$$\int\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{x}}\, dx$$
A few days ago I asked a similar (looking) question where all the pluses here were minuses. It could be much more easily manipulated than this one using difference of $2$ squares. The other method used to solve the previous problem is replacing $x$ with $\arccos^2x$. Out of curiosity I have attempted to use the latter method to solve this integral, and made a little progress with it.
Just to make sure this integral has a solution, Wolframalpha found a solution that was not too bad.
Replacing $x$ with $\arctan^2t$, the integral becomes
$$\int\frac{2a \sec^3 t \tan t}{1+\tan t}\, dt$$
From here I tried to perfrom integration by parts, where $2a$ is taken outside of the integral, and numberator split into $sec x\tan x$ times $\sec^2x$, as $u$ and $v'$ respectively. $v$ can easily be found but $u'$ is messy. Integration by parts needed to be done again, also, which would have ended up really messy.
So I tried to convert everything into sines and cosines, then letting $u$ equal $\cos t$, and $du$ equal to $-\sin tdt$.
$$\int\frac{2a\sin t}{\cos^3 t \cos t +\sin t}\, dt$$
$$\int\frac{-2au}{u^3 u+\sqrt{1-u^2}}\, dt$$
However that doesn't look very pretty either. I don't think it is possible to perform Partial Fraction Decomposition on it.
What are the right paths to take to solve this integral? Are there any 'tricks of the trade' I have missed along the way?