series for this function $f(x) = 1+\frac{x}{2}+\frac{x^2}{2}+\frac{2x^3}{3}+\frac{25x^4}{24}+\frac{9x^5}{5}+O(x^6)$ and then find the inverse power series, which is $f^{-1}(x) = 2(x-1) + 4(x-1)^2 + \frac{16}{3}(x-1)^3 - \frac{20}{3}(x-1)^4 + \frac{112}{15}(x-1)^5 + O(x^6)$, but I can't imagine function which may has it as series expansion at $x = 1$.
Since there is the Lambert W function, it seems to be some kind of $x = y\cdot\exp(y)$ equation. Can anyone help me with the inverse function to given one?
In fact, for my purposes it will be sufficient to guess it and then check on computer, but any analytical solution hints also will be very appreciative.
$$y=f(x) = (1+\frac{W(-x)}{2})\exp(-W(-x))$$ $2y = (2+W(-x))\exp(-W(-x))$
$2y\,e^{-2} = (2+W(-x))\exp(-W(-x)-2)$
$-2y\,e^{-2} = (-W(-x)-2)\exp(-W(-x)-2)$
Let $\quad t=(-W(-x)-2) \quad\to\quad -2y\,e^{-2} = te^t$
$te^t=-2y\,e^{-2} \quad\to\quad t=W(-2y\,e^{-2})$
$t=(-W(-x)-2) \quad\to\quad W(-x)=-t-2 \quad\to\quad -x=(-t-2)e^{-t-2}$
$x=(t+2)e^{-t-2}= (W(-2y\,e^{-2})+2)e^{-W(-2y\,e^{-2})-2} $
The inverse function is : $$x(y)=\left(W(-2e^{-2}y)+2\right)\exp\left(-W(-2e^{-2}y)-2\right)$$