Evaluating $\int_{0}^{\infty}\sin^3(x)\cos[a\tan(x)]\frac{dx}{x}$ $$I(a)=\int_{0}^{\infty}\sin^3(x)\cos[a\tan(x)]\frac{dx}{x}$$ I'd like to evaluate the integral by differentiating with respect to  parameter $a$ but no success yet. Seems impossible. What would the other options?  
Edit: A hypothetical closed form solution:  
$$I(a)=\frac{\pi}{4}(1-a)e^{-a}$$

A special case is easy, using $4 \sin^3(x) = 3 \sin(x) - \sin(3x)$:
$$
    I(0) = \int_0^\infty \frac{\sin^3(x)}{x} \mathrm{d} x = \frac{3}{4} \int_0^\infty \frac{\sin(x)}{x} \mathrm{d} x - \frac{1}{4} \int_0^\infty \frac{\sin(3x)}{x} \mathrm{d}x = \frac{1}{2} \int_0^\infty \frac{\sin(x)}{x} \mathrm{d} x = \frac{\pi}{4}
$$
It is easy to see that $I^\prime(a)$ is singular at $a=0$.
In order to evaluate $I(a)$ numerically some massaging is needed: 
$$ \begin{eqnarray}
  I(a) &=& \int_0^\infty \sin^3(x) \cos(a \tan x) \frac{\mathrm{d}x}{x} \\ &=& \frac{1}{2} \int_{-\infty}^\infty \sin^3(x) \cos(a \tan x) \frac{\mathrm{d}x}{x} \\ &=& \frac{1}{2} \sum_{n=-\infty}^\infty \int_{-\frac{\pi}{2} + \pi n }^{\frac{\pi}{2} + \pi n} \sin^3(x) \cos(a \tan x) \frac{\mathrm{d}x}{x} 
\\ &=& \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sum_{n=-\infty}^\infty (-1)^n \sin^3(y) \cos(a \tan y) \frac{\mathrm{d}y}{y+\pi n} \\
 &\stackrel{t =\tan(u)}{=} & \frac{1}{2} \int_{-\infty}^\infty \frac{u^3}{(1+u^2)^{5/2}} \cos(a u) f(u) \mathrm{d} u = \int_{0}^\infty \frac{u^3}{(1+u^2)^{5/2}} \cos(a u) f(u) \mathrm{d} u
\end{eqnarray}
$$
where $f(u)$ is defined by the sum:
$$
\begin{eqnarray}
  f(u) &=& \sum_{n=-\infty}^\infty \frac{(-1)^n}{\pi n + \arctan(u)} = \frac{1}{\arctan(u)} - 2 \arctan(u) \sum_{n=1}^\infty \frac{(-1)^n}{\pi^2 n^2 - \arctan(u)^2} \\
 &=& \frac{1}{\pi} \left( \Phi \left(-1,1,\frac{\arctan(u)}{\pi }\right)+\Phi
   \left(-1,1,1-\frac{\arctan(u)}{\pi }\right) \right)
\end{eqnarray}
$$
where $\Phi(z,s,a)$ denotes the Lerch transcendent. With this we can plot $I(a)$ using numerical methods:

