Ways to simplify arctan() in integral results?

Question

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!

Answer 1

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 $$ \int\frac{\sqrt{2}}{1+\sqrt{2}\sqrt{1-u^2}}\,du $$

Answer 2

Such simplifications are possible thanks to the identities

$$\sin x=\sqrt{1-\cos^2x},\\\tan x=\sqrt{1+\sec^2x}$$ and many similar, also written

$$\arccos t=\arcsin\sqrt{1-t^2},\\\arccos t=\arctan\sqrt{1-\frac1{t^2}}$$

and so on.

But it doesn't seem possible to predict the simplifications that their use can yield without trying them.

It is likely that Alpha used a substitution with a sine rather than with a tangent, which gave a simpler expression "by chance" (though here the "simpler" character of the solution is questionable).