Using the inert form makes this a lot faster and helps a lot with this kind of geometric problem.
Normally, I only need cos(alpha) and sin(alpha) from alpha which I can get directly from exp_x, exp_y of my example above (for rotation matrizes in kinematics). The angle alpha itself is only needed as a time derivative.
The time derivative of the inert form is calculated as fast as the time derivative of the full expression, so I guess the inert arctan expression does not have to be evaluated any more for time differentiation:
> alpha1:=%arctan(exp_y, exp_x): # fast
> alphaD:=diff(alpha1, t): # fast
> alpha2:=arctan(exp_y, exp_x): # slow
> alphaD:=diff(alpha2, t): # fast