Integrals of a particular form Recently, I was given a question sheet with a lot of integrals, and I could solve all of them except for a particular type of them:
$$I_1=\int \frac{dx}{(x+1)^2\sqrt{x^2+2x+2}}=\int \frac{dx}{x^2\sqrt{x^2+1}}=\frac{-\sqrt{x^2+1}}{x}+C$$
$$I_2=\int \frac{dx}{(x-1)^2\sqrt{x^2-x+1}}$$
$$I_3=\int \frac{dx}{(2x+1)\sqrt{x^2-8}}$$
$$I_4=\int \frac{dx}{x\sqrt{x^2+x+1}}=\log x + \log\left(x+2+2\sqrt{x^2+x+1}\right)+C$$
And the like($I_2,I_3$ also are expressible in elementary functions, but they are not that cute). The method I see that wolfram is using(trial pro subscription) is to complete the square inside the roots to transform them to $\sqrt{ax^2+b}$ and then substitute $x=\sqrt{\frac{-b}{a}}\sec x$ if $b$ is negative, and $x=\sqrt{\frac{b}{a}}\tan x$ if positive. Finally, it uses the tangent half-angle subsitution. I really don't think this is the best way.
In particular, the trigonometric substitutions in $I_1,I_4$ seem to be unnecessary, since at the end we transform back the resulting composition of trigonometric functions into a function that doesn't need trigonometry at all (e.g. $-\csc(\tan^{-1}x)=-\frac{\sqrt{x^2+1}}{x}$)
So my question is: Is there some easy way to deal with the integrals:
$$\int\frac{dx}{(ax+b)\sqrt{x^2+c}}, \int\frac{dx}{(ax+b)^2\sqrt{x^2+c}}$$
Without having to do the secant-Weiertrass substitution?
 A: Usually more direct, with $\sqrt {u^2 + 1},$ to take $u = \sinh t.$ With $\sqrt {u^2 - 1},$ to take $u = \cosh t.$ Tends to save several steps.
Relevant: it will still be necessary, to integrate a rational function of $\sinh x$ and $\cosh x,$ to use a stereographic projection. I worked it out:
$$  v = \tanh \frac{x}{2} = \frac{e^x - 1}{e^x + 1}; \; \; \; \mbox{so} \; |v| < 1.  $$
$$  x = 2 \operatorname{argtanh} x = \log \left( \frac{1+v}{1-v}\right)   $$
$$  \sinh x =  \frac{2v}{1-v^2} $$
$$  \cosh x =  \frac{1+v^2}{1-v^2} $$
$$ dx =   \frac{2}{1-v^2} \, dv  $$
A: for the type of $\displaystyle \int \frac{1}{(ax+b)^2 \sqrt{cx^2 + dx + e}}dx$, $\displaystyle \frac{1}{ax+b} = t$ transforms into $\displaystyle \frac{at}{\sqrt{bt^2 + ct + d}}$ which can be reduced into standard integrals.
The same seems to work in linear case too.
$$\frac{1}{(ax+b) \sqrt{cx^2+ dx + e}} = \frac{at}{t \sqrt{bt^2+ct+d}} = \frac{a}{\sqrt{bt^2+ct+d}}$$
A: The Euler substitution $t=\sqrt{ax^2+bx+c}-\sqrt{a}x$ for positive $a$ will reduce all these integrals to rational functions.
