Would IBP work?

Ohh, it seems the answer has already fixed it.

Here is my solution. Consider $\int_0^\infty\frac{x^{n-1}}{1+x^m}\,dx=\frac{\pi}{m\sin\frac{n\pi}{m}}\qquad\qquad(\text{*})$ It can easily be proven by taking substitution $\displaystyle y=\frac{1}{1+x^m}$ and the integral becomes Beta function $\frac{1}{m}\int_0^1 y^{\large 1-\frac{n}{m}-1}\ (1-y)^{\large \frac{n}{m}-1}\,dy=\frac{\Gamma\left(1-\frac{n}{m}\right)\Gamma\left(\frac{n}{m}\right)}{m}=\frac{\pi}{m\sin\frac{n\pi}{m}}$ where the last part comes from Euler's reflection formula for the gamma function . Differentiating (*) with respect to $n$ twice yields $\int_0^\infty\frac{x^{n-1}\ln^2x}{1+x^m}\,dx=\frac{\pi^3}{m^3}\left({\frac{1+2\cot^2\frac{n\pi}{m}}{\sin\frac{n\pi}{m}}}\right)$ Substituting $n=4$ and $m=6$ yields $\int_0^\infty\frac{x^{3}\ln^2x}{1+x^6}\,dx=\frac{\pi^3}{6^3}\left({\frac{1+2\cot^2\frac{2\pi}{3}}{\sin\frac{2\pi}{3}}}\right)=\frac{5\pi^3}{324\sqrt{3}}$ So, $a+b+c+d=335$ .