Inverse function of $y=W(e^{ax+b})-W(e^{cx+d})+zx$ I have a simple question for which I am looking for a closed form expression (If there exits one). In other words, given:
$$y=W(e^{ax+b})-W(e^{cx+d})+zx$$
where $W$ is the Lambert $W$ function and $a,b,c,d,z$ are some constants, what is the function $f$, such that
$$x=f(y)$$
Thanks alot in advance.  
EDIT : If there exists no closed form solution, I will be happy to see nice arguments supporting this.
$\rightarrow$EDIT2 : As can be seen, we have a solution for the simplified version of this problem. If there exists a solution I have the following ideas to resolve the full version of the problem:
$1$- Is it possible to write $$W(e^{f(x)})=W(e^{ax+b})-W(e^{cx+d})$$ 
$2$- Having $$y_1=W(e^{ax+b})+z_1x$$ and  $$y_2=-W(e^{cx+d})+z_2x$$
where $z=z_1+z_2$, $y=y_1+y_2$ and $f^{-1}(y_1)$ and $f^{-1}(y_2)$ are known functions as already found. Can we say that $f^{-1}(y)=f^{-1}(y_1)+f^{-1}(y_2)$? or can we modify this idea to get somethign useful?
 A: I tried to solve the simplified version of the problem. Here are what I've done so far.
I am given: $$y=W(e^x)+zx\quad\quad(1)$$
From the definition of $W$, for
$$x=re^r$$
We have $$r=W(x)$$
Having $e^{x}=e^{re^r}=te^t\rightarrow t=W(e^x)$ I also have $$y=t+zx$$ Using $x=t+\log(t)$, I have $$y=t+z(t+\log(t))=t(1+z)+z\log(t)$$ taking the exponential of both sides, $$e^y=e^{t(1+z)+z\log(t)}=e^{t(1+z)}t^z$$ Taking the $(1/z)$th power of both sides we have $$e^{y/z}=e^{\frac{t(1+z)}{z}}t.$$ Let $t^{'}=t\frac{1+z}{z}$ then we have $$e^{y/z}=e^{t'}t^{'}\frac{z}{1+z}$$ Accordingly we have $$e^{t{'}}t^{'}=e^{y/z}\frac{1+z}{z}$$ Again using the definition of Lambert $W$ function we get $$t{'}=W\left(e^{y/z}\frac{1+z}{z}\right)$$ Going back to the relation $t^{'}=t\frac{1+z}{z}$, we get $$t=W\left(e^{y/z}\frac{1+z}{z}\right)\frac{z}{1+z}$$ Using the relation $t=W(e^x)=W\left(e^{y/z}\frac{1+z}{z}\right)\frac{z}{1+z}$ in equation $(1)$,$$y=W\left(e^{y/z}\frac{1+z}{z}\right)\frac{z}{1+z}+zx$$ which can be rewritten as $$x=\frac{y}{z}-W\left(e^{y/z}\frac{1+z}{z}\right)\frac{1}{1+z}$$ I checked for some numerical results and it seems that the solution for this result is correct. 
A: Let's do some special cases.  
Is it true that the solution of
$$
y = \operatorname{W} \bigl(\operatorname{e} ^{x}\bigr) + z x
$$
is
$$
x = \frac{y - \operatorname{exp}{\left(y/z - \operatorname{W} \left({(z + 1) \operatorname{e} ^{y/z}}/{z}\right)\right)}}{z}
$$
I state this as a question because Maple doesn't do this by itself.  
added 
simplified:
$$
x = \frac{y}{z} - \frac{\operatorname{W} \left({(z + 1) \operatorname{e} ^{{y}/{z}}}/{z}\right)}{z + 1}
$$  
added 
SG has written this in an answer.  Here is my version, not much different:
Let $\operatorname{W}$ be the Lambert W function, so that (formally)
$a=\operatorname{W}(b) \Longleftrightarrow ae^a = b$.
Question: solve $y=\operatorname{W}(e^x)+zx$ for $x$.  Answer
(at least formally):
$$
x = \frac{y}{z} - \frac{\operatorname{W} \left({(z + 1) \operatorname{e} ^{{y}/{z}}}/{z}\right)}{z + 1}
$$
Explanation.  This solution works as long as these steps are reversible:
\begin{gather*}
y = \operatorname{W}(e^x)+zx
\\
\operatorname{W}(e^x) = y-zx
\\
e^x = (y-zx)e^{y-zx}
\\
1 = (y-zx)e^{y-xz-x}
\\
e^{y/z} =(y-xz)e^{y-xz-x+y/z}
\\
\frac{e^{y/z}}{z} = \left(\frac{y}{z}-x\right)e^{(y/z-x)(z+1)}
\\
\frac{(z+1)e^{y/z}}{z} = \left(\frac{y}{z}-x\right)(z+1)e^{(y/z-x)(z+1)}
\\
\operatorname{W}\left(\frac{(z+1)e^{y/z}}{z}\right) = \left(\frac{y}{z}-x\right)(z+1)
\\
x = \frac{y}{z} - \frac{\operatorname{W}\big((z+1)e^{y/z}/z\big)}{z+1}
\end{gather*}
A: $y=W(e^{ax+b})-W(e^{cx+d})+zx$
$W(e^{cx+d})+y-zx=W(e^{ax+b})$
$(W(e^{cx+d})+y-zx)e^{W(e^{cx+d})+y-zx}=e^{ax+b}$
$(W(e^{cx+d})+y-zx)e^{W(e^{cx+d})}e^{y-zx}=e^{ax+b}$
$(W(e^{cx+d})+y-zx)\dfrac{e^{cx+d}}{W(e^{cx+d})}=e^{(a+z)x+b-y}$
$W(e^{cx+d})+y-zx=e^{(a-c+z)x+b-d-y}W(e^{cx+d})$
$(e^{(a-c+z)x+b-d-y}-1)W(e^{cx+d})=y-zx$
$W(e^{cx+d})=\dfrac{y-zx}{e^{(a-c+z)x+b-d-y}-1}$
$e^{cx+d}=\left(\dfrac{y-zx}{e^{(a-c+z)x+b-d-y}-1}\right)e^{\frac{y-zx}{e^{(a-c+z)x+b-d-y}-1}}$
Then you can only force to use Lagrange inversion theorem or Lagrange reversion theorem unless the special cases when $a=0$ or $c=0$ or $a-c+z=0$ .
