For what $p$, $\int_0^{\infty} \frac{\ln (1+x^{3p})}{(x+x^3)^{4p}arctg\sqrt{x}}$ converges

Question

I have to see for what values of $p$ the following integral converges. $$\int_0^{\infty} \frac{\ln (1+x^{3p})}{(x+x^3)^{4p}arctg\sqrt{x}}=\int_0^1 \frac{\ln (1+x^{3p})}{(x+x^3)^{4p}arctg\sqrt{x}} + \int_1^{\infty} \frac{\ln (1+x^{3p})}{(x+x^3)^{4p}arctg\sqrt{x}}$$ So the first part is easy to estimate because it's convergent with $$ \int_0^1 \frac{1}{x^{p+\frac{1}{2}}}$$, so here $p<\frac{1}{2}$.But with the second part I have no idea what to do apart from $arctg\sqrt{x}$ being bounded by $\frac{\pi}{2}$. Any ideas and solutions are welcomed.