How to deal with $n\sqrt n$ in master theorem?
You have to remember that $\sqrt[x]{y} = y^{\frac{1}{x}}$.

Then the rest should follow easily.

Don't look at the following unless you're genuinely stuck.


   $n\sqrt{n} = n^{1}\cdot n^{\frac{1}{2}} = n ^{\frac{3}{2}}$, hence $d = \frac{3}{2}$. We also have $a = 8$ and $b = 4$, so $\log_{b}(a) = \log_{4}(8) = \frac{3}{2}$. Hence we have the second case of the Master Theorem where $f(n) \in \Theta(n^{d}\log^{0}n) = \Theta(n^{d})$ with $d = \log_{b}(a)$. Therefore $T(n) \in \Theta(n^{d}\log n) = \Theta(n^{\frac{3}{2}}\log n)$.