How to evaluate $\int_0^{\frac{\pi}{2}}x^3(\sqrt{\tan x}-\sqrt{\cot x}) d x$?

Question

The substitution $x\mapsto \frac{\pi}{2}-x$ is always helpful for evaluating integral consisting of polynomials and trigonometric function. $$I_0=\int_0^{\frac{\pi}{2}}(\sqrt{\tan x}-\sqrt{\cot x}) d x = \int_0^{\frac{\pi}{2}}(\sqrt{\cot x}-\sqrt{\tan x}) d x \Rightarrow I_0 =0$$ However, when I multiply the integrand by $x$ and use the same substitution, I fails to evaluate $$I_1= \int_0^{\frac{\pi}{2}}x(\sqrt{\tan x}-\sqrt{\cot x}) d x$$ since $$ \begin{aligned} I_1 & =\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right)(\sqrt{\cot x}-\sqrt{\tan x}) d x \\ & =\frac{\pi}{2} \int_0^{\frac{\pi}{2}}(\sqrt{\cot x}-\sqrt{\tan x}) d x+I_1\end{aligned} $$ in which $I_1$ on both sides cancel each other.

Then I started to investigate $I_1$ by letting $t=\tan x$ and get $$ \begin{aligned} I_1& =\int_0^{\infty} \tan ^{-1} t\left(\sqrt{t}-\frac{1}{\sqrt{t}}\right) \frac{d t}{1+t^2} \\ & =\int_0^{\infty} \frac{t-1}{\sqrt{t}\left(1+t^2\right)} \tan ^{-1} t d t \\ & =2 \int_0^{\infty} \frac{t^2-1}{1+t^4} \tan ^{-1} (t^2) d t \text {, where } t \mapsto t^2\\&=2J(1) \end{aligned} $$ where $J(a)= \int_0^{\infty} \frac{t^2-1}{1+t^4} \tan ^{-1} (at^2) d t .$

Differentiating $J(a)$ w.r.t. $a$ yields $$ \begin{aligned} J^{\prime}(a)&=\int_0^{\infty} \frac{t^2\left(t^2-1\right)}{\left(1+t^4\right)\left(1+a^2 t^4\right)} d t \\ & = \frac{1}{a^2-1}\left[\int_0^{\infty} \frac{t^2+1}{1+t^4} d t-\int_0^{\infty} \frac{1+a^2 t^2}{1+a^2 t^4} d t\right]\\ & =\frac{1}{a^2-1}\left[\frac{\pi}{\sqrt{2}}-\frac{\pi(a+1)}{2\sqrt{2a}}\right] \\ & =\frac{\pi}{\sqrt{2}\left(a^2-1\right)}\left(1-\frac{a+1}{2 \sqrt{a}}\right) \end{aligned} $$ Integrating back yields $$ \begin{aligned} J(1) & =\frac{\pi}{\sqrt{2}} \int_0^1 \frac{1}{a^2-1}\left(1-\frac{a+1}{2 \sqrt{a}}\right) d a \\ & =\frac{\pi\ln 2}{2\sqrt 2} \end{aligned} $$ Hence $$\boxed{I_1=\frac{\pi\ln 2}{2} }$$ Putting $x\mapsto \frac{\pi}{2}-x$ again to $I_2$ gives $$ \begin{aligned} I_2 & =\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right)^2(\sqrt{\cot x}-\sqrt{\tan x}) d x \\ &= \frac{\pi^2}{4} \int_0^{\frac{\pi}{2}}(\sqrt{\cot x}-\sqrt{\tan x}) d x-\pi \int_0^{\frac{\pi}{2}} x(\sqrt{\cot x} -\sqrt{\tan x}) -I_2 \\ \end{aligned} $$ Rearranging gives $$\boxed{I_2 =\frac{\pi}{2} \int_0^{\frac{\pi}{2}} x(\sqrt{\tan x}-\sqrt{\cot x}) d x =\frac{\pi}{2} \cdot \frac{\pi \ln 2}{\sqrt{2}}=\frac{\pi^2 \ln 2}{2 \sqrt{2}}}$$

To continue, I am stuck by the cancelling of $I_3$ in the equation $$ \begin{aligned} I_3 & =\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right)^3(\sqrt{\cot x}-\sqrt{\tan x}) d x \\ & =\frac{3 \pi^2}{4} I_1-\frac{3 \pi}{2} I_2+I_3 \end{aligned} $$ which reveals that it is harder to find $I_{2n-1}$ for any $n\in \mathbb N$.

My question:

How to evaluate $\int_0^{\frac{\pi}{2}}x^3(\sqrt{\tan x}-\sqrt{\cot x}) d x$?

Your comments and solutions are highly appreciated.