I am trying to find the $m$th derivative with respect to both $x$ and $p$ of
\begin{align*} e^{-x^2-p^2}L_n(2x^2+2p^2)\,, \end{align*}
where $L_n$ is a Laguerre polynomial and $n$ is a non-negative integer. If $n=0$, then
\begin{align*} \frac{\partial^m}{\partial p^m}\frac{\partial^m}{\partial x^m}e^{-x^2-p^2}=\frac{\partial^m}{\partial p^m}H_m(-x)e^{-x^2-p^2}=H_m(-x)H_m(-p)e^{-x^2-p^2}\,. \end{align*}
If $n\neq0$, I run into problems. I know that
\begin{align*} \frac{\partial^m}{\partial z^m}L_n(z)=(-1)^mL_{n-m}^m(z)\,, \end{align*}
but I am having trouble applying this to $e^{-x^2-p^2}L_n(2x^2+2p^2)$. From the properties of derivatives,
\begin{align*} \frac{\partial^m}{\partial p^m}\frac{\partial^m}{\partial x^m}f(x,p)g(x,p)=\sum_{a=0}^m\sum_{b=0}^m {a \choose m}{b \choose m}\frac{\partial^{2m-a-b}f(x,p)}{\partial x^{m-a}\partial p^{m-b}}\frac{\partial^{a+b}g(x,p)}{\partial x^{a}\partial p^{b}}\,, \end{align*}
so I was thinking of finding the $m$th derivative of $L_n(2x^2+2p^2)$ with respect to both $x$ and $p$ and then evaluating the sum.
I tried to compute successive derivatives of $p$ of $L_n(2x^2+2p^2)$ and I calculated: \begin{align*} \frac{\partial L_n(2x^2+2p^2) }{\partial p}&=-4 p L_{n-1}^1\left(2 p^2+2 x^2\right)\,,\\ \frac{\partial^2 L_n(2x^2+2p^2) }{\partial p^2}&=16 p^2 L_{n-2}^2\left(2 p^2+2 x^2\right)-4 L_{n-1}^1\left(2 p^2+2 x^2\right)\,,\\ \frac{\partial^3 L_n(2x^2+2p^2) }{\partial p^3}&=48 p L_{n-2}^2\left(2 p^2+2 x^2\right)-64 p^3 L_{n-3}^3\left(2 p^2+2 x^2\right)\,,\\ \frac{\partial^4 L_n(2x^2+2p^2) }{\partial p^4}&=-384 p^2 L_{n-3}^3\left(2 p^2+2 x^2\right)+48 L_{n-2}^2\left(2 p^2+2 x^2\right)+256 p^4 L_{n-4}^4\left(2 p^2+2 x^2\right)\,. \end{align*}
However, I do not see any general pattern. Though this way should ultimately give me the $m$th derivative of $e^{-x^2-p^2}L_n(2x^2+2p^2)$, is there an easier way to proceed? I feel like the answer to the $mth$ derivative of either $L_n(2x^2+2p^2)$ or $e^{-x^2-p^2}L_n(2x^2+2p^2)$ should be in a table of derivatives, but I have not seen it.
Any assistance would be greatly appreciated.