I'm trying to solve this recursion equation:
$T(n)=n\sqrt{n}T(\sqrt{n})+n^3\log^2{(n\log{(\log{n})})}$
I tried to substitute $m=\log{n}$ but it doesn't seem to help:
$T(2^m)=2^{\frac{3m}{2}}T(2^{\frac{m}{2}})+2^{3m}\log^2{(2^m\log{(m)})}$
How can I proceed from here?
As
$$ T\left(2^{\log_2 n}\right)=n\sqrt{n}T\left(2^{\log_2 \sqrt{n}}\right)+n^3\log_2^2(n\log_2(\log_2 n)) $$
making $\mathcal{T}(\cdot)=T(2^{(\cdot)})$ and $z = \log_2 n$ we have
$$ \mathcal{T}(z)=2^z 2^{\frac z2}\mathcal{T}\left(\frac z2\right)+2^{3z}\log_2^2(2^z\log_2z) $$
now calling $\mathbb{T}(\cdot)=\mathcal{T}(2^{(\cdot)})$ and making $u = \log_2 z$ we have
$$ \mathbb{T}(u)=2^{\frac 32 2^u}\mathbb{T}(u-1)+2^{3\times 2^u}\log_2^2(2^{u}u) $$
with solution
$$ \mathbb{T}(u)=8^{2^u-2} \left(c_1+\sum _{k=1}^{u} 2^{6-2^{k}} \log_2 ^2\left(2^{k} k\right)\right) $$
finally we should return to $T(n)$ with $\mathbb{T}_u\to\mathcal{T}_z\to T_n$