Solving recurrence with logarithm squared $T(n)=2T(n/2) + n \log^2n$ $T(n)=2T(n/2) + n\log^2(n)$.

If I try to substitute $m = \log(n)$ I end up with 

$T(2^m)=2 T(2^{m-1}) + 2^m\log^{2}(2^m)$.

Which isn't helpful to me. Any clues?

PS. hope this isn't too localized. I specified that the problem was a squared logarithm which should make it possible to find for others wondering about the same thing.
This is indeed the second case in the Master Theorem. For the standard recursion form $$T(n)=a\;T(n/b)+f(n),$$
you get $a=b=2$, and therefore  $f(n)=\Theta(n^{\log_b a} \log^2 n)=\Theta(n  \log^2 n)$.

Applying the Master theorem yields $T(n)=\Theta(n\log^3 n)$.