Does $\int x\sqrt{x-1}\arctan(x)\, \text{d}x$ have a closed form?

Question

I misread an integral on my school maths book coming up with this indefinite integral $$\int x \sqrt{x-1}\arctan x\, \text{d}x$$ that revealed to be quite nasty. The first thing I thought has been to substitute $x-1=t^2$ and go by parts deriving $\arctan x$ obtaining $$\frac2{15}\left((3t^5+5t^3)\arctan (t^2+1)-2\int\frac{3t^6+5t^4}{t^4+2t^2+2}\, \text{d}t \right)$$ The problem now is that rational integral that looks even worse than the original one. In fact, throught sinthtic division the integrand becomes $3t^2-1-2\frac{2t^2-1}{t^4+2t^2+2}$ that still keeps the quartic denominator with no chance to express the numerator as the derivative or to apply add-and-subtract something… Then, I think that my initial substitution isn't the best but also trying with $x=\tan u$ I have no great results as well as with something more. Any idea?

Answer 1

Hint. In the last rational function, the denominator is quadratic in $t^2.$ So you can factor into two quadratics and express the fraction in parts. This would require manipulating complex entities, but the procedures are the same.

Another way may be to try a substitution of the form $$x=at+\frac bt,$$ where $a$ and $b$ are suitable constants.