Seeking Methods to solve $\int_{0}^{\frac{\pi}{2}} \ln\left|\sec^2(x) + \tan^4(x) \right|\:dx $ After weeks of going back and forth I've been able to solve the following definite integral:
$$I = \int_{0}^{\frac{\pi}{2}} \ln\left|\sec^2(x) + \tan^4(x) \right|\:dx $$
To solve this I employ Feynman's Trick with Glasser's Master Theorom but I'm excited to learn of other methods that can be employed. Are there any other 'tricks' that can be used? or alternatively series based solutions? or transformations? (or anything for that matter). 
For those who may be interested my process was:
(1) First make the substitution: $u = \tan(x)$
$$I = \int_{0}^{\infty} \frac{\ln\left|u^2 + 1 + u^4 \right|}{u^2 + 1}\:du = \int_{0}^{\infty} \frac{\ln\left|1 + u^2\left(u^2 + 1\right) \right|}{u^2 + 1}\:du$$
(2) Now employ Feynman's Trick by introducing a new parameter:
$$I(t) = \int_{0}^{\infty} \frac{\ln\left|1 + t^2u^2\left(u^2 + 1\right) \right|}{u^2 + 1}\:du$$
Note here that $I = I(1)$ and $I(0) = 0$
(3) Take the derivative w.r.t 't' 
$$I'(t) = \int_{0}^{\infty} \frac{2tu^2\left(u^2 + 1\right)}{1 + t^2u^2\left(u^2 + 1\right)}\frac{1}{u^2 + 1}\:du = \frac{1}{t} \int_{-\infty}^{\infty} \frac{1}{\left(u - \frac{1}{tu}\right)^2 + \frac{2}{
t} + 1}\:du$$
(4) Employ Glasser's Master Theorem:
$$I'(t) =  \frac{1}{t} \int_{-\infty}^{\infty} \frac{1}{\left(u - \frac{1}{tu}\right)^2 + \frac{2}{t} + 1} \:du= \frac{1}{t}\int_{-\infty}^{\infty}\frac{1}{u^2 + \frac{2}{t} + 1} \:du$$
As: $\frac{2}{t} + 1 > 0 $ we arrive at
$$I'(t) = \frac{1}{t}\left[\frac{1}{\sqrt{\frac{2}{t} + 1}}\arctan\left(\frac{u}{\frac{2}{t} + 1}\right)\right]_{-\infty}^{\infty}= \frac{\pi}{\sqrt{t\left(t + 2\right)}}$$
(5) We now integrate w.r.t 't' 
$$I(t) = \int \frac{\pi}{\sqrt{t\left(t + 2\right)}}\:dt = 2\pi\sinh^{-1}\left(\frac{t}{\sqrt{2}} \right) + C$$
Where $C$ is the constant of integration. As above $I(0) = 0 \rightarrow C = 0$ and so, our final solution is given by:
$$I = I(1) = 2\pi\sinh^{-1}\left(\frac{1}{\sqrt{2}} \right)$$

Here's an alternative solution. First, notice that if $I(a,b) = \int_0^\pi \ln\left(a + b\cos(x)\right) \mathrm{d}x$ then $I(a,0) = \pi \ln(a)$. Thus
\begin{align}
\frac{\partial I}{\partial a}
&=\int_0^\pi \frac{\mathrm{d}x}{a + b\cos(x)} \overset{t=\tan\left(\frac{x}{2}\right)}{=}\frac{2}{a-b}\int_0^\infty \frac{\mathrm{d}t}{t^2 + \frac{a+b}{a-b}} \overset{t =\sqrt{\frac{a+b}{a-b}}\tan(u)}{=} \frac{\pi }{\sqrt{a^2-b^2}}
\end{align}
So integrating the above equation w.r.t. $a$ we get
$$
I(a,b) =  \pi\int\frac{\mathrm{d}a}{\sqrt{a^2-b^2}} = \pi\int\frac{1 + \frac{a}{\sqrt{a^2-b^2}}}{a+\sqrt{a^2-b^2}} \,\mathrm{d}a \overset{u =a+\sqrt{a^2-b^2} }{=} \pi\ln\left(a+\sqrt{a^2-b^2}\right) + C
$$
Recalling our initial condition, by plugging in $b=0$ into the previous equation we get $\pi \ln(a) = I(a,0) = \pi \ln(2a) + C$. Hence $C =  - \pi \ln(2)$ and
$$\int_0^\pi \ln\left(a + b\cos(x)\right) \,\mathrm{d}x = \pi\ln\left(\frac{a+\sqrt{a^2-b^2}}{2}\right) \tag{1}$$
Lastly, using basic integral manipulations we can obtain $\int_0^{\frac{\pi}{2}}\ln(\cos(x)) \mathrm{d}x = - \frac{\pi}{2}\ln(2)$. Combining everything we conclude the problem as follows
\begin{align}
\int_0^{\frac{\pi}{2}} \ln\left(\sec^2(x) + \tan^4(x)\right) \mathrm{d}x & =\int_0^{\frac{\pi}{2}} \ln\left(\frac{7+\cos(4x)}{8\cos^4(x)}\right) \mathrm{d}x \\
& = \frac{1}{2}\int_0^{\pi}\ln(7+\cos(u))\, \mathrm{d}u - 4\int_0^{\frac{\pi}{2}} \ln(\cos(x)) \mathrm{d}x - \frac{\pi}{2}\ln(8)\\
& \overset{(1)}{=}\frac{\pi}{2}\ln\left( \frac{7+4 \sqrt{3}}{2}\right) + 2\pi \ln(2) - \frac{3\pi}{2}\ln(2)\\
& = \boxed{\frac{\pi}{2} \ln\left(7 + 4 \sqrt{3}\right)}
\end{align}
which agrees with the other answers since $\sqrt{7 + 4 \sqrt{3}} = 2 + \sqrt{3}$.
