My thirteenth integration problem

We simplify the given expression using trigonometric identities:

$\displaystyle \Rightarrow \frac{1}{2} \displaystyle \int_{0}^{\pi} x^3 \Big( 2\sin(x)\cos(x) \Big) \Big( \sin^2(\frac{\pi}{4})\cos^2(x) - \cos^2(\frac{\pi}{4})\sin^2(x) \Big) dx$

$\displaystyle \Rightarrow \frac{1}{8} \displaystyle \int_{0}^{\pi} x^3 \Big( 2\sin(2x)\cos(2x) \Big) dx$

$\displaystyle \Rightarrow \frac{1}{8} \displaystyle \int_{0}^{\pi} x^3 \sin(4x) dx$

Applying Integration by parts repeatedly, we get:

$\displaystyle \Rightarrow \bigg( \Big( \frac{3x}{256} - \frac{x^3}{32} \Big) \cos(4x) + \Big( \frac{3x^2}{128} - \frac{3}{1024} \Big) \sin(4x) \bigg)_{0} ^{\pi}$

$\displaystyle \Rightarrow \frac{3\pi}{256} - \frac{\pi^3}{32}$

$\displaystyle \Rightarrow a + b + c + d + e = 3 + 256 + 1 + 3 + 32 = \boxed{295}$