Evaluate $\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$ 
I want to find a quick way of evaluating $$\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$$

This problem appeared on the qualifying round of MIT's 2014 Integration Bee, which leads me to think there should be a shortish way (no more than three minutes by hand) to solve it. Examining the indefinite integral (thanks to WolframAlpha) hasn't particularly helped me:
$$\int\sqrt{x+\sqrt{x^2+1}}\,dx=-\frac{2}{3} \left(\sqrt{x^2+1}-2x\right) \sqrt{x+\sqrt{x^2+1}}+C$$
The bounds on the integral hint at trigonometric substitution, but I got nowhere by trying $x=\tan u$.
I also noticed that we can transform the integral by multiplying $\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x}$ in the first square root, but there didn't seem to be anything to do after that either.

Using formula:
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{1}{2} \left(a+\sqrt{a^2-b}\right)}+\sqrt{\frac{1}{2} \left(a-\sqrt{a^2-b}\right)}$$
so:
$$\sqrt{x+\sqrt{x^2+1}}=\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}$$
and then:
$$\int \left(\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}\right) \, dx=\frac{1}{3} \sqrt{2} (-i+x)^{3/2}+\frac{1}{3} \sqrt{2} (i+x)^{3/2}+C$$
