Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$

Question

To solve the integral $$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$ I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change $$ I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t} $$ but if it were correct, how could I go on?

Answer 1

$$\int\frac{dx}{2+x+\sqrt{1-x^2}}$$ Substitute $x= \sin 2u;\;dx=2\cos 2u$

$$\int \frac{2\cos (2 u)}{2+\sin (2 u)+\cos (2 u)} \, du$$ $$\int \frac{2\cos ^2 u-2\sin ^2 u}{2-\sin ^2 u+\cos ^2 u+2 \sin u \cos u}\,du$$ $$\int \frac{2\cos ^2 u-2\sin ^2 u}{\sin ^2 u+3 \cos ^2 u+2 \sin u \cos u}\,du$$ divide numerator and denominator by $\cos^2 u$ $$\int \frac{2-2\tan ^2 u}{\tan ^2 u +2 \tan u +3}\,du$$ substitute $\tan u = t\to dt=\frac{du}{1+u^2}$ $$\int \frac{2-2t^2}{\left(t^2+1\right) \left(t^2+2 t+3\right)}\,dt$$ using partial fraction $$\int \left(\frac{1-t}{t^2+1}+\frac{t-1}{t^2+2 t+3}\right)\,dt$$ $$\frac{1}{2} \left(-\log \left(t^2+1\right)+\log \left(t^2+2 t+3\right)+2 \arctan t-2 \sqrt{2} \arctan\left(\frac{t+1}{\sqrt{2}}\right)\right)+C$$ $t=\tan u$ and $x=\sin 2u$ we have $t=\frac{1-\sqrt{1-x^2}}{x}$ $$\frac{1}{2} \log \left(\sqrt{1-x^2}+x+2\right)+\arctan\left(\frac{1-\sqrt{1-x^2}}{x}\right)-\sqrt{2} \arctan\left(\frac{-\sqrt{1-x^2}+x+1}{\sqrt{2} x}\right)+C$$

Answer 2

Bioche's rules say you should set $\;u=\tan \frac t2,\enspace \mathrm du=\frac12(1+u^2)\,\mathrm dt$, whence the integral of a rational function $$\int\frac{2(1-u^2)\,\mathrm du}{(1+u^2)(u^2+2u+3)},$$ which is easily calculated using partial fractions decomposition.

Some more details: using the half angle formulæ, we get \begin{align} \frac{\cos t}{2 + \sin t + \cos t}\,\mathrm dt&=\frac{\cfrac{1-u^2}{1+u^2}}{2+\cfrac{2u}{1+u^2}+\cfrac{1-u^2}{1+u^2}}\,\frac{2\,\mathrm du}{1+u^2} \\ &=\frac{2(1-u^2)}{(1+u^2)\bigl(2(1+u^2)+2u+1-u^2\bigr)} \\ &=\frac{2(1-u^2)}{(1+u^2)\bigl(u^2+2u+3\bigr)}. \\ \end{align}

Answer 3

Mathematica gives:

$$\frac{1}{8} \left(4 \sin ^{-1}(x)+4 \sqrt{2} \tan ^{-1}\left(\sqrt{2} (x+1)\right)+\frac{2 i \left(i+2 \sqrt{2}\right) \tan ^{-1}\left(\frac{\left(-16+6 i \sqrt{2}\right) x^4-2 i \left(9 \sqrt{1-2 i \sqrt{2}} \sqrt{1-x^2}+10 \left(-3 i+\sqrt{2}\right)\right) x^3-i \left(36 \sqrt{1-2 i \sqrt{2}} \sqrt{1-x^2}+79 \sqrt{2}-20 i\right) x^2+3 \left(-9 i \sqrt{1-2 i \sqrt{2}} \sqrt{1-x^2}-20 i \sqrt{2}+20\right) x-9 i \sqrt{2}+36}{\left(66 i+32 \sqrt{2}\right) x^4+16 \left(14 i+3 \sqrt{2}\right) x^3+\left(253 i-14 \sqrt{2}\right) x^2-48 \left(-2 i+\sqrt{2}\right) x-18 \sqrt{2}+9 i}\right)}{\sqrt{1-2 i \sqrt{2}}}+2 \sqrt{1+2 i \sqrt{2}} \tan ^{-1}\left(\frac{2 \left(8+3 i \sqrt{2}\right) x^4+2 \left(-9 i \sqrt{1+2 i \sqrt{2}} \sqrt{1-x^2}-10 i \sqrt{2}+30\right) x^3+\left(-36 i \sqrt{1+2 i \sqrt{2}} \sqrt{1-x^2}-79 i \sqrt{2}+20\right) x^2-3 i \left(9 \sqrt{1+2 i \sqrt{2}} \sqrt{1-x^2}+20 \left(-i+\sqrt{2}\right)\right) x-9 i \sqrt{2}-36}{\left(-66 i+32 \sqrt{2}\right) x^4+16 \left(-14 i+3 \sqrt{2}\right) x^3-\left(253 i+14 \sqrt{2}\right) x^2-48 \left(2 i+\sqrt{2}\right) x-9 \left(i+2 \sqrt{2}\right)}\right)+2 \log \left(2 x^2+4 x+3\right)-\frac{\left(i+2 \sqrt{2}\right) \log \left(4 \left(2 x^2+4 x+3\right)^2\right)}{\sqrt{1-2 i \sqrt{2}}}-\frac{\left(-i+2 \sqrt{2}\right) \log \left(4 \left(2 x^2+4 x+3\right)^2\right)}{\sqrt{1+2 i \sqrt{2}}}+\frac{\left(i+2 \sqrt{2}\right) \log \left(\left(2 x^2+4 x+3\right) \left(-2 \left(-i+\sqrt{2}\right) x^2+2 i \left(\sqrt{2-4 i \sqrt{2}} \sqrt{1-x^2}+2\right) x+2 i \sqrt{2-4 i \sqrt{2}} \sqrt{1-x^2}+2 \sqrt{2}+3 i\right)\right)}{\sqrt{1-2 i \sqrt{2}}}+\frac{\left(-i+2 \sqrt{2}\right) \log \left(-\left(2 x^2+4 x+3\right) \left(2 \left(i+\sqrt{2}\right) x^2+2 i \left(\sqrt{2+4 i \sqrt{2}} \sqrt{1-x^2}+2\right) x+2 i \sqrt{2+4 i \sqrt{2}} \sqrt{1-x^2}-2 \sqrt{2}+3 i\right)\right)}{\sqrt{1+2 i \sqrt{2}}}\right)$$

which strongly suggests no clean simplification.

Answer 4

Continue with $x=\sin t$

\begin{align} &\int \frac{1}{2 + x + \sqrt{1 - x^2}} dx=\int \frac{\cos t}{2 + \sin t + \cos t}dt\\ = &\ \frac12\int \left(1+ \frac{\cos t-\sin t}{2 + \sin t + \cos t}- \frac{2}{2 + \sin t + \cos t} \right)dt\\ =& \ \frac12t +\frac12 \ln(2 + \sin t + \cos t) -\sqrt2\tan^{-1} \frac{\tan\frac t2+1}{\sqrt2}\\ =&\ \frac12\sin^{-1}x+\frac12 \ln\left(2 + x+ \sqrt{1-x^2}\right) -\sqrt2\tan^{-1} \frac{1+x-\sqrt{1-x^2}}{x \sqrt2} \end{align}

Answer 5

Use Euler substitution $$tx-1=\sqrt{1-x^2}$$ $$t=\frac{1+\sqrt{1-x^2}}{x}$$ $$x=\frac{2t}{t^2+1}$$ $$dx=\frac{2(1-t^2)}{(t^2+1)^2}dt$$ $$\int \frac{dx}{2+x+\sqrt{1-x^2}}=\int\frac{1}{2+(t+1)\frac{2t}{t^2+1}-1}\frac{2(1-t^2)}{(t^2+1)^2}dt=\int\frac{2(1-t^2)dt}{(3t^2+2t+1)(t^2+1)}$$$$=-\int\frac{t+1}{t^2+1}dt+\frac{1}{2}\int\frac{6t+2}{3t^2+2t+1}dt+\frac{2}{3}\int\frac{dt}{(t+\frac{1}{3})^2+\frac{2}{9}}$$$$=-\frac{1}{2}\ln(t^2+1)-\arctan(t)+\frac{1}{2}\ln(3t^2+2t+1)+\frac{2}{3}\frac{1}{\frac{2}{9}}\arctan\left(\frac{t+\frac{1}{3}}{\frac{2}{9}}\right)+C$$ $$=\frac{1}{2}\ln\left(\frac{3t^2+2t+1}{t^2+1}\right)-\arctan(t)+3\arctan\left(\frac{9t+3}{2}\right)+C$$ where $t=\frac{1+\sqrt{1-x^2}}{x}$

Answer 6

Using @Bernard solution $$y=\frac{2(1-u^2)}{(1+u^2)\bigl(u^2+2u+3\bigr)}$$ using partial fractions $$y=-\frac{1+i}{2(u-i)}-\frac{1-i}{2 (u+ i)}+\frac{\sqrt{2}+2 i}{2 \left(\sqrt{2} u+\sqrt{2}-2 i\right)}+\frac{\sqrt{2}-2 i}{2 \left(\sqrt{2} u+\sqrt{2}+2 i\right)}$$ $$\int y\,du=\left(-\frac{1}{2}-\frac{i}{2}\right) \log (-2 u+2 i)\left(-\frac{1}{2}+\frac{i}{2}\right) \log (2 u+2 i)+$$ $$\frac{\left(\sqrt{2}+2 i\right) \log \left(\sqrt{2} u+\sqrt{2}-2 i\right)}{2 \sqrt{2}}+\frac{\left(\sqrt{2}-2 i\right) \log \left(\sqrt{2} u+\sqrt{2}+2 i\right)}{2 \sqrt{2}}$$

Now, recombining everything $$\int y\,du=\tan ^{-1}(u)-\sqrt{2} \tan ^{-1}\left(\frac{u+1}{\sqrt{2}}\right)+\frac 12 \log \left(\frac{u^2+2 u+3}{u^2+1}\right)$$