<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] To prove that $I = J$ , note that
$$I-J = \int^{\pi/2}_0 \frac{\tan x-\tan^3 x}{\sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx = \int^{\pi/4}_{-\pi/4} \frac{\tan (\pi/4-x)-\tan^3 (\pi/4-x)}{\sqrt{(1+\tan^2 (\pi/4-x))(1+\tan^6 (\pi/4-x))}}\mathrm dx$$
We need to prove the follwing function is odd 
$$f(x) = \frac{\tan (\pi/4-x)-\tan^3 (\pi/4-x)}{\sqrt{(1+\tan^2 (\pi/4-x))(1+\tan^6 (\pi/4-x))}}$$
Now use that
$$\tan\left(\frac{\pi}{4} -x\right) = \frac{\cos x - \sin x}{\cos x + \sin x}$$
After expanding the denominator and nominator  we realize 
$$\tan (\pi/4-x)-\tan^3 (\pi/4-x)=\frac{-\cos(x) + \cos(3 x) + \sin(x) + \sin(3 x)}{(\cos x + \sin x)^3}$$
$$(1+\tan^2 (\pi/4-x))(1+\tan^6 (\pi/4-x))=\frac{10 - 6 \cos(4 x)}{(\cos x + \sin x )^8}$$
Hence 
$$f(x) = \frac{(-\cos(x) + \cos(3 x) + \sin(x) + \sin(3 x))(\sin x+ \cos x)}{\sqrt{10 - 6 \cos(4 x)}}$$
Interestingly this simplifies to 
$$f(x) = \frac{\sin(4x)}{\sqrt{10 - 6 \cos(4 x)}}$$
Hence $f(x)$ is odd which imlies

$$I-J = \int^{\pi/4}_{-\pi/4} f(x) \,dx = 0$$

Now using lab bhattacharjee result and $I=J$
$$4I=\int_0^{\infty}\dfrac{4du}{(3+2u-1)\sqrt{3+(2u-1)^2}} = {\log(7+4\sqrt{3})\over \sqrt3}$$
which implies 
$$J=I= {\log(7+4\sqrt{3})\over 4\sqrt3}$$
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