Its too hard to come up with cool names for integrals

For $a\geq 0$ define : $f(a)= \int_0^{\infty} \frac{(ax)^3-\sin^3 (ax)}{x^5} \ \mathrm{d}x$ Differentiate it 4 times (note that the third derivative is absolutely convergent, then we can differentiate) to get : $f^{(4)}(a)= \int_0^{\infty} \frac{81\sin(3 ax)}{4x} -\frac{3\sin (ax)}{4x} \ \mathrm{d}x = \frac{\pi}{2}\left(\frac{81}{4}-\frac{3}{4} \right)= \frac{39\pi}{4}.$ Note that

$f(0)=f'(0)=f^{(2)}(0)=f^{(3)}(0)= 0$ then : $f(a) = \frac{13a^4\pi }{32}$

Easily we get $f(1)= \frac{13\pi}{32}$ , then answer is $13+1+32=\boxed{46}$ .

Another method : we have $I=\frac{1}{4} \int_0^{\infty} \frac{4x^3 +\sin(3x)-3\sin x}{x^5} \ \mathrm{d}x$ Integrate by parts to get : $4I= \int_0^{\infty} \frac{4x^3 +\sin(3x)-3\sin x}{x^5} \ \mathrm{d}x =\frac{-1}{4} \underbrace{\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}}_{=0} +\frac{1}{4} \int_0^{\infty} \frac{12x^2 -3\cos x +3\cos(3x)}{x^4} \ \mathrm{d}x$ The brackets gives zero since near zero we have $4x^3+\sin (3x)-3\sin x=2x^5+o(x^5)$ by Taylor series, and the limit at $\infty$ is obviously $0$ , we repeat the same thing $16 I= \int_0^{\infty} \frac{12x^2 -3\cos x +3\cos(3x)}{x^4} \ \mathrm{d}x =\frac{1}{3} \int_0^{\infty} \frac{ 24x+3\sin x-9\sin 3x}{x^3} \ \mathrm{d}x$ And again two times : $48I = \frac{1}{2} \int_0^{\infty}\frac{24+3\cos x -27\cos(3x)}{x^2} \ \mathrm{d}x = \frac{1}{2} \int_0^{\infty} \frac{81\sin(3x)}{x} - \frac{3\sin x}{x} \ \mathrm{d}x$ We know that for any $a\neq 0$ $\int_0^{\infty} \frac{\sin a x}{x} \ \mathrm{d}x = \frac{\pi}{2}$ (For a proof Google for Dirichlet integral), then : $96 I = \left(81-3\right)\frac{\pi}{2} \Longrightarrow I=\boxed{\frac{13\pi}{32}} .$