Solving recurrence with logarithm squared $T(n)=2T(n/2) + n \log^2n$
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 $T(n)=\Theta(n\log^3 n)$.