It's a kind of follow up of Show that : $\frac{1}{\zeta(3)}<2C-1$:
Does the following integral have a closed form or a nice representation (series,equation,integral...)?:
$$\int_{0}^{1}\left(\left(\frac{2\arctan\left(x\right)}{x}-\frac{3}{2}x^{\frac{1}{2}}\right)\left(\left(\frac{6}{5}-x^{\frac{4}{5}}-\frac{1}{10}\right)^{\frac{6}{5}}\right)\right)^{\frac{1}{2}}dx=I=?$$
Background :
This integral comes from an attempt to use Cauchy-Schwarz to evaluate the product in the linked question above
Motivation :
It would be benefic to understand the problem with Catalan's constant and $\zeta(3)$ in evaluating this integral above knowing we can use Callebaut's inequality (in continuous form) to show the inequality .
Attempt :
$I$ have some similarities with the integral :
$$\int_{0}^{1}\left(\left(\frac{2\tanh\left(x\right)}{x}-\frac{3}{2}x^{\frac{1}{2}}\right)\left(\left(\frac{6}{5}-x^{\frac{4}{5}}-\frac{1}{10}\right)^{\frac{6}{5}}\right)\right)^{\frac{1}{2}}dx$$
So using the series of hyperbolic tangent we have :
$$\int_{0}^{1}\left(\left(\frac{2\left(x-\frac{x^{3}}{3}+\frac{2}{15}x^{5}\right)}{x}-\frac{3}{2}x^{\frac{1}{2}}\right)\left(\left(\frac{6}{5}-x^{\frac{4}{5}}-\frac{1}{10}\right)^{\frac{6}{5}}\right)\right)^{\frac{1}{2}}dx\simeq I$$
As you can see It's already a monster as integral and I cannot find an issue .
Question :
What is $I$ ? And if there is no closed form what would be interesting to say about it ?