Nasty integral

Alright this was a fun problem to solve! Anyway to the solution. $\begin{aligned} I&=\int^{a/2}_0\sqrt{x^2+ax}\;dx=\int^{a/2}_0\sqrt{(x+a/2)^2-a^2/4}\;dx\\ x&\mapsto x+a/2\\ \implies I&=\int^a_{a/2}\sqrt{x^2-a^2/4}\;dx&\\ x&\mapsto\frac a2\sec\theta\implies dx=\frac a2\sec\theta\tan\theta d\theta\\ \implies I&=\frac a2\int^{\pi/3}_0\sqrt{\frac {a^2}4\left(\sec^2\theta-1\right)}\sec\theta\tan\theta\;d\theta=\frac{a^2}4\int^{\pi/3}_0\sec\theta\tan^2\theta d\theta\\ &=\frac{a^2}4\left[\int^{\pi/3}_0\sec^3\theta d\theta-\int^{\pi/3}_0\sec\theta d\theta\right]=\frac{a^2}{4}\left[I_1-I_2\right]\\ I_1&=\int^{\pi/3}_0\sec^3\theta d\theta=\left|\sec\theta\tan\theta\right|^{\pi/3}_0-\int^{\pi/3}_0\sec\theta\tan^2\theta d\theta=2\sqrt3-\frac{4}{a^2}I\\ I_2&=\int^{\pi/3}_0\sec\theta d\theta=\left|\vphantom{\Large |}\log|\sec\theta+\tan\theta|\right|^{\pi/3}_0=\log(2+\sqrt3)\\ \implies I&=\frac{a^2}4\left[2\sqrt3-\frac4{a^2}I-\log(2+\sqrt{3})\right]=\frac{a^2}{4}\left(2\sqrt3-\log(2+\sqrt3)\right)-I\\ \implies2I&=\frac{a^2}{4}\left(2\sqrt3-\log(2+\sqrt3)\right)\\ \implies I&=\frac{a^2}4\left(\sqrt3-\frac12\log(2+\sqrt3)\right)\\ \implies I&=\frac{a^2}4\left(\sqrt3-\log\left(\sqrt{2+\sqrt3}\right)\right)\end{aligned}$ Therefore finally we have $n_1=3$ , $n_2=2$ and $n_3=3$ giving us $n_1+n_2+n_3=\boxed{8}$ .