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

Because the second integral remains unevaluated without the substitution. Did you try it out yourself?

Where I find this Bierens de Haan Table 148?

Edited:

Ok I found this Book.

Regards M.I.

Here is another integral from Bierens de Haan Equation (13) Table 147

What is the question?