Didn't expect this to be that long!

The substitution $\sin u = e^{-x}$ converts the integral as follows: $I \; = \; \int_0^\infty x^2 e^{-3x} \tan^{-1}\left(\frac{e^{-x}}{\sqrt{1-e^{-2x}}}\right)\,dx \; = \; \int_0^{\frac12\pi} u \,\sin^2u \,\cos u \,\ln(\sin u)^2\,du \; =\; F''(2)$ where $F(a) \; = \; \int_0^{\frac12\pi} u \sin^a u \cos u\,du \; = \;\frac{\pi}{2(a+1)} - \frac{1}{a+1}\int_0^{\frac12\pi} \sin^{a+1}u\,du \; = \; \frac{\pi - B\big(1 + \tfrac12a,\tfrac12\big)}{2(a+1)}$ After some differentiation and simplification using polygamma functions, we obtain $I \; = \; \tfrac{1}{27}\pi - \tfrac{14}{27} + \tfrac{1}{54}\pi^2 - \tfrac29\ln^22 +\tfrac{14}{27}\ln2$ making the answer $27 + 14 + 2 + 54 + 9 = \boxed{106}$ .

The Answer is: $\dfrac{\pi}{27}-\dfrac{14}{27}+\dfrac{\pi^{2}}{54}-\dfrac{2}{9}\ln^{2}(2)+\dfrac{14}{27}\ln(2)$ Hints:-
- Substitute $e^{x}$ as $t$
- Express the Changed Integral as the Differential of another integral(easy to solve)
- Plug the required values( requires use of digamma and trigamma functions )
- Simplify