Put $$x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta $$

$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^3}\theta \,.\,{{\sec }^2}\theta \,d\theta } \over {\sqrt {1 + {{\tan }^2}\theta + \sqrt {{{\sec }^6}\theta } } }}} $$

$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^2}\theta {{\sec }^2}\theta \,d\theta } \over {\sec \theta \sqrt {1 + \sec \theta } }}} $$

$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15({{\sec }^2}\theta - 1)\sec \theta \tan \theta \,d\theta } \over {\left( {\sqrt {1 + \sec \theta } } \right)}}} $$

Now put $$1 + \sec \theta = {t^2}$$

$$ \Rightarrow \sec \theta \tan \theta \,d\theta = 2tdt$$

$$ \Rightarrow I = \int\limits_{\sqrt 2 }^{\sqrt 3 } {{{15\left( {{{({t^2} - 1)}^2} - 1} \right)2t\,dt} \over t}} $$

$$ \Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2} + 1 - 1)\,dt} $$

$$ \Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2})\,dt} $$

$$ \Rightarrow I = \left. {30\left( {{{{t^5}} \over 5} - {{2{t^3}} \over 3}} \right)} \right|_{\sqrt 2 }^{\sqrt 3 }$$

$$ = 30\left[ {\left( {{9 \over 5}\sqrt 3 - 2\sqrt 3 } \right) - \left( {{{4\sqrt 2 } \over 5} - {{4\sqrt 2 } \over 3}} \right)} \right]$$

$$ = \left( {54\sqrt 3 - 60\sqrt 3 } \right) - \left( {24\sqrt 2 - 40\sqrt 2 } \right)$$

$$ = 16\sqrt 2 - 6\sqrt 3 $$

$$\therefore$$ $$\alpha = 16$$ and $$\beta = - 6$$

$$\alpha + \beta = 10.$$