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:
$$I = I(1) = 2\pi\sinh^{-1}\left(\frac{1}{\sqrt{2}} \right)$$
 A: $$I = \int_{0}^{\frac{\pi}{2}} \ln\left(\sec^2(x) + \tan^4(x) \right)dx=\int_0^\infty \frac{\ln(1+x^2+x^4)}{1+x^2}dx$$Consider: $$I(a)=\int_0^\infty \frac{\ln((1+x^2)a+x^4)}{1+x^2}dx$$
Derivating under the integral sign with respect to $a$ gives: $$I'(a)=\int_0^\infty \frac{1+x^2}{(1+x^2)a+x^4}\frac{dx}{1+x^2}=\int_0^\infty \frac{1}{x^4+ax^2+a}dx\overset{\large{x=\frac{\sqrt a}{t}}}=\int_0^\infty \frac{\frac{t^2}{\sqrt a}}{t^4+at^2+a}dt$$
$$2I'(a)=\int_0^\infty \frac{\frac{t^2}{\sqrt a}+1}{t^4+at^2+a}dt\Rightarrow I'(a)=\frac{1}{2\sqrt a}\int_0^\infty \frac{t^2+\sqrt a}{t^4+at^2+a}dt$$
$$=\frac{1}{2\sqrt a}\int_0^\infty \frac{1+\frac{\sqrt a}{t^2}}{\left(t-\frac{\sqrt a}{t}\right)^2+a+2\sqrt a}dt=\frac{1}{2\sqrt a}\int_0^\infty \frac{d\left(t-\frac{\sqrt a}{t}\right)}{\left(t-\frac{\sqrt a}{t}\right)^2+\left(\sqrt{a+2\sqrt a}\,\right)^2}$$
$$=\frac{1}{2\sqrt a}\frac{1}{\sqrt{a+2\sqrt a}}\arctan\left(\frac{t-\frac{\sqrt a}{t}}{\sqrt{a+2\sqrt a}}\right)\bigg|_0^\infty \Rightarrow I'(a)=\frac{\pi}{2\sqrt a}\frac{1}{\sqrt{a+2\sqrt a}}$$
And noticing that $I(0)=4\int_0^\infty \frac{\ln x}{1+x^2} dx=0$. By the fundamental theorem of Calculus we have: $$I=I(1)-I(0)=\int_0^1 I'(a)da=\frac{\pi}{2}\int_0^1 \frac{1}{\sqrt a \sqrt {a+2\sqrt a}}da$$ Finally setting  $\sqrt a =x$ gives:
$$I=\pi \int_0^1 \frac{1}{\sqrt{(x+1)^2-1}}dx=\pi\ln(2+\sqrt 3)$$
A: We use the representation

$$
I=\int_0^{\infty}\frac{\log(g(x))}{1+x^2}dx
$$

derived by OP.
Here $g(z)=1+z^2+z^4$. Note that $\log(g(z))$ has four branch points at $z_n=e^{i n \pi/3}$, $n={1,2,4,5}$ 
of which $z_{1,2}$ lie in the upper half of the complex plane.
Let us define
$$
f(z)=\frac{\log(g(z))}{1+z^2}
$$
By parity we have also that $2\int_0^{\infty}\frac{\log(g(x))}{1+x^2}dx=\int_{-\infty}^{\infty}\frac{\log(g(x))}{1+x^2}dx$. We furthermore note that since $\log(g(z))\sim_i-2i(x-i)$ the residue at $i$ vanishs.
Last but not least, $|f(z)
|\sim C\log(R)/R^2$ so integrals over large semicirles of this function vanish in the limit of $R\rightarrow \infty$. 
We can therefore state that twice our integral of interest equals the two integrals encirceling the two branchcuts in the upper half of the complex plane ($\delta\rightarrow 0_+$).
$$
2I=\color{blue}{\int_{e^{i \pi(1/3-\delta)}}^{e^{i \pi(1/3-\delta)}\infty}f(z)dz-\int_{e^{i \pi(1/3+\delta)}}^{e^{i \pi(1/3+\delta)}\infty}f(z)dz}-\\
\color{red}{\int_{e^{i \pi(2/3-\delta)}}^{e^{i \pi(2/3-\delta)}\infty}f(z)dz-\int_{e^{i \pi(2/3+\delta)}}^{e^{i \pi(2/3+\delta)}\infty}f(z)dz}$$
It is a well known fact that such pairs of integrals collapse into integrals over the discontinuity  of the integrand which is given in both cases by $2 \pi i\times(1+z^2)^{-1}$ and therefore:
$$
2I=2\pi i\left[\color{blue}{z_1\int_1^{\infty}\frac{dq}{1+(z_1q)^2}}-\color{red}{z_2\int_1^{\infty}\frac{dq}{1+(z_2q)^2}}\right]=\\
2\pi i[\color{blue}{\text{arccot}(z_1)}-\color{red}{\text{arccot}(z_2)}]
$$
Annoying algebra yields ($\text{arccot}(z_{1,2})=\mp i\log(2+\sqrt{3})+\frac{\pi}{4}$) the pleasantly simple end result:

$$
I=\pi(\log(2+\sqrt{3}))
$$

A: 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 $\sqrt{7 + 4 \sqrt{3}} = 2 + \sqrt{3}$.
