How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as $x^2+\left(1-\frac{1}{3}+\frac{1}{5}\right)\frac{x^6}{3}...$ How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as
$x^2+\left(1-\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{13}\right)\dfrac{x^{14}}{7}+...$
I have tried to use Cauchy product or long multiplication, but the result, although of course, are equivalent, is not as elegantly expressed as this expression. This is taken from Ferrar "A textbook of convergence" page 123.
For example, for $x^6$, I have $\dfrac{2}{5}-\dfrac{1}{9}$, while for $x^{10}$, I have $\dfrac{2}{9}+\dfrac{1}{25}-\dfrac{2}{21}$. I don't know how to make it pop out the partial Leibnitz series for the coefficients.

Hint:
$f(x)=[\tan^{-1}(x)][\tanh^{-1}(x)]$
$f'(x)=\dfrac{ \tan^{-1}(x)}{1-x^2}+\dfrac{ \tanh^{-1}(x)}{1+x^2} $
If $|x|<1$:
$\displaystyle f'(x)=\left(\sum_{n=0}^{+\infty}  \dfrac{(-1)^n}{2n+1} x^{2n+1}   \right) \left(\sum_{n=0}^{+\infty}   x^{2n}  \right)   +  \left(\sum_{n=0}^{+\infty}  \dfrac{1}{2n+1} x^{2n+1}   \right) \left(\sum_{n=0}^{+\infty} (-1)^n  x^{2n}  \right)    $
Edit to show the idea
$f′(x)=\left(x−\dfrac{x^3}{3}+\dfrac{x^5}{5}−\dfrac{x^7}{7}+\dfrac{x^9}{9}...\right)(1+x^2+x^4+x^6+x^8...)+\left(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+\dfrac{x^7}{7}+\dfrac{x^9}{9}...\right)(1−x^2+x^4−x^6+x^8...) $
By using Cauchy product:
$f′(x)=2x+ 2\left(1−\dfrac{1}{3}+\dfrac{1}{5}\right)x^5+2\left(1 −\dfrac{1}{3}+\dfrac{1}{5}−\dfrac{1}{7}+\dfrac{1}{9}\right)x^9+...$
Then we integrate:
$f(x)=x^2+\left(1−\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3} +\left(1 −\dfrac{1}{3}+\dfrac{1}{5}−\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5} +...  $
