How to evaluate $\int \frac{1}{\sqrt{x^{4}+3x^{2}+2}}dx$?

Question

How to integrate $$\int \frac{1}{\sqrt{x^{4}+3x^{2}+2}}dx$$ ? I am trying to evaluate this integral by factorizing $x^{4}+3x^{2}+2$ . Now we can write $x^{4}+3x^{2}+2=(x^{2}+1)(x^{2}+2)$ . So, we can write the above integral as $$\int \frac{1}{\sqrt{(x^{2}+1)(x^{2}+2)}}dx$$ . Now we can write $1$ as $(x^{2}+2)-(x^{2}+1)$ . So, the next step of the integral will become $$\int \sqrt{\frac{x^{2}+2}{x^{2}+1}}dx$$ $-$ $$\int \sqrt{\frac{x^{2}+1}{x^{2}+2}}dx$$ . But after this step, the integral is looking messy. But for integrating $$\int \sqrt{\frac{x^{2}+2}{x^{2}+1}}dx$$ , I am thinking of substituting $x$ $=$ $tan\theta$ . So, $dx=(sec^{2}\theta) d\theta$ . Now we can write $$\int \sqrt{\frac{x^{2}+2}{x^{2}+1}}dx$$ $=$ $$\int \sqrt{2+tan^{2}\theta}×sec\theta×d\theta$$ . But now after this process I can't integrate further. Also, I can't integrate $$\int \sqrt{\frac{x^{2}+1}{x^{2}+2}}dx$$ . Please help me out.

Answer 1

$$I:= \int \frac{1}{\sqrt{x^4+3x^2 +1}}dx=\int \frac{1}{\sqrt{(x^2+1)^2+(x^2+1)}}dx$$ let $x= \tan(t)$ $$I=\int \frac{\sec^2(t)}{\sqrt{\sec^4(t)+\sec^2(t)}}dt=\int \frac{\sec(t)}{\sqrt{\sec^2(t)+1}}dt= \int \frac{1}{\sqrt{\cos^2(t)+1}}dt =\int \frac{1}{\sqrt{2-\sin^2(x)}}dt= \frac{1}{\sqrt 2}\int \frac{1}{\sqrt{1-\frac{1}{2}\sin^2(x)}}dt$$ this is an incomplete elliptical integral of the first kind $ \ \ F(t; k) := \int_0^t \frac{du}{\sqrt{1 - k^2 \sin^2 u}}$ $$I=\frac{1}{\sqrt 2} F\left( t ;\frac{1}{\sqrt 2} \right)+C = \frac{1}{\sqrt 2} F\left( \arctan(x) ;\frac{1}{\sqrt 2} \right)+C$$