Integrals $\int\frac{dx}{(ax+b)\sqrt{x^2+c}}, \int\frac{dx}{(ax+b)^2\sqrt{x^2+c}}$

Question

Recently, I was given a question sheet with a lot of integrals. I could solve all of them except for a particular type of them: \begin{align} &I_1=\int \frac{dx}{(x+1)^2\sqrt{x^2+2x+2}}=-\frac{\sqrt{x^2+2x+2}}{x+1}\\ &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) \end{align}

($I_2,I_3$ also are expressible in elementary functions, but they are not that cute.)

The method 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 $x=\sqrt{{-b/a}}\sec x$ if $b$ is negative, and $x=\sqrt{{b/a}}\tan x$ if positive. Finally, it uses the tangent half-angle subsitution. I really don't think this is the best way.

The trigonometric substitutions seem to be unnecessary, since at the end we transform back the resulting trigonometric functions into a function that does not rely on trigonometry at all, e.g. $-\csc(\tan^{-1}x)=-{\sqrt{x^2+1}/x}$.

Question: 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 the trigonometric and Weiertrass substitution?

Answer 1

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. $$\frac{1}{(ax+b) \sqrt{cx^2+ dx + e}} = \frac{at}{t \sqrt{bt^2+ct+d}} = \frac{a}{\sqrt{bt^2+ct+d}}$$

Answer 2

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: $$ v = \tanh \frac{x}{2} = \frac{e^x - 1}{e^x + 1}; \; \; \; \mbox{so} \; |v| < 1. $$ $$ x = 2 \operatorname{argtanh} x = \log \left( \frac{1+v}{1-v}\right) $$ $$ \sinh x = \frac{2v}{1-v^2} $$ $$ \cosh x = \frac{1+v^2}{1-v^2} $$ $$ dx = \frac{2}{1-v^2} \, dv $$

Answer 3

The Euler substitution $t=\sqrt{ax^2+bx+c}-\sqrt{a}x$ for positive $a$ will reduce all these integrals to rational functions.

Answer 4

Examples for Euler's first substitution, one here suggests that $\large\sqrt{x^2+c}=-x+t$ with: $$x=\frac{t-c^2}{2t},\hspace{0.2cm}dx=\frac{t^2+c}{2t^2} \hspace{0.5cm}\text{and}\hspace{0.5cm} \sqrt{x^2+c}=\frac{t^2+c}{2t}$$ Then $$\int\frac{dx}{(ax+b)\sqrt{x^2+c}}=\int\frac{2\;dt}{at^2+2bt-ac}$$ $$\int\frac{dx}{(ax+b)^2\sqrt{x^2+c}}=\int\frac{4t\;dt}{(at^2+2bt-ac)^2}$$

Answer 5

Note that the second integral can be expressed in terms of the first \begin{align} &\int\frac{1}{(ax+b)^2\sqrt{x^2+c}}dx\\ =& -\frac{a \sqrt{x^2+c}}{(b^2+a^2c)(ax+b)} +\frac b{b^2+a^2c} \int\frac{1}{(ax+b)\sqrt{x^2+c}}dx \end{align} So, it boils down to evaluate $$I= \int\frac{1}{(ax+b)\sqrt{x^2+c}}dx$$ for non-zero but otherwise arbitrary $a$, $b$ and $c$. To avoid domain dependent substitutions, let $t=\frac{ac-bx}{\sqrt{x^2+c}}$ instead to integrate for the distinctive cases below \begin{align} I&= \int \frac{1}{t^2-(b^2+a^2c)}dt\\ &=\begin{cases} \frac{1}{\sqrt{-(b^2+a^2c)}} \tan^{-1}\frac{ac-bx}{\sqrt{-(b^2+a^2c)(x^2+c)}},& c< -\frac{b^2}{a^2}\\ \frac{a}{b(ax+b)} \sqrt{x^2-\frac{b^2}{a^2}} \,& c=-\frac{b^2}{a^2}\\ -\frac{1}{\sqrt{(b^2+a^2c)}} \coth^{-1}\frac{ac-bx}{\sqrt{(b^2+a^2c)(x^2+c)}},& -\frac{b^2}{a^2}<c<0\\ -\frac{1}{\sqrt{(b^2+a^2c)}} \tanh^{-1}\frac{ac-bx}{\sqrt{(b^2+a^2c)(x^2+c)}},& c>0 \end{cases} \end{align}

Now, apply the solutions above to the four listed integrals

\begin{align} &I_1\overset{y=x+1}=\int \frac{dy}{y^2\sqrt{y^2+1}}=-\frac{\sqrt{y^2+1}}{y}\\ &I_2 \overset{y=x-\frac12}=\int \frac{dy}{(y-\frac12)^2\sqrt{y^2+\frac34}}\overset{c>0}=-\frac{\sqrt{4y^2+3}}{2y-1}+\frac12\tanh^{-1}\frac{3+2y}{2 \sqrt{4y^2+3}}\\ &I_3=\int \frac{dx}{(2x+1)\sqrt{x^2-8}} \overset{c<-{b^2}/{a^2}}=-\frac1{\sqrt{31}}\tan^{-1}\frac{16+x}{\sqrt{31}\sqrt{x^2-8}}\\ &I_4\overset{y=x+\frac12}=\int \frac{dy}{(y-\frac12)\sqrt{y^2+\frac34}}\overset{c>0}=-\tanh^{-1}\frac{3+2y}{2\sqrt{4y^2+3}} \end{align}