Factoring?

$\begin{aligned} I & = \int_0^\infty \frac {dx}{1-x^3} & \small \color{#3D99F6} \text{Using identity: }\int_0^\infty f(x)\ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2} dx \\ & = \frac 12 \int_0^\infty \left(\frac 1{1-x^3} + \frac 1{x^2\left(1-\frac 1{x^3}\right)} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac 1{1-x^3} + \frac x{x^3-1} \right) dx \\ & = \frac 12 \int_0^\infty \frac {1-x}{1-x^3} dx \\ & = \frac 12 \int_0^\infty \frac 1{x^2+x+1} dx \\ & = \frac 12 \int_0^\infty \frac 1{\left(x+\frac 12 \right)^2 + \frac 34} dx \\ & = \frac 23 \int_0^\infty \frac 1{\left(\color{#3D99F6}\frac {2x+1}{\sqrt 3} \right)^2 + 1} dx & \small \color{#3D99F6} \text{Let }u = \frac {2x+1}{\sqrt 3} \implies du = \frac 2{\sqrt 3} dx \\ & = \frac 1{\sqrt 3} \int_{\frac 1{\sqrt 3}}^\infty \frac 1{{\color{#3D99F6}u^2} + 1} du \\ & = \frac 1{\sqrt 3} \left[\tan^{-1} u \right]_{\frac 1{\sqrt 3}}^\infty \\ & = \frac 1{\sqrt 3} \left[\frac \pi 2 - \frac \pi 6\right] \\ & = \frac \pi {3\sqrt 3} = \frac {\sqrt 3 \pi}9 \end{aligned}$

Therefore, $a+b+c = 1 + 3 + 9 = \boxed{13}$ .

$I=\displaystyle\int_{0}^{\infty}\frac{dx}{1-x^3}= \displaystyle\int_{0}^{1}\frac{dx}{1-x^3}+\underbrace{\displaystyle\int_{1}^{\infty}\frac{dx}{1-x^3}}_{J}$

$\implies J=\displaystyle\int_{0}^{1}\frac{-x}{1-x^3}dx$ $\implies I=\displaystyle\int_{0}^{1}\frac{1-x}{1-x^3}dx = \displaystyle\int_{0}^{1}\frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}$ $\implies I=\frac{2}{\sqrt{3}}\arctan({\frac{3}{\sqrt{3}}})-\frac{2}{\sqrt{3}}\arctan({\frac{1}{\sqrt{3}}})$ $\implies I=\frac{\sqrt{3}\pi}{9} \implies a+b+c=\boxed{13}$