Lagrange inversion theorem of $x^r(x+k)$ to generalize the W Lambert function Motivation:
$2$ branches of Lambert $\text W_k(z)$ is a limit of the inverse of $x^n(x+c)$ which is probably expressible in terms of FoxH in Mathematica or through another general hypergeometric function.
$\text W_0(x)=\text W(x):$
$$-\lim_{a\to0}e^{\frac{(-x)^a((a+1)x+1)-1}a-1}=xe^x$$
Using the inverse function of $-e^{\frac{(-x)^a((a+1)x+1)-1}a-1}$:

Blue:$\text W(x)$
Red:inverse at $a=\frac12$
Yellow:inverse at $a=1$
The Goal:
$\text W_{-1}(x):$
$$\lim_{b\to\infty}\frac{(x+b+1)^bx}{eb^b}=xe^x$$
Using the inverse of $\frac{(x+b+1)^bx}{eb^b}$:

Blue:$\text W_{-1}(x)$
Red:inverse at $b=10$
Purple:inverse at $b=5$
However, is there a way to invert $x^r (x+k)$ using the Lagrange Inversion theorem? This way, we can put $2$ branches of the W-lambert function in terms of a hypergeometric function and potentially find other identities of them. Please correct me and give me feedback!
The following $n$th derivative at $x=1$ must be evaluated to use the Lagrange Inversion theorem:
$$\frac{d^{n-1}}{dx^{n-1}}\left[\left(\frac{x-1}{x^r(x+k)-(1+k)}\right)^n\right]\bigg|_{x=1}$$
 A: Excuse for causing trouble. I am able to reproduce the given plot:

I had to uses two definitions to make this plot. Only the orange part is the LambertW in the original definition. That works because the function is real there and has no imaginary part. The blue part is just the real part of the LambertW function. But at most this is just a $W(x)$ without an index and the point of the cusp is $(\frac{1}{e},-1)$.
I too use Mathematica.
My references are
LambertW Function,
Lagrange Inversion Theorem
The second one appears in the first. Furthermore my first reference solves the question but not in the direct suggested by the question.
The Lambert W-function has the series expansion
$W(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n^{n-2}}{(n-1)!}x^n$
The Lagrange Inversion Theorem gives it the equivalent series expansion
$W(z)=\sum_{n=1}^{\infty}\frac{(-n)^{n-1}}{n!}z^n$
where $n!$ is the factorial. They warn on that reference that the later series is not nice for practical numerical computation and thereby for visualization.
There are several further series expansions possible.
Form this search LambertW on wolfram.com there is taken the promoted representation for the LambertW function in Mathematica.
ProductLog!!! with is the principal solution of the Lambert equation. This page confirms my result above in the Details section.
Furthermore this page defines the named
$W_{k}(z)=ProductLog(k,z)$
But they do not the index $0$ for the LambertW function at all.
For $W_{1}(x)$ the graph over the real axis interval from above is

again this is just the real part but this time for both parts of the interval.
$W_{-1}(x)$" />
Evaluating the limit for $W_{-1}(x)$ I get
$\lim_{b->\infty}\frac{(x+b+1)^{b} x}{e b^{b}}=e^{x} x$
There is another referenceProductLog. That confirms the given notation on representation through other functions, more general functions.
The result stems from using Mathematica or WolframAlpha for the task.
This can be proved by separating what is not needed in the limit and only there to get the desired result and then make use of the definition of the
$\lim_{b->\infty}\frac{(x+b+1)^{b} x}{b^{b}}=e^{x+1}$
for every real $b$ taking the natural logarithm before taking the limit.
$-b \ln(b)+b \ln(1+b+x)$
There is a constant factor then visible that can be factor out and then the inverse operation to the natural logarithm the exponentiation and then the limit again. There is the result without Lagrange at all.
$\lim_{b->\infty}-b \ln(b)+b \ln(1+b+x)=1+x$
for positive real $b$ and $x$. This is clear by remembering that the two logarithms approach each other and the $x$ remains. This is math from the introductory lessons in analysis or school maths that depends on the career. The can be viewed as a differential quotient that can be resolved for the denominator.
I hope that helped. I excuse not to make use of the FoxH representation. The ProductLog does not appear in Mathematica to be related to FoxH. It is like the HeunG in FoxHReduce.
