Estimating $\int_0^{2\pi}\left(\sum_{n=1}^{N} \cos (n^2x) \right)^4\ dx$ for large $N$

Question

In a paper I am reading, it is claimed without proof that for every $\varepsilon > 0$ and $N > N_0(\varepsilon)$ $$ \int_0^{2\pi}\left(\sum_{n=1}^{N} \cos (n^2x) \right)^4\ dx < N^{2 + \varepsilon} $$ I do not quite see how this follows. The Cauchy-Schwarz inequality shows that $$ \left(\sum_{n=1}^{N} \cos (n^2x) \right)^4 \le N^2\left(\sum_{n=1}^{N} \cos^2(n^2x) \right)^2 \le N^3\left(\sum_{n=1}^{N} \cos^4(n^2x) \right) $$ and hence $$ \int_0^{2\pi}\left(\sum_{n=1}^{N} \cos (n^2x) \right)^4\ dx \le N^3\int_0^{2\pi}\sum_{n=1}^{N} \cos^4(n^2x) = \frac{3}{4}\pi N^4, $$ but this is evidently not good enough. Any help proving the claim is appreciated.