What is the primitive of $\tan(x)\arctan(x)$? Can you help me find the primitive of $$\tan(x)\arctan(x)$$
Thanks in advance :)
 A: I am quite sure (and Mathematica supports my intuition) that there is no primitive in terms of standard functions. $\\ $
However, on any disk in the complex plane centred around the origin with radius < 1,  each of the two factors is a holomorphic function and can be represented by their Taylor expansion. (c.f. Wikipedia, Taylor series for the individual coefficients and the radii of convergence). Because, within their radius of convergence, power series are absolutely convergent, the product of the two Taylor series is equal to their Cauchy product:
$\tan(x)\times \arctan(x) = \left( \sum\limits_{n=0}^\infty \underbrace{\frac{B_{2n}(-4)^n(1-4^n)}{(2n)!}}_{\equiv a_n} x^n\right)\times \left( \sum\limits_{m=0}^\infty \underbrace{\frac{(-1)^n}{2n+1}}_{\equiv b_m} x^m\right) =\sum\limits_{j=0}^\infty \left( \sum\limits_{k=0}^{\infty}a_k b_{j-k} \right)x^j$.
This can be integrated term by term (uniform convergence of power series) to yield:
$\int \tan(x)\times \arctan(x) dx= \sum\limits_{j=0}^\infty \frac{ \sum\limits_{k=0}^{\infty}a_k b_{j-k} }{j+1}x^{j+1}+const.$.
This is a series representation of the primitive that is valid for all $x, |x|<1$.
