How to integrate $\int (\sqrt{x^{3}+6x^{2}+11x+6})dx$?

Question

How to integrate $$\int (\sqrt{x^{3}+6x^{2}+11x+6})dx$$ ? I am trying to evaluate this integral by factorizing $x^{3}+6x^{2}+11x+6$. Therefore, we can write that $x^{3}+6x^{2}+11x+6=(x+1)(x+2)(x+3)$. Therefore the above integral will become $$\int (\sqrt{(x+1)(x+2)(x+3)})dx$$. Now here if we substitute $(x+1)=t$, then the above integral will become $$\int (\sqrt{t(t+1)(t+2)})dt$$. Now again here I am thinking of substituting $(t+1)=u$. Therefore after this step we can write the above integral as $$\int (\sqrt{(u-1)u(u+1)})du$$. Therefore we can write this integral as $$\int (\sqrt{u(u^{2}-1)})du$$. Now here I am thinking of substituting $u=\sec(\theta)$. Therefore, $du=\sec(\theta)\tan(\theta)d\theta$. Therefore the above integral will become $$\int (\sec(\theta))^{\frac{3}{2}}|\tan(\theta)|(\tan(\theta))d\theta$$. Now here $3$ cases will arise. $1^{st}$ case- $|\tan(\theta)|=\tan(\theta)$[when $\tan(\theta)>0$] , $2^{nd}$ case- $|\tan(\theta)|=-\tan(\theta)$[when $\tan(\theta)<0$] , $3^{rd}$ case- $|\tan(\theta)|=0$[when $\tan(\theta)=0$]. Therefore for the $1^{st}$ case the above integral will become $$\int \frac{\sin^{2}(\theta)}{(\cos(\theta))^{\frac{7}{2}}}d\theta$$. For the $2^{nd}$ case the above integral will become $$-\int \frac{\sin^{2}(\theta)}{(\cos(\theta))^{\frac{7}{2}}}d\theta$$. And lastly for the $3^{rd}$ case the result of the above integral will be $0$. Now I am facing problem to integrate the $1^{st}$ and the $2^{nd}$ cases. Please help me out with this integral.

Answer 1

If you do not want (or cannot) use hypergeometric functions or elliptic integrals, the simplest would probably to use the series expansion $$\sqrt{(x+1)(x+2)(x+3)}=\sum_{n=0}^\infty a_n\, x^n$$ where $$a_n=-\frac{11 (2 n-3)}{12 n}\,a_{n-1}-\frac{n-3}{n}\,a_{n-2}-\frac{2 n-9}{12 n}\,a_{n-3}$$ where $$a_0=\sqrt{6} \qquad \qquad a_1=\frac{11}{2 \sqrt{6}} \qquad \qquad a_2=\frac{23}{48 \sqrt{6}}$$ and integrate termwise.

The problem would probably be a quite slow convergence.

Edit

In terms of series, you could let $(x+2)=t$ and write $$\sqrt{(t-1)t(t+1)}=\sum_{n=0}^\infty b_n\, t^n$$ where $$b_n=\frac{(n-3)}{n\,t^2} b_{n-2}\,\qquad b_0=t^{3/2}\,\qquad b_1=0$$

If you use the hypergeometric function $$I=\int \sqrt{(t-1)t(t+1)}\,dt=\frac{2 t \sqrt{t \left(t^2-1\right)}}{3 \sqrt{1-t^2}} \, _2F_1\left(-\frac{1}{2},\frac{3}{4};\frac{7}{4};t^2\right)$$ $$I=-\frac{t \sqrt{t \left(t^2-1\right)}}{\sqrt{\pi } \sqrt{1-t^2}}\sum_{n=0}^\infty \frac{\Gamma \left(n-\frac{1}{2}\right)}{(4 n+3) \Gamma (n+1)}\, t^{2n}$$