If you try to write it down for several recursion cycles, you get this :
2*n^(1/2) [2*n^(1/4) (2*n^(1/8) . T(n^(1/16) + c log n) + c log n] + c log n
If you try to count it, it would be (assymptoticaly) :
2^log n * n^(1/2 + 1/4 + 1/8 + ... + 1/log n) + 2^(log n) * n(1/2 + 1/4 + 1/8 + ... + 1/log n) * c * log n
By sumation of series and thanks to that 2^log_2 n = n
you get (assymptoticaly) :
n^2 + c * n^2 * log n
Which actually is assymptoticaly : n^2(1 + c * log n) = n^2(c * log n) = n^2 * c * log n
Result : T(n) = O(c * n^2 * log n
)