Let $A, B$ and $C$ be commuting linear operators such that $[A, A^\dagger]=[B,B^\dagger]=[C, C^\dagger]=1$, where the dagger denotes the adjoint of the operator. Consider the problem of finding $Z=\log(e^X e^Y$), given $$X=i\beta(A^\dagger C + C^\dagger A) \quad Y=i(\alpha A^\dagger B+\overline\alpha B^\dagger A) $$ for $\beta\in \mathbb R$ and $\alpha\in \mathbb C$. Immediately I find $$\tag{1}[X,Y]=-\beta(\alpha C^\dagger B-\overline\alpha B^\dagger C)\ne 0 $$ which means that in order to find $Z$ we need to apply the Baker-Hausdorff-Campbell formula.
I think it is reasonable to expect that there exists a closed expression for $Z$, based on the following observation: every time a commutator is computed, the repeated operator (in the case of $(1)$ this was $A$) is contracted away and you get something proportional to either $A^\dagger C+C^\dagger A$, $A^\dagger B+B^\dagger A$, or $C^\dagger B-B^\dagger C$. Most terms will then be zero because they will be commutators of an operator with itself. So for example since $[X, [X,Y]]$ contains $A$ and $B$ it commutes with $Y$, and so all terms of the form $$[Y,X, [X, ...,[X,[X, Y]]]] $$ for an odd number of $X$'s are zero.
Using Dynkin's formula it should be possible to derive $Z$. However I'm finding it hard to keep track of all contributing terms with the correct coefficients. Would anyone here help me out a bit?