May be too lengthy

Integration by parts gives $\begin{array}{rcl} I & = & \displaystyle \int_0^\pi \tan^{-1}\left(\frac{\sin x}{3 - \cos x}\right)\sin 3x\,dx \\ & = & \displaystyle \left[-\tfrac13\tan^{-1}\left(\frac{\sin x}{3 - \cos x}\right)\cos3x\right]_0^\pi + \tfrac13\int_0^\pi \frac{3\cos x - 1}{10 - 6\cos x} \cos 3x\,dx \\ & = & \displaystyle \tfrac13\int_0^\pi \frac{3\cos x - 1}{10 - 6\cos x} (4\cos^3x - \cos x)\,dx \\ & = & \displaystyle \tfrac13\int_0^\pi \left[ -\tfrac{146}{27} - \tfrac{53}{18}\cos x - \tfrac83\cos^2x - 2\cos^3x + \frac{730}{27(5 - 3\cos x)}\right]\,dx \\ & = & \displaystyle \tfrac13\left[ -\tfrac{146}{27}\pi + 0 - \tfrac83 \times \tfrac12\pi - 0 + \tfrac{730}{27}\times\tfrac14\pi\right] \\ & = & \tfrac{1}{162}\pi \end{array}$ making the answer $1 + 162 + 1 \,=\, \boxed{164}$ .