Let $D=\frac{d}{dx}$ be the derivative operator and $f(x)$ be a cubic polynomial. Is it possible to simplify the following differential operator? $T=e^{D^2}f(x)e^{-D^2}$.

I tried to use [Hadamard's formula](https://ncatlab.org/nlab/show/Hadamard%27s+formula):

$$\begin{equation}
T=e^{D^2}f(x)e^{-D^2}=f(x)+[D^2,f(x)]+\frac{1}{2}[D^2,[D^2,f(x)]]+...
\end{equation}\tag{1}$$

where $$[D^2,f(x)]=\frac{d^2f}{dx^2}-f(x)\frac{d^2}{dx^2}.\tag{2}$$ Since $f(x)$ is a cubic polynomial, $$[D^2,[D^2,f(x)]]=\frac{d^4f}{dx^4}-\frac{d^2}{dx^2}(f(x)\frac{d^2}{dx^2})-\frac{d^2f}{dx^2}\frac{d^2}{dx^2}-f(x)\frac{d^4}{dx^4}=0-\frac{d^2}{dx^2}(f(x)\frac{d^2}{dx^2})-\frac{d^2f}{dx^2}\frac{d^2}{dx^2}+f(x)\frac{d^4}{dx^4}.\tag{3}$$

Continuting this approach will just add more terms which do not cancel out each other. Any hint in what other approach that I can try?