OK, first instead of using $A$, I will use $\kappa = D/(A-1)$ which is a more natural parameter for the problem. (If $D$ is small, $\kappa$ is essentially a bound on the logarithmic derivative.) Now the natural guess is that the optimiser will saturate the constraint at every point. Indeed, if it fails in the neighbourhoods of two points, I can transfer a bit of mass from the one further from the origin to the one closer to it and I improve the expectation of $|x|$. This means that I can restrict myself to piecewise constant functions, which I can assume to be symmetric.
Take now $f$ such that $f(x) = b$ for $|x| < a$. Then, in order to saturate the constraint everywhere, I have no more choice: I have to set $f(x) = b/A$ for $a \le |x| < a+D$, $f(x) = b/A^2$ for $a+D \le |x| < a+2D$, etc. Now if I want to get a probability measure, it turns out that I have to choose $$ b = {1\over 2(a+\kappa)}\;. $$ It remains to optimise over $a$ to minimise the expectation. The answer to this exercise is $$ a = \sqrt{\kappa(D+\kappa)}-\kappa\;. $$ With this choice, one then obtains $$ \int |x| f(x)\,dx = {1\over 2}\sqrt{\kappa(D+\kappa)}\;. $$ I haven't double-checked the calculations, but this should be about right...