If $$\int_0^{\sqrt{3}} \frac{15x^3}{\sqrt{(1+x^2)} + \sqrt{(1+x^2)^3}} dx = \alpha\sqrt{2} + \beta\sqrt{3}$$, where $$\alpha, \beta$$ are integers, then $$\alpha + \beta$$ is equal to

Given,

$$I=\int_0^{\sqrt3}\frac{15x^3}{\sqrt{1+x^2}+\sqrt{(1+x^2)^3}}\,dx$$

Put

$$x=\tan\theta$$

Then,

$$dx=\sec^2\theta\,d\theta$$

Also,

$$\sqrt{1+x^2}=\sec\theta,\qquad \sqrt{(1+x^2)^3}=\sec^3\theta$$

Limits become:

$$x=0\Rightarrow\theta=0,\qquad x=\sqrt3\Rightarrow\theta=\frac{\pi}{3}$$

Hence,

$$I=\int_0^{\pi/3}\frac{15\tan^3\theta\sec^2\theta}{\sec\theta+\sec^3\theta}\,d\theta$$

$$=\int_0^{\pi/3}\frac{15\tan^3\theta\sec\theta}{1+\sec^2\theta}\,d\theta$$

Using

$$\tan^2\theta=\sec^2\theta-1,$$

$$I=\int_0^{\pi/3}\frac{15(\sec^2\theta-1)\sec\theta\tan\theta}{1+\sec^2\theta}\,d\theta$$

Now put

$$1+\sec\theta=t^2$$

Then,

$$2t\,dt=\sec\theta\tan\theta\,d\theta$$

When

$$\theta=0,\qquad t=\sqrt2$$

and when

$$\theta=\frac{\pi}{3},\qquad t=\sqrt3$$

Also,

$$\sec\theta=t^2-1$$

Hence,

$$I=30\int_{\sqrt2}^{\sqrt3}\frac{(t^2-1)^2-1}{t^2}\,dt$$

$$=30\int_{\sqrt2}^{\sqrt3}(t^2-2)\,dt$$

$$=30\left[\frac{t^3}{3}-2t\right]_{\sqrt2}^{\sqrt3}$$

$$=30\left[\left(\frac{3\sqrt3}{3}-2\sqrt3\right)-\left(\frac{2\sqrt2}{3}-2\sqrt2\right)\right]$$

$$=30\left[-\sqrt3+\frac{4\sqrt2}{3}\right]$$

$$=40\sqrt2-30\sqrt3$$

Therefore,

$$\alpha=40,\qquad \beta=-30$$

Hence,

$$\alpha+\beta=10$$

Therefore, the required answer is

$$\boxed{10}$$.