I'd like to note that with a little handholding, Mathematica can be made to spit out this closed form.
Split the integral as
$$\int_0^1 \frac{\log\log t}{1+t^2} \mathrm dt+\int_1^\infty \frac{\log\log t}{1+t^2} \mathrm dt$$
The first readily evaluates:
f1 = Integrate[Log[Log[t]]/(1 + t^2), {t, 0, 1}]
1/4 π (I π + Log[(4 π^3)/Gamma[1/4]^4])
The second one is a little recalcitrant, but it succumbs to the substitution
$t\mapsto1/u$:
f2 = Integrate[Log[Log[1/u]]/(1 + u^2), {u, 0, 1}]
1/4 π Log[(4 π^3)/Gamma[1/4]^4]
It should be noted at this juncture that f2 is also the real part of f1. Adding these two results up should give an equivalent result:
π/4 (I π + 2 Log[(4 π^3)/Gamma[1/4]^4]) ==
π/4 (I π + Log[4 π^2 (Gamma[3/4]/Gamma[1/4])^4]) // FullSimplify
True
