<s>[INST] 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:
$$I = I(1) = 2\pi\sinh^{-1}\left(\frac{1}{\sqrt{2}} \right)$$
 [/INST] $$I = \int_{0}^{\frac{\pi}{2}} \ln\left(\sec^2(x) + \tan^4(x) \right)dx=\int_0^\infty \frac{\ln(1+x^2+x^4)}{1+x^2}dx$$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:
$$I=\pi \int_0^1 \frac{1}{\sqrt{(x+1)^2-1}}dx=\pi\ln(2+\sqrt 3)$$
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