<s>[INST] 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:  
$$I(a)=\frac{\pi}{4}(1-a)e^{-a}$$
 [/INST] Let us assume that $f(z)$ has a Maclaurin series expansion with real coefficients that converges on the unit circle on the complex plane.
Then
$$ \text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx = \int_{0}^{\infty} \frac{\sin^3 x}{x} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \cos(2nx)  \, dx .$$
Now assuming we can switch the order of summation and integration (which would be more difficult to justify if the series doesn't converge absolutely),
$$ \begin{align} &\text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx \\ &= \sum_{n=0}^{\infty}  \frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(2nx) \, dx \\ &= f(0) \int_{0}^{\infty} \frac{\sin^{3} x}{x} \, dx + f'(0) \int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(2x) \, dx + \sum_{n=2}^{\infty}\frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(2nx) \, dx. \end{align}$$
I will now show that all the integrals vanish except for the first two.
Notice that for $a \ge 0$,
$$ \begin{align} &\int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(ax) \, dx \\ &= \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin(x) - \sin (3x)}{x} \cos(ax) \, dx \\ &= \frac{1}{8} \int_{0}^{\infty} \frac{3 \sin [(1+a)x] + 3 \sin [(1-a)x] - \sin [(3+a)x] - \sin [(3-a)x]}{x} \, dx \\ &= \frac{\pi}{16} \Big(3 \, \text{sgn}(1+a) + 3 \, \text{sgn}(1-a) - \text{sgn}(3+a) - \text{sgn}(3-a)  \Big) \tag{1} \\ &= \begin{cases} 0 & \text{if} \ a >3 \\ - \frac{\pi}{16} & \text{if} \  a = 3 \\ - \frac{\pi}{8} & \text{if} \ 1<a <3 \\ \frac{\pi}{16} & \text{if} \ a = 1 \\ \frac{\pi}{4} & \text{if} \ a <1 \end{cases} \end{align}$$
Therefore,
$$ \begin{align} \text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx &= f(0) \left(\frac{\pi}{4} \right) + f'(0) \left(-\frac{\pi}{8} \right) +  \sum_{n=2}^{\infty}\frac{f^{n}(0)}{n!} (0) \\ &= \frac{\pi}{8} \Big(2 f(0) - f'(0) \Big). \end{align}$$
$(1)$ $\int_{0}^{\infty} \frac{\sin (ax)}{x} \, dx = \frac{\pi}{2} \, \text{sgn}(a)$

As a simple example, let us consider the entire function $f(z) = e^{z}$.
Then $$\text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} \, e^{e^{2ix}} \, dx = \int_{0}^{\infty} \frac{\sin^{3} x}{x} \, e^{\cos (2x)} \cos (\sin 2x) \, dx  = \frac{\pi}{8} \Big( 2(1)-1 \Big) = \frac{\pi}{8}.$$

Now let us consider the function $$ f(z) = \exp \left(a \, \frac{z-1}{z+1} \right)$$ for $a\ge 0$.
Since $\exp \left(a \, \frac{z-1}{z+1} \right)$ has an essential singularity at $z=-1$, we know that its Maclaurin series has a radius of convergence of $1$.
What we don't immediately know, and what is proving hard to ascertain, is if the series converges on the unit circle (excluding the point $z=-1$).
But assuming that it does we have
$$ \begin{align} \text{Re} \int_{0}^{\infty} \frac{\sin^{3}(x)}{x} \, \exp \left(a \, \frac{e^{2ix}-1}{e^{2ix}+1} \right) \, dx &= \text{Re} \int_{0}^{\infty} \frac{\sin^{3}(x)}{x} \, \exp (ia \tan x) \, dx   \\ &= \int_{0}^{\infty} \frac{\sin^{3}(x)}{x} \cos(a \tan x) \, dx \\ &= \frac{\pi}{8} \Big(2 (e^{-a}) -2ae^{-a} \Big) \\ &= \frac{\pi}{4} (1-a) e^{-a}. \end{align}$$
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