An interesting indefinite integral Here's an interesting integral that I'm struggling with.
$$ \int \frac{x\ dx}{\sqrt{x^4 + 4x^3 - 6x^2 + 4x + 1}} $$
MY EFFORTS:
I'm very very close in reaching a closed form.
\begin{align}
&= \frac12\int \frac{2\ dx}{\sqrt{\left(x^2+\dfrac{1}{x^2}\right) + 4\left(x+\dfrac{1}{x}\right)-6}} \\
&= \frac12\int \frac{\left(1 - \dfrac{1}{x^2}\right)\ dx}{\sqrt{\left(x+\dfrac{1}{x}\right)^2 + 4\left(x+\dfrac{1}{x}\right)-8}} + \frac12\int \frac{\left(1 + \dfrac{1}{x^2}\right)\ dx}{\sqrt{\left(x-\dfrac{1}{x}\right)^2 + 4\sqrt{\left(x-\dfrac{1}{x}\right)^2+4}-4}} \\
&= \frac12 \int \frac{du}{\sqrt{u^2+4u-8}} + \underbrace{\frac12 \int \frac{dt}{\sqrt{t^2 + 4\sqrt{t^2+4}-4}}}_{\text{How to solve this?}}
\end{align}
As you all can see, I'm able to break integral by 2 parts and I'm struck in evaluating 2nd part. The first part can easily be solved by completing the square and trigonometric substitutions.
NOTE: In my effort pic. the variable of integration is x and should not be confused with letter n.
([1]: https://i.stack.imgur.com/k2Vhu.jpg)
 A: I don't think that your effort leads to a simpler integral that the original one. They are two radicals in your last integral. This makes it more difficult to integrate. So, you are not close to solve it.
HINT :
$$x^4+4x^3-6x^2+x^4+1= (x-r_1)(x-r_2)(x-r_3)(x-r_4)$$
$r_1=\frac12\left(1+\sqrt{2}+\sqrt{7-4\sqrt{2}} \right)$
$r_2=\frac12\left(1-\sqrt{2}+\sqrt{7-4\sqrt{2}} \right)$
$r_3=\frac12\left(1+\sqrt{2}-\sqrt{7-4\sqrt{2}} \right)$
$r_4=\frac12\left(1-\sqrt{2}-\sqrt{7-4\sqrt{2}} \right)$
$$I= \int \frac{x\ dx}{\sqrt{x^4 + 4x^3 - 6x^2 + 4x + 1}} $$
It is easier to integrate on this form :
$$I= \int \frac{x\ dx}{\sqrt{(x-r_1)(x-r_2)(x-r_3)(x-r_4)}} $$
The result involves an elliptic integral as a term of a complicated formula. From WolframAlpha :
https://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt((x-r_1)(x-r_2)(x-r_3)(x-r_4))
NOTE :
If the integral comes from an academic exercise, there is probably a typo in it, because solving it requires an hight level of knowledge of special functions. I will not loose time with a problem suspected of mistake in the wording.
Please check your calculus which leads to the integral. Eventually edit the original version of the problem in your question.
