Runtime for algorithm
Let $f(n) = n^2\log n$, with the $\log$ taken to base 2. Then
$$
\frac{f(2n)}{f(n)} = 4 \frac{\log(2n)}{\log n} = 4 \left(1 + \frac{1}{\log n}\right).
$$
That is, for large $n$ you would expect $f(2n) \approx 4f(n)$. This shows that this test isn't sensitive enough to detect logarithmic factors.