Campbell-Baker-Hausdorff formula for $\log(\exp(X+Y)\exp(X-Y))$ Given $X,Y\in \mathfrak g\mathfrak l_{\mathbb R}(n)$, and the CBH formula for $\log(\exp X\exp Y)$ (wiki), is it possible to derive the general term in the series of $\log(\exp(X+Y)\exp(X-Y))$ that involve only an odd number of $X$? The even term should vanish since $f(X):=\log(\exp(X+Y)\exp(X-Y))$ is an odd function. 
Additional question:
If this can be worked out in a nice way, I would also like to have a similar result for $\log(\exp(X+Y)\exp(Z)\exp(X-Y))$. In this case, only terms involving an odd number of $X$'s and/or $Z$'s remain.
 A: See 4.29 (on page 54) 


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*Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008.(pdf)
for complete formulas involving the the general terms of the BCH formula. You can use this to derive some expressions for the formula you want. 
A: The systematics of the CBH expansion comport with the structure of the Lie algebra involved. There is no "nice" full expansion beyond Dynkin's unwieldy expression, except in special cases. To fix language, you might consult the bare minimum review for theoretical physicists. 
To account for the "number of commutators" in the expansion, introduce the gratuitous parameter t to set equal to 1 in the end, if you wish. it is the number of nested commutators minus 1, so $Z(t;X,Y)\equiv \log( ~ \exp (t X) \exp( tY)~) $. Thus $Z(t;X,Y)=-Z(-t;Y,X)$, that is,  odd powers of t are X-Y symmetric, and even powers of t are antisymmetric. 
Your $W(t;X,Y)=Z(t;X+Y,X-y)= \log(~\exp(t(X+Y))\exp(t(X−Y))~) $ will then have the symmetry 
$$W(t;X,Y)=-Z(-t;X-Y,X+Y)=-W(-t;X,-Y),$$ 
so that odd powers of t will be even in Y and hence odd in X; likewise, even powers of t are odd in Y and thus odd in X, as you already observed. 
But that is as far as the train goes: for further simplifications you need special features of the algebra, some of which the cited reference illustrates.
