Prove $\int_{\mathbb{R^{+}}} \frac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\frac{\pi}{32} (3\pi-4)^2$ How do you arrive at the result $$I=\displaystyle\int_{\mathbb{R^{+}}} \dfrac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\dfrac{\pi}{32} (3\pi-4)^2\ ?$$
Wolfram Alpha agrees numerically.
I tried replacing $\sin p$ by $\dfrac{e^{ip}-e^{-ip}}{2i}$ and similarly for $\cos p$, but in vain.

The following is an approach using contour integration.
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$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{3}(\pi x^{2}) \cos(4x^{2})}{x^{5}} \ dx &= \frac{1}{2}\int_0^{\infty} \frac{\sin^3(\pi t)\cos(4t)}{t^3} \ dt \\ &= \frac{1}{4} \int_{-\infty}^{\infty} \frac{\sin^{3} (\pi t) \cos(4t)}{t^{3}} \ dt  \\ &= \frac{1}{4} \text{Re} \int_{-\infty}^{\infty} \frac{\sin^{3} (\pi t) e^{4it}}{t^{3}} \ dt \\ &= \frac{1}{16} \text{Re} \int_{-\infty}^{\infty} \frac{[3 \sin (\pi t) - \sin(3\pi t)]e^{4it}}{t^{3}} \ dt\\ &= \frac{1}{32}  \text{Re} \frac{1}{i} \int_{-\infty}^{\infty} \frac{(3e^{\pi i t} - 3e^{ -\pi i t} - e^{3 \pi i t} +  e^{-3 \pi i t})e^{4it}}{t^{3}} \ dt \\ &= \frac{1}{32} \text{Re} \frac{1}{i} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{3e^{(4+\pi) i t} - 3 e^{(4- \pi) i t} - e^{(4+3 \pi) i t} +  e^{(4-3 \pi) i t}}{(t-i \epsilon)^{3}} \ dt \\ &= \frac{1}{32} \text{Re} \frac{1}{i} \lim_{\epsilon \to 0^{+}} 2 \pi i \  \text{Res} \left[ \frac{3e^{(4+\pi)iz} - 3 e^{(4- \pi)iz} - e^{(4+ 3 \pi)iz}}{(z- i \epsilon)^{3}},i \epsilon \right] \\ &= \frac{\pi }{16} \text{Re} \lim_{\epsilon \to 0^{+}} \frac{1}{2!} \lim_{z \to i \epsilon} \frac{d^{2}}{dz^{2}} \left( 3e^{(4+ \pi)iz} - 3 e^{(4-\pi)iz} - e^{(4+ 3\pi)iz}\right) \\ &= \frac{\pi}{32} \left(-3( 4+ \pi)^{2} + 3 (4- \pi)^{2} +(4 + 3 \pi)^{2}\right) \\ &=  \frac{\pi}{32} \left(3 \pi - 4 \right)^{2} \end{align}$$
EDIT:
Notice that on line 6, $4 - 3 \pi <0$. 
So by Jordan's lemma, $\displaystyle  \int_{-\infty}^{\infty}\frac{e^{(4- 3\pi)it}}{(z-i \epsilon)^{3}} dt$ would need to be broken off and evaluated separately by closing the contour with the lower half of $|z|=R$ as opposed to the upper half of $|z|=R$. 
But since the pole is in the upper half-plane, $\displaystyle \int_{-\infty}^{\infty} \frac{e^{(4- 3\pi)it}}{(t-i \epsilon)^{3}} \ dt $ evaluates to $0$.
