$$\frac{1}{2\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + \tan y)}{(\cos y + \sqrt{2} \sin(y + \frac{\pi}{4})) \sqrt{1 + \sqrt{2} \sin(2y + \frac{\pi}{4})}} \, dy = G$$
Recall that $$\sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} (\cos x + \sin x) \text{ and } \cos(2x) = 2\cos^2 x - 1$$
$$J = \frac{1}{2\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + \tan x)}{(\cos x + \sqrt{2} \sin(x + \frac{\pi}{4})) \sqrt{1 + \sqrt{2} \sin(2x + \frac{\pi}{4})}} \, dx$$
$$= \frac{1}{2\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + \tan y)}{(2\cos y + \sin y) \sqrt{2\cos^2 y + 2\sin y \cos y}} \, dy$$.
$$= \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + \tan y)}{(2 + \tan y) \sqrt{1 + \tan y}} \cdot \frac{1}{\cos^2 y} \, dy$$
Continue with $z=1+\tan y$.
$$I = \int_0^\tfrac\pi2 \frac{\log(1+\tan y)}{(2+\tan y) \sqrt{1+\tan y}} \, \sec^2y\,dy = \int_1^\infty \frac{\log z}{\sqrt z(1+z)} \, dz$$
Upon replacing $z\mapsto\dfrac1{z^2}$, we can use Taylor series and integration by parts to finish the job.
$$I = -4 \int_0^1 \frac{\log z}{1+z^2} \, dz = -4 \sum_{n\ge0} (-1)^n \int_0^1 z^{2n} \log z \, dz = 4 \sum_{n\ge0} \frac{(-1)^n}{(2n+1)^2} = 4G$$