Evaluate an integral in terms of elliptic integral

Question

I'm trying to evaluate the integral $$\int \sqrt{\frac{x^2-a^2}{1+x^2}}\,\mathrm{d}x$$ ($a>0$) in terms of an elliptic integral of the second kind, with modulus $\kappa =1/\sqrt{1+a^2}$. I have tried the usual changes of variable $x=\tan t$, $x=\cot t$, and they lead almost to the canonical form, but with $\sin x$, $\cos x$ in the denominator. Is there any other change of variables worth trying?

Answer 1

Denote $k=\sqrt{1+a^2}$. For $t>a$,

$$\newcommand{\arccot}{\operatorname{arccot}} \begin{align*} I(t) &= \int_a^t \frac{\sqrt{x^2-a^2}}{\sqrt{1+x^2}} \, dx \\ &= t \frac{\sqrt{t^2-a^2}}{\sqrt{1+t^2}} - k^2 \int_a^t \frac{x^2\,dx}{\sqrt{x^2-a^2}\left(1+x^2\right)^{3/2}} & \text{by parts} \\ &= t \frac{\sqrt{t^2-a^2}}{\sqrt{1+t^2}} + k^2 \int_{\operatorname{arccot}a}^{\operatorname{arccot}t} \frac{\csc y-\sin y}{\sqrt{\cot^2y-a^2}} \, dy & x=\cot y \\ &= t \frac{\sqrt{t^2-a^2}}{\sqrt{1+t^2}} - \int_{\operatorname{arccot}t}^{\operatorname{arccot}a} \sqrt{1-k^2\sin^2y} \, dy \\ &\qquad + \left(1-k^2\right) \int_{\operatorname{arccot}t}^{\operatorname{arccot}a} \frac{dy}{\sqrt{1-k^2\sin^2y}} \\ &= t \frac{\sqrt{t^2-a^2}}{\sqrt{1+t^2}} + E\left(\arccot t,k\right) - E\left(\arccot a,k\right) \\ &\qquad + a^2 \left[F\left(\arccot t,k\right) - F\left(\arccot a,k\right)\right] \end{align*}$$

Using the identities listed here we can rewrite $I(t)$ in terms of $\kappa=\dfrac1k$:

$$\begin{align*} F(\arccot t,k) &= \frac1{\sqrt k} F\left(\arcsin\sqrt{\frac k{1+t^2}},\frac1k\right) \\[2ex] E(\arccot t,k) &= \sqrt k \, E\left(\arcsin\sqrt{\frac k{1+t^2}}, \frac1k\right) + \frac{1-k}{\sqrt k} F\left(\arcsin\sqrt{\frac k{1+t^2}}, \frac1k\right) \end{align*}$$

Answer 2

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Define the function $\mathcal{J}$ via the elliptic integral,

$$\mathcal{J}{\left(a,b,z\right)}:=\int_{a}^{z}\mathrm{d}t\,\sqrt{\frac{t^{2}-a^{2}}{t^{2}+b^{2}}};~~~\small{z>a>0\land b>0}.$$

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For $z>a>0\land b>0$, we find

$$\begin{align} \mathcal{J}{\left(a,b,z\right)} &=\int_{a}^{z}\mathrm{d}t\,\sqrt{\frac{t^{2}-a^{2}}{t^{2}+b^{2}}}\\ &=\int_{\frac{1}{z}}^{\frac{1}{a}}\mathrm{d}u\,\frac{1}{u^{2}}\sqrt{\frac{1-a^{2}u^{2}}{1+b^{2}u^{2}}};~~~\small{\left[t=\frac{1}{u}\right]}\\ &=\int_{\frac{a}{z}}^{1}\mathrm{d}v\,\frac{a^{2}}{v^{2}}\sqrt{\frac{1-v^{2}}{a^{2}+b^{2}v^{2}}};~~~\small{\left[au=v\right]}\\ &=\int_{\frac{a}{z}}^{1}\mathrm{d}v\,\frac{a^{2}-a^{2}v^{2}}{v^{2}\sqrt{(1-v^{2})(a^{2}+b^{2}v^{2})}}\\ &=\int_{\frac{a}{z}}^{1}\mathrm{d}v\,\frac{a^{2}+b^{2}v^{4}-b^{2}v^{4}-a^{2}v^{2}}{v^{2}\sqrt{(1-v^{2})(a^{2}+b^{2}v^{2})}}\\ &=\int_{\frac{a}{z}}^{1}\mathrm{d}v\,\frac{a^{2}+b^{2}v^{4}}{v^{2}\sqrt{(1-v^{2})(a^{2}+b^{2}v^{2})}}-\int_{\frac{a}{z}}^{1}\mathrm{d}v\,\frac{a^{2}+b^{2}v^{2}}{\sqrt{(1-v^{2})(a^{2}+b^{2}v^{2})}}\\ &=\int_{\frac{a}{z}}^{1}\mathrm{d}v\,\frac{d}{dv}\left[-\frac{\sqrt{(1-v^{2})(a^{2}+b^{2}v^{2})}}{v}\right]-\int_{\frac{a}{z}}^{1}\mathrm{d}v\,\frac{\sqrt{a^{2}+b^{2}v^{2}}}{\sqrt{1-v^{2}}}\\ &=\frac{z}{a}\sqrt{\left(1-\frac{a^{2}}{z^{2}}\right)\left(a^{2}+\frac{a^{2}b^{2}}{z^{2}}\right)}\\ &~~~~~-\int_{\arcsin{\left(\frac{a}{z}\right)}}^{\arcsin{\left(1\right)}}\mathrm{d}\theta\,\frac{\cos{\left(\theta\right)}\sqrt{a^{2}+b^{2}\sin^{2}{\left(\theta\right)}}}{\sqrt{1-\sin^{2}{\left(\theta\right)}}};~~~\small{\left[\arcsin{\left(v\right)}=\theta\right]}\\ &=\frac{\sqrt{\left(z^{2}-a^{2}\right)\left(z^{2}+b^{2}\right)}}{z}-\int_{\arcsin{\left(\frac{a}{z}\right)}}^{\frac{\pi}{2}}\mathrm{d}\theta\,\sqrt{a^{2}+b^{2}\sin^{2}{\left(\theta\right)}}\\ &=\frac{\sqrt{\left(z^{2}-a^{2}\right)\left(z^{2}+b^{2}\right)}}{z}\\ &~~~~~-\int_{0}^{\frac{\pi}{2}-\arcsin{\left(\frac{a}{z}\right)}}\mathrm{d}\varphi\,\sqrt{a^{2}+b^{2}\sin^{2}{\left(\frac{\pi}{2}-\varphi\right)}};~~~\small{\left[\theta=\frac{\pi}{2}-\varphi\right]}\\ &=\frac{\sqrt{\left(z^{2}-a^{2}\right)\left(z^{2}+b^{2}\right)}}{z}-\int_{0}^{\arccos{\left(\frac{a}{z}\right)}}\mathrm{d}\varphi\,\sqrt{a^{2}+b^{2}\cos^{2}{\left(\varphi\right)}}\\ &=\frac{\sqrt{\left(z^{2}-a^{2}\right)\left(z^{2}+b^{2}\right)}}{z}-\int_{0}^{\arccos{\left(\frac{a}{z}\right)}}\mathrm{d}\varphi\,\sqrt{a^{2}+b^{2}-b^{2}\sin^{2}{\left(\varphi\right)}}\\ &=\frac{\sqrt{\left(z^{2}-a^{2}\right)\left(z^{2}+b^{2}\right)}}{z}-\sqrt{a^{2}+b^{2}}\int_{0}^{\arccos{\left(\frac{a}{z}\right)}}\mathrm{d}\varphi\,\sqrt{1-\frac{b^{2}}{a^{2}+b^{2}}\sin^{2}{\left(\varphi\right)}}\\ &=\frac{\sqrt{\left(z^{2}-a^{2}\right)\left(z^{2}+b^{2}\right)}}{z}-\sqrt{a^{2}+b^{2}} E{\left(\arccos{\left(\frac{a}{z}\right)},\frac{b}{\sqrt{a^{2}+b^{2}}}\right)},\\ \end{align}$$

where the incomplete elliptic integral of the second kind $E$ is given here by

$$E{\left(\alpha,\kappa\right)}:=\int_{0}^{\alpha}\mathrm{d}\varphi\,\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}};~~~\small{0<\kappa<1\land0<\alpha\le\frac{\pi}{2}}.$$

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