Bounds for Akra-Bazzi versus Master theorem I had my final exam today in an advanced algorithms course. In it there were two recurrence relations and I used the Akra-Bazzi theorem to solve them. After the examination I was discussing the problem with one of my colleagues and they apparently got a different bound using Master theorem. My question is which of this two bounds are tighter?

The recurrences were: 

$1)\quad T(n) = 8T\left(\frac{(2n-1)}{4}\right)+n^3+n\log n$

$2)\quad T(n) = 3T(n/2)+n\log^2 n $

Now, correct me if I'm wrong but Akra-Bazzi gives 1) as $\Theta(n^3\log n )$ and 2) as $\Theta(n\log^2 n )$
Can somebody please verify 2) by Akra-Bazzi? I checked and regularilty condition doesn't hold for 2) .
Both methods give $\Theta$ type estimates. That is, the master theorem gives some estimate $T = \Theta(f)$, and the Akra&ndash;Bazzi theorem gives some estimate $T = \Theta(g)$. From the definition of $\Theta$ it immediately follows that $f = \Theta(g)$ and $g = \Theta(f)$, that is, the estimates are equivalent. I suspect either one of you was mistaken, or both bounds that you got, while seemingly different, are in fact equivalent.