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.

Here I present another way to solve/generalize the problem. It is much easier. Define
$$ I_q(p)=\frac{1}{2}\int_0^\infty\frac{\sin^3(pt)\cos(qt)}{t^3}dt,p\ge0,q\ge0. $$
Clearly $I_q(0)=I_q'(0)=I_q''(0)=0$ and
\begin{eqnarray}
I_q''(p)&=&\frac{3}{8}\int_0^\infty\frac{(3\sin(3pt)-\sin(pt))\cos(qt)}{t}dt\\
&=&\frac{3}{16}\int_0^\infty\frac{3\sin((3p+q)t)+3\sin((3p-q)t)-\sin((p+q)t)-\sin((p-q)t)}{t}dt\\
&=&\left\{\begin{array}{l}
0, \text{ if }p<\frac{q}{3},\\
\frac{9\pi}{32}, \text{ if }p=\frac q3,\\
\frac{9\pi}{16}, \text{ if }\frac{q}{3}<p<q,\\
\frac{15\pi}{32}, \text{ if }p=q,\\
\frac{3\pi}{8}, \text{ if }p>q.
\end{array}\right.
\end{eqnarray}
Here we used 
$$ \int_0^\infty\frac{\sin(\alpha x)}{x}dx=\text{sgn}(\alpha)\frac{\pi}{2}. $$
Thus
$$ I_q'(p)=\left\{\begin{array}{l}
0, \text{ if }p\le\frac{q}{3},\\
\frac{9\pi}{16}(p-\frac{q}{3}), \text{ if }\frac{q}{3}<p\le q,\\
\frac{3p\pi}{8}, \text{ if }p>q,
\end{array}\right. $$
and hence
$$ I_q(p)=\left\{\begin{array}{l}
0, \text{ if }p\le\frac{q}{3},\\
\frac{\pi}{32}(3p-q)^2, \text{ if }\frac{q}{3}<p\le q,\\
\frac{(3p^2-q^2)\pi}{16}, \text{ if }p>q.
\end{array}\right. $$
For $p=\pi,q=4$, since $\frac{q}{3}<p<q$, we have
$$ I_4(\pi)=\frac{\pi}{32}(3\pi-4)^2.$$
