Ways to simplify arctan() in integral results? Lately when I was computing $$\int\frac{\mathrm{d}x}{1+\sqrt{1-2x-x^2}},$$I got the result$$-2\arctan{\frac{\sqrt{1-2x-x^2}-1}{x}}-\ln\left(1-\frac{\sqrt{1-2x-x^2}-1}{x}\right)+\mathbf{C}.$$However, Wolfram|Alpha gives me its result in a different form, which may be written as$$\arcsin{\frac{x+1}{\sqrt{2}}}+\frac{1}{2}\ln{\frac{\sqrt{1-2x-x^2}-x+1}{\sqrt{1-2x-x^2}+x+3}}+\mathbf{C'},$$and I accordingly found that actually$$-2\arctan{\frac{\sqrt{1-2x-x^2}-1}{x}}=\arcsin{\frac{x+1}{\sqrt{2}}}+\frac{\pi}{4},$$a much neater expression! But apparently, without Wolfram|Alpha, I couldn't have found this neat relation that turns a complicated $2\arctan$ into a neat $\arcsin$ at all, so I've been wondering if there are any general way to simplify these $\arctan$s in integral expressions. Thanks!

Note that
$$
1-2x-x^2=2-(1+2x+x^2)=2-(1+x)^2
$$
so you can do the substitution $u\sqrt{2}=1+x$ and the integral becomes
$$
\int\frac{\sqrt{2}}{1+\sqrt{2}\sqrt{1-u^2}}\,du
$$
