Can $\sum_{\ell=1}^{k-1} \binom{k}{\ell} \frac{x^{\ell}y^{k-\ell}}{\ell} $ be written more simply?

Question

Question: Can the expression $$\sum_{\ell=1}^{k-1} \binom{k}{\ell} \frac{x^{\ell}y^{k-\ell}}{\ell} $$ be written as a polynomial in or other elementary function of $(x+y)$,$x$,$y$ with a fixed number of terms for all $k\ge 2$? (I.e. in a way such that the number of terms does not increase with $k$, but is constant irrespective of the value of $k$.)

As a warning, I'm fairly certain that the answer is no, but I have no idea how one could even hypothetically prove that it's impossible. Is it easier to prove that it's impossible to do so in terms of polynomials or rational functions?

I know that for $k \ge 3$ one can write

$$\sum_{\ell=1}^{k-1} \binom{k}{\ell} \frac{x^{\ell}y^{k-\ell}}{\ell} = (x+y)^k - x^k - y^k -\sum_{\ell=2}^{k-1} \binom{k}{\ell}\frac{\ell -1}{\ell}x^{\ell}y^{k-\ell} $$

but that obviously doesn't give a solution or even simplify the solution, since that has $(k-2)+3 = k+1$ terms, and thus the number of terms also goes to infinity with $k$, just like for the original expression.

Motivation: I was trying to write a bivariate power series related to the exponential integral function as a single summation instead of a double summation. While $\sum_{k=1}^{\infty} \frac{x^k}{k \cdot k!}$ can be written in terms of an antiderivative for $\frac{e^x}{x}$, because of the unsolvability of the above problem it seems there is no way to write the "bivariate analogue" $$\sum_{k=1}^{\infty} \frac{1}{k!} \sum_{\ell=1}^{k-1} \frac{1}{\ell}\binom{k}{\ell}x^{\ell}y^{k-\ell} $$ simply, i.e. it does not seem to simplify to a single summation power series, not even to something that could be written in terms of $$\sum_{k=1}^{\infty} \frac{1}{k!}(x+y)^k $$ (which itself can be written in terms of an antiderivative for $\frac{e^{z}}{z}$, defining $z=x+y$).

EDIT: This question corresponds well to my username (i.e. it is a really stupid question), because I just realized that if one re-indexes the above power series by setting $k=p+q$, $\ell=p$ (and so $k-\ell=q$), then the bivariate power series is really easy to write in terms of the univariate: $$\sum_{p=1}^{\infty}\sum_{q=1}^{\infty} \frac{1}{(p+q)!} \frac{1}{p} \binom{p+q}{p} x^p y^q = \sum_{p=1}^{\infty} \frac{x^p}{p \cdot p!} \sum_{q=1}^{\infty} \frac{y^q}{q!} = (e^y - 1) \sum_{k=1}^{\infty} \frac{x^k}{k \cdot k!}\,. $$ (The key observation is that the $(p+q)!$'s cancel, $\binom{p+q}{p}=\frac{(p+q)!}{p!q!}$ after all, and then after that it's possible to separate everything between the $p$'s and $q$'s.)

Answer 1

Just playing around, I get this:

$\begin{array}\\ f_k(x, y) &=\sum_{j=1}^{k-1} \binom{k}{j} \frac{x^{j}y^{k-j}}{j}\\ &=y^k\sum_{j=1}^{k-1} \binom{k}{j} \frac{(x/y)^{j}}{j}\\ &=y^kg_k(x/y)\\ g_k(x) &=\sum_{j=1}^{k-1} \binom{k}{j} \frac{x^{j}}{j}\\ g_k'(x) &=\sum_{j=1}^{k-1} \binom{k}{j} x^{j-1}\\ &=\dfrac1{x}\sum_{j=1}^{k-1} \binom{k}{j} x^{j}\\ &=\dfrac1{x}((1+x)^k-1-x^k)\\ \end{array} $

But integrating this last gives, according to Wolfy, a hypergeometric function, so this is not much help.

Answer 2

Let $ k $ be a positive integer, we have :

\begin{aligned}\sum_{\ell=1}^{k}{\binom{k}{\ell}\frac{x^{\ell}y^{k-\ell}}{\ell}}&=\int_{0}^{x}{\sum_{\ell=1}^{k}{\binom{k}{\ell}t^{\ell-1}y^{k-\ell}}\,\mathrm{d}t}\\ &=\int_{0}^{x}{\frac{\left(y+t\right)^{k}-y^{k}}{t}\,\mathrm{d}t}\\ &=\sum_{j=0}^{k-1}{y^{j}\int_{0}^{x}{\left(y+t\right)^{k-1-j}\,\mathrm{d}t}}\\ &=\sum_{j=0}^{k-1}{\frac{y^{j}\left(y+x\right)^{k-j}}{k-j}}\\ \sum_{\ell=1}^{k}{\binom{k}{\ell}\frac{x^{\ell}y^{k-\ell}}{\ell}}&=\sum_{\ell=1}^{k}{\frac{y^{k-\ell}\left(y+x\right)^{\ell}}{\ell}}\end{aligned}

I don't see how we can do further simplifications.