<s>[INST] Show that $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$ Consider the integrals $(1)$ and $(2)$, how does on show that
(1): $I=J$
(2): and $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$

$$\int_{0}^{\pi/2}{\tan x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx=I\tag1$$
$$\int_{0}^{\pi/2}{\tan^3 x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx=J\tag2$$

An attempt:
$u=\tan x$ $\implies du=\sec^2 x dx$
$(1)$ becomes
$$\int_{0}^{\infty}{u\over\sqrt{(1+u^2)(1+u^6)}}\cdot{\mathrm du\over 1+u^2}\tag3$$
$v=1+u^2$ then
$${1\over 2}\int_{1}^{\infty}{\mathrm dv\over \sqrt{v[1+(v-1)^3]}}\tag4$$

$u=\tan^3 x$ $\implies du=3\tan^2 x\sec^2 x dx$
$(2)$ becomes
$${1\over 3}\int_{0}^{\infty}{u^{1/3}\over\sqrt{(1+u^2)(1+u^6)}}\cdot{\mathrm du\over 1+u^{2/3}}\tag5$$
We are not sure how to continue from here...
 [/INST] On the path of Lab Bhattacharjee,
$J=\displaystyle \int_0^{+\infty} \dfrac{1}{\sqrt{(1+x)(1+x^3)}}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}\displaystyle J&=\int_{-1}^{1} \dfrac{1}{\sqrt{1+3x^2}}dx\\
\end{align}$
Perform the change of variable $y=\sqrt{3}x$,
$\begin{align}
J&=\dfrac{1}{\sqrt{3}} \int_{-\sqrt{3}}^{\sqrt{3}}\dfrac{1}{\sqrt{1+x^2}}dx\\
&=\dfrac{1}{\sqrt{3}}\Big[\text{arcsinh}(x)\Big]_{-\sqrt{3}}^{+\sqrt{3}}\\
&=\dfrac{1}{\sqrt{3}}\Big[\ln\left(x+\sqrt{1+x^2}\right)\Big]_{-\sqrt{3}}^{+\sqrt{3}}\\
&=\dfrac{1}{\sqrt{3}}\ln\left(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\right)\\
&=\boxed{\dfrac{1}{\sqrt{3}}\ln\left(7+4\sqrt{3}\right)}
\end{align}$
</s>