Not my calculus problem #2

The integral can be solved using integration by parts as follows:

$\begin{aligned} I & = \int_0^\pi x^3 \cos x \ dx \\ & = \int_0^\pi \frac {x^3(1 + \cos 2x )}2 \ dx \\ & = \int_0^\pi \frac {x^3}2 \ dx + \int_0^\pi \frac {x^3 \cos 2x}2 \ dx \\ & = \frac {x^4}8 \ \bigg|_0^\pi + \frac {x^3 \sin 2x}4 \ \bigg|_0^\pi - \frac 34 \int_0^\pi x^2 \sin 2x \ dx \\ & = \frac {\pi^4}8 + 0 + \frac {3x^2 \cos 2x}8 \ \bigg|_0^\pi - \frac 34 \int_0^\pi x \cos 2x \ dx \\ & = \frac {\pi^4}8 + \frac {3\pi^2}8 - \frac {3x \sin 2x}8 \ \bigg|_0^\pi + \frac 38 \int_0^\pi \sin 2x \ dx \\ & = \frac {\pi^4}8 + \frac {3\pi^2}8 - 0 - \frac {3\cos 2x}{16} \ \bigg|_0^\pi \\ & = \frac {\pi^4}8 + \frac {3\pi^2}8 - 0 - 0 \\ & = \frac {3\pi^2 + \pi^4}8 \end{aligned}$

Therefore $a+b+c+d = 3+2+4+8 = \boxed {17}$ .