Another integral $\int \frac{3 x^2+2 x+1}{ \left(x^3+x^2+x+2\right) \sqrt{1+\sqrt{x^3+x^2+x+2}}} \, dx$

Question

Here is an indefinite integral that is similar to an integral I wanna propose for a contest. Apart from using CAS, do you see any very easy way of calculating it?

$$\int \frac{1+2x +3 x^2}{\left(2+x+x^2+x^3\right) \sqrt{1+\sqrt{2+x+x^2+x^3}}} \, dx$$

EDIT: It's a part from the generalization

$$\int \frac{1+2x +3 x^2+\cdots n x^{n-1}}{\left(2+x+x^2+\cdots+ x^n\right) \sqrt{1\pm\sqrt{2+x+x^2+\cdots +x^n}}} \, dx$$

Supplementary question: How would you calculate the following integral using the generalization above? Would you prefer another way?

$$\int_0^{1/2} \frac{1}{\left(x^2-3 x+2\right)\sqrt{\sqrt{\frac{x-2}{x-1}}+1} } \, dx$$

As a note, the generalization like the one you see above and slightly modified versions can be wisely used for calculating very hard integrals.

Answer 1

HINT:

As $\dfrac{d(2+x+x^2+x^3)}{dx}=1+2x+3x^2,$

let $\sqrt{1+\sqrt{2+x+x^2+x^3}}=u\implies 2+x+x^2+x^3=(u^2-1)^2$

and $(1+2x+3x^2)dx=(u^2-1)2u\ du$

Now use Partial Fraction Decomposition,

$\dfrac1{(u^2-1)^2}=\dfrac A{u-1}+\dfrac B{(u-1)^2}+\dfrac C{u+1}+\dfrac D{(u+1)^2}$

Answer 2

Let $$2+x+x^2+x^3=t^2\implies (1+2x+3x^2)dx=2tdt$$ $$\int \frac{2tdt}{t^2\sqrt{1+t}}$$ $$=2\int \frac{dt}{t\sqrt{1+t}}$$ Let $1+t=x^2\implies dt=2xdx$ $$=2\int \frac{2xdx}{(x^2-1)x}$$ $$=4\int \frac{dx}{x^2-1}$$ $$=4\int \frac{dx}{(x-1)(x+1)}$$ $$=4\frac{1}{2}\int \left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx$$ $$=4\frac{1}{2}\left(\int\frac{dx}{x-1}-\int\frac{dx}{x+1}\right)$$ $$=2\left(\ln|x-1|-\ln|x+1|\right)+c$$

$$=2\ln\left|\frac{x-1}{x+1}\right|+c$$ $$=2\ln\left|\frac{\sqrt{1+t}-1}{\sqrt{1+t}+1}\right|+c$$ $$=2\ln\left|\frac{\sqrt{1+\sqrt{1+2+x+x^2+x^3}}-1}{\sqrt{1+\sqrt{1+2+x+x^2+x^3}}+1}\right|+c$$

Answer 3

Let $\sinh^4(t)=x^3+x^2+x+2$. Then $\cosh(t)=\sqrt{1+\sqrt{x^3+x^2+x+2}}$ and $$ \begin{align} &\int\frac{3x^2+2x+1}{\left(x^3+x^2+x+2\right)\sqrt{1+\sqrt{x^3+x^2+x+2}}} \,\mathrm{d}x\\ &=\int\frac{\mathrm{d}\sinh^4(t)}{\sinh^4(t)\cosh(t)}\\ &=4\int\frac{\mathrm{d}t}{\sinh(t)}\\ &=4\int\frac{\mathrm{d}\cosh(t)}{\cosh^2(t)-1}\\ &=2\int\left(\frac1{\cosh(t)-1}-\frac1{\cosh(t)+1}\right)\mathrm{d}\cosh(t)\\[2pt] &=2\log\left(\frac{\cosh(t)-1}{\cosh(t)+1}\right)+C\\[6pt] &=-4\operatorname{arccoth}(\cosh(t))+C\\[6pt] &=-4\operatorname{arccoth}\left(\sqrt{1+\sqrt{x^3+x^2+x+2}}\right)+C \end{align} $$