What is the adjoint of $e^{\alpha \psi} \partial_{x}^4 e^{-\alpha \psi}$?

Question

I would like to compute the adjoint with respect to the standard $L^2$ inner product on $\mathbb{R}$ of the operator $Sf:=e^{\alpha \psi} \partial_{x}^4 (e^{-\alpha \psi}f)$, where $f\in C^\infty_0(\mathbb{R})$ ( so any boundary terms while doing integration parts vanish), $\psi \in C^\infty(\mathbb{R})$ and $\alpha\in\mathbb{R}$. I would like to split up $S$ into symmetric and anti-symmetric parts for an application in my research, and thinking of $S$ as a product of three operators (multiplication by $e^{-\alpha\psi}$, followed by $\partial_x^4$, finally followed by multiplication of $e^{\alpha\psi}$, each of these operations being symmetric, I am tempted to write $S^\dagger f= e^{-\alpha\psi} \partial_x^4 (e^{\alpha\psi}f)$, i.e. using $(ABC)^\dagger=CBA$, if $A,B, C$ are all symmetric. Is this correct?