Can anybody help with the integral: $\int \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}} dx$ please? I have searched online. I have got a solution from Wolfram but I don't understand it, or how to reach it. If anyone has a method that would be fantastic, thanks!
 A: This is an elliptic integral, but it is interesting to point out what is the explicit value of the integral over $(0,1)$.
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx = \frac{1}{2}\int_{0}^{1}\sqrt{\frac{1-x}{x(1+x)}}\,dx=\frac{1}{2}\int_{0}^{1}\sqrt{\frac{x}{(1-x)(2-x)}}\,dx$$
equals
$$ \int_{0}^{1}\frac{x^2}{\sqrt{(1-x^2)(2-x^2)}}\,dx =\int_{0}^{\pi/2}\frac{\sin^2\theta}{\sqrt{2-\sin^2\theta}}\,d\theta$$
or
$$2\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{2-\sin^2\theta}}-\int_{0}^{\pi/2}\sqrt{2-\sin^2\theta}\,d\theta=\sqrt{2}\left[K\left(\tfrac{1}{2}\right)-E\left(\tfrac{1}{2}\right)\right]$$
by representing the complete elliptic integrals according to Mathematica's notation, such that the argument is the elliptic modulus. A series representation for the RHS is given by
$$ \frac{\pi}{\sqrt{2}}\sum_{n\geq 1}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{2n}{1-2n}\cdot\frac{1}{2^n}$$
which due to $\frac{\binom{2n}{n}}{4^n}\approx\frac{1}{\sqrt{\pi n}}$ for large values of $n$, can be approximated by
$$ \frac{\pi}{4\sqrt{2}}+\sqrt{2}\sum_{n\geq 2}\frac{n}{2^n(2n-1)}=\frac{\pi-4}{4\sqrt{2}}+\log(1+\sqrt{2}).$$
The Cauchy-Schwarz inequality provides a simpler (and tighter) upper bound, namely
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx<\sqrt{\int_{0}^{1}(1-x^2)\,dx\int_{0}^{1}\frac{dx}{1+x^2}}=\sqrt{\frac{\pi}{6}}.$$
On the other hand both $K\left(\tfrac{1}{2}\right)$ and $E\left(\tfrac{1}{2}\right)$ can be computed through the $\text{AGM}$ and are explicitly related to the $\Gamma$ function. Indeed the former is given by a multiple of $B\left(\tfrac{1}{4},\tfrac{1}{4}\right)$ and the latter is related to the former via Legendre's relation. This gives
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx = \color{blue}{\frac{\Gamma\left(\tfrac{1}{4}\right)^2}{4\sqrt{2\pi}}-\frac{\pi\sqrt{2\pi}}{\Gamma\left(\tfrac{1}{4}\right)^2}}=\color{red}{\frac{\pi}{2\,\text{AGM}(1,\sqrt{2})}-\frac{\text{AGM}(1,\sqrt{2})}{2}}$$
which can be proved through Fourier-Legendre expansions, too.
A: If the integral is a definite integral with the interval of integration being from $0$ to $1$, here is an approach that makes use of the following form for the Beta function $\operatorname{B} (x,y)$:
$$\operatorname{B} (x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt, \qquad x,y > 0. \tag1$$
Let
$$I = \int_0^1 \sqrt{\frac{1 - x^2}{1 + x^2}} \, dx.$$
Enforcing a substitution of $x \mapsto \sqrt{x}$ yields
$$I = \frac{1}{2} \int_0^1 \sqrt{\frac{1 - x}{1 + x}} \, \frac{dx}{\sqrt{x}}.$$
Rearranging the integrand we have
\begin{align}
I &= \frac{1}{2} \int_0^1 \sqrt{\frac{1 - x}{1 + x}} \cdot \sqrt{\frac{1 - x}{1 - x}} \, \frac{dx}{\sqrt{x}}\\
&= \frac{1}{2} \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} \, \frac{dx}{\sqrt{x}}\\
&= \frac{1}{2} \int_0^1 \frac{dx}{\sqrt{1 - x^2} \sqrt{x}} - \frac{1}{2} \int_0^1 \frac{\sqrt{x}}{\sqrt{1 - x^2}} \, dx. \qquad \qquad \qquad \qquad (*)
\end{align}
Enforcing a substitution of $x \mapsto \sqrt{x}$ in each of the above two integrals leads to
\begin{align}
I &= \frac{1}{4} \int_0^1 x^{-3/4} (1 - x)^{-1/2} \, dx - \frac{1}{4} \int_0^1 x^{-1/4} (1 - x)^{-1/2} \, dx\\
&= \frac{1}{4} \int_0^1 x^{1/4 - 1} (1 - x)^{1/2 - 1} \, dx - \frac{1}{4} \int_0^1 x^{3/4 - 1} (1 - x)^{1/2 - 1} \, dx\\
&= \frac{1}{4} \operatorname{B} \left (\frac{1}{4}, \frac{1}{2} \right ) - \frac{1}{4} \operatorname{B} \left (\frac{3}{4}, \frac{1}{2} \right ) \tag2\\
&= \frac{1}{4} \left [\frac{\Gamma (\frac{1}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{3}{4})} - \frac{\Gamma (\frac{3}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{1}{4})} \right ] \tag3\\
&= \frac{\Gamma^2 (\frac{1}{4})}{4 \sqrt{2 \pi}} - \frac{\pi \sqrt{2 \pi}}{\Gamma^2 (\frac{1}{4})}.\tag4
\end{align}
Explanation
(2) Each integral is of the form of the Beta function given in (1).
(3) Using the result $\operatorname{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$. 
(3) Using $\Gamma (\frac{1}{2}) = \sqrt{\pi}$ and $\Gamma (\frac{3}{4}) = \dfrac{\pi \sqrt{2}}{\Gamma (\frac{1}{4})}$, a result that comes immediately from Euler's reflexion formula for the Gamma function.

If it is the indefinite integral you are really after, then as suggested in the comments, your integral can be expressed in terms of elliptic functions of the first and second kinds.
From ($*$), namely
$$I = \frac{1}{2} \int \frac{du}{\sqrt{1 - u^2} \sqrt{u}} - \frac{1}{2} \int \frac{\sqrt{u}}{\sqrt{1 - u^2}} \, du,$$
where $u = x^2$, substituting $u = \sin \theta$ into each of the above integrals one arrives at
\begin{align}
I &= \frac{1}{2} \int \sqrt{\operatorname{cosec} x} \, d\theta - \frac{1}{2} \int \sqrt{\sin \theta} \, d\theta\\
&= - F \left (\frac{\pi - 2 \theta}{4} \Big{|} 2 \right ) + E \left (\frac{\pi - 2 \theta}{4} \Big{|} 2 \right ) + C\\
&= E \left (\frac{\pi - 2 \sin^{-1} (x^2)}{4} \Big{|} 2 \right ) - F \left (\frac{\pi - 2 \sin^{-1} (x^2)}{4} \Big{|} 2 \right ) + C, \qquad -1 \leqslant x \leqslant 1.
\end{align}
Here $F( \phi | k^2)$ is an elliptic integral of the first kind while $E(\phi | k^2)$ is an elliptic integral of the second kind.
A: $$I=\int \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}}\, dx$$ Let $x=\sinh(t)$ which makes
$$I=\int \sqrt{1-\sinh ^2(t)}\,dt$$ Now, use the fact that $\sin(iz)=i\sinh(z)$; in other words, make $t=i u$ to make
$$I=i \int  \sqrt{1+\sin ^2(u)}\,du$$ Have a look here.
Edit
May be, as I wrote in a comment, you could consider a series expansion of the integrand
$$\frac{\sqrt{1-x^2}}{\sqrt{1+x^2}}=\sum_{n=0}^\infty a_n\, x^{2n}$$ in which yhe $a_n$ are defined by
$$a_{n+4}=-\frac{2a_{n+2}-n\,a_n}{n} \qquad \text{with} \qquad a_0=1,\,\,a_1=0,\,\,a_2=-1,\,\,a_3=0$$ making all odd coeffcients equal to $0$. Starting for $n=0$, the sequence for even values of $n$ is
$$\left\{1,-1,\frac{1}{2},-\frac{1}{2},\frac{3}{8},-\frac{3}{8},\frac{5}{16},-\frac{5
   }{16},\frac{35}{128},-\frac{35}{128},\frac{63}{256},-\frac{63}{256},\frac{231}{1
   024},-\frac{231}{1024},\frac{429}{2048},-\frac{429}{2048}\right\}$$
Now, integrate termwise to get, as a truncated series,
$$I=x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{14}+\frac{x^9}{24}-\frac{3
   x^{11}}{88}+\frac{5 x^{13}}{208}-\frac{x^{15}}{48}+\frac{35
   x^{17}}{2176}-\frac{35 x^{19}}{2432}+\frac{3 x^{21}}{256}+O\left(x^{23}\right)$$
