Does this integral have a closed-form solution?
$$\int\limits_{0}^{\infty}\dfrac{x^4\left(2x^{22}+\left(11\sqrt{5}+11\right)x^{17}-20\right)}{2\left(x^{22}+1\right)\sqrt{4\left(\sqrt{x^{22}+1}+x^{10}+(\sqrt{5}+1)x^{5}\right )+6+2\sqrt{5}}}dx$$
EDIT
Sorry for the lack of context, I was late for a compromise. I saw this integral in my university, in the notes of a professor. I asked him what it was, and as response I got "integrals that computers do not solve."
I plotted the integrand chart on the desmos site. The result was this:
Clearly, the integral is convergent and probably has a value assigned to it.
After much effort, I asked my teacher to solve this. Here's his solution:
You need to notice that $$4(\sqrt{x^{22}+1}+x^{10}+(\sqrt{5}+1)x^{5})+6+2\sqrt{5}=4\left (\sqrt{x^{22}+1}+x^{10}+(\sqrt{5}+1)x^{5}+\frac{6+2\sqrt{5}}{4} \right )=4\left (\sqrt{x^{22}+1}+(x^{5}+\varphi)^{2} \right )$$
where $\varphi=\frac{\sqrt{5}+1}{2}$ is the golden ratio. Now the integrand assumes the following form:
$$\int\limits_{0}^{\infty}\dfrac{x^4\left(2x^{22}+\left(11\sqrt{5}+11\right)x^{17}-20\right)}{2\left(x^{22}+1\right)\sqrt{4\left(\sqrt{x^{22}+1}+(x^{5}+\varphi)^{2}\right )}}dx=\int\limits_{0}^{\infty}\dfrac{x^4\left(2x^{22}+\left(11\sqrt{5}+11\right)x^{17}-20\right)}{4\left(x^{22}+1\right)^{\frac{5}{4}}\sqrt{1+\left(\frac{x^{5}+\varphi}{\sqrt[4]{x^{22}+1}} \right)^{2}}}dx$$
Next, subs. $u=\frac{x^{5}+\varphi}{\sqrt[4]{x^{22}+1}}$, therefore
$$du=x^{4}\left[5(x^{22}+1)^{-\frac{1}{4}}-\frac{11}{2}x^{17}(x^{5}+\varphi)(x^{22}+1)^{-\frac{5}{4}} \right]dx$$
After some algebra, the integral will take the following - very simple - form:
$$\int_{0}^{\varphi}\frac{du}{\sqrt{1+u^{2}}}=\boxed{\textrm{arcsinh}(\varphi)}$$