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...
 A: 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 
$$J=I= {\log(7+4\sqrt{3})\over 4\sqrt3}$$
A: HINT:
$$2(I+J)=\int_0^{\pi/2}\dfrac{2\tan x(1+\tan^2x)}{\sqrt{(1+\tan^2x)(1+\tan^6x)}}dx$$
Set $\tan^2x=u$
$$2(I+J)=\int_0^{\infty}\dfrac{du}{\sqrt{(1+u)(1+u^3)}}$$
$$=\int_0^{\infty}\dfrac{4du}{(3+2u-1)\sqrt{3+(2u-1)^2}}$$
Set $2u-1=\sqrt3\tan v$
$$2(I+J)=\int_{-\pi/6}^{\pi/2}\dfrac{2\sqrt3\sec v\ dv}{3+\sqrt3\tan v}=\int_{-\pi/6}^{\pi/2}\dfrac{2\ dv}{\sqrt3\cos v+\sin v}=\int_{-\pi/6}^{\pi/2}\csc(v+\pi/3)dv$$
A: 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 $y=\sqrt{3}x$,
$\begin{align}
J&=\dfrac{1}{\sqrt{3}} \int_{-\sqrt{3}}^{\sqrt{3}}\dfrac{1}{\sqrt{1+x^2}}dx\\
&=\dfrac{1}{\sqrt{3}}\Big[\text{arcsinh}(x)\Big]_{-\sqrt{3}}^{+\sqrt{3}}\\
&=\dfrac{1}{\sqrt{3}}\Big[\ln\left(x+\sqrt{1+x^2}\right)\Big]_{-\sqrt{3}}^{+\sqrt{3}}\\
&=\dfrac{1}{\sqrt{3}}\ln\left(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\right)\\
&=\boxed{\dfrac{1}{\sqrt{3}}\ln\left(7+4\sqrt{3}\right)}
\end{align}$
A: Note that, under $\frac{\pi}{2}-x\to x$, 
\begin{eqnarray}
I-J&=&\int_{0}^{\pi/2}{\tan x-\tan^3x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx\\
&=&\int_{0}^{\pi/2}{\cot x-\cot^3x\over \sqrt{(1+\cot^2 x)(1+\tan^6 x)}}\mathrm dx\\
&=&\int_{0}^{\pi/2}{\tan^4(\cot x-\cot^3x)\over \sqrt{\tan^8x(1+\cot^2 x)(1+\cot^6 x)}}\mathrm dx\\
&=&\int_{0}^{\pi/2}{\tan^3 x-\tan x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx\\
&=&-(J-I),
\end{eqnarray}
and hence $I-J=0$ or $I=J$.
