The integral equals $\sqrt{2+\sqrt{8}} \cdot \Omega$, where $\Omega$
is the real half-period $\Omega = 1.3736768699491\ldots$
of the elliptic curve
$$
E : y^2 = x^3 - 4 x^2 - 4 x,
$$
i.e. the complete elliptic integral
$$
\Omega = \int_{2-\sqrt{8}}^0 \frac{dx}{\sqrt{x^3-4x^2-4x}}
= \int_{2+\sqrt{8}}^\infty \frac{dx}{\sqrt{x^3-4x^2-4x}}
$$
(the integrand can also be brought to the classical form
${\bf K}(k) = \int_0^1 dz \, / \sqrt{(1-z^2) (1-k^2 z^2)}$,
but with a more complicated $k$ and probably also
an elementary factor more complicated than our $\sqrt{2+\sqrt{8}}$).
Here's gp code for this formula:
sqrt(2+sqrt(8)) * ellinit([0,-4,0,-4,0])[15]
The curve $E$ is reasonably nice, with conductor $128=2^7$ and
$j$-invariant $10976 = 2^5 7^3 = 1728 + 2^5 17^2$;
but $E$ does not have complex multiplication (CM), so we
do not expect to get a simpler form as would be possible for a CM curve
[e.g. $\int_1^\infty dx/\sqrt{x^3-1}$ is a Beta integral, and
$\int_0^\infty dx/\sqrt{x^3+4x^2+2x}
= \Gamma(1/8) \Gamma(3/8) / (4\sqrt{\pi})$].
Harry Peter already used the trigonometric substitution
$$
(\cos \phi, \sin \phi, d\phi) =
\left( \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}, \frac{2 \, dt}{1+t^2} \right)
$$
(which I guess is the "Weierstrass substitution" suggested in the comment of
Steven Stadnicki) to write $I$ as
$$
\int_0^1 \frac{(1+t)dt}{\sqrt{(1+2t-t^2) (t-t^3)}},
$$
which is a half-period of the holomorphic differential
$(1+t) dt/u$ on the hyperelliptic curve $C: u^2 = (1+2t-t^2) (t-t^3)$
of genus $2$. Most such periods cannot be simplified further,
but this one is special because the curve has more symmetry than
just the "hyperelliptic involution" $(t,u) \leftrightarrow (t,-u)$.
In particular $C$ has an involution
$$
\iota: (t,u) \leftrightarrow
\left( \frac{1-t}{1+t}, \frac{2^{3/2}}{(1+t)^3} u \right)
$$
which also sends the interval $(0,1)$ to itself, reversing
the orientation. This suggests splitting the integral
at the midpoint $t_0 := \sqrt{2} - 1$ and applying
the change of variable $(t,dt) \leftarrow ((1-t)/(1+t), -2\,dt/(1+t)^2)$
to the integral over $(t_0,1)$ to obtain $\sqrt{2} \int_0^{t_0} dt/u$.
Hence
$$
I = \int_0^{t_0} \frac{(\sqrt{2}+1+t)dt}{\sqrt{(1+2t-t^2) (t-t^3)}}
$$
and now the change of variable $X = t + (1-t)/(1+t)$ transforms $I$ to an
elliptic integral corresponding to the quotient curve $C\,/\langle\iota\rangle$.
While $C\,/\langle\iota\rangle$ has irrational coefficients involving $\sqrt{2}$,
it has rational $j$-invariant, so we can find coordinates that identify
$C\,/\langle\iota\rangle$ with our curve $E$ with rational coefficients,
though at the cost of introducing the factor $\sqrt{2+\sqrt{8}}$
into the formula for $I$ given at the start of this answer.
By replacing $\phi$ with $\arctan(t)$, then using integration by parts, we have:
$$ I = \int_{0}^{1}\frac{1}{1+t^2}\,\arctan\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)\,dt =\frac{\pi^2}{8}-\int_{0}^{1}\frac{3\sqrt{2}\, t \arctan(t)}{(3-t^2)\sqrt{1-t^2}}\,dt.$$
Now comes the magic. Since:
$$\int \frac{3\sqrt{2}\,t}{(3-t^2)\sqrt{1-t^2}}\,dt = -3\arctan\sqrt{\frac{1-t^2}{2}}\tag{1}$$
integrating by parts once again we get:
$$ I = \frac{\pi^2}{8}-3\int_{0}^{1}\frac{1}{1+t^2}\arctan\sqrt{\frac{1-t^2}{2}}\,dt \tag{2}$$
hence we just need to prove that:
$$ \int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\arctan\sqrt{1-2t^2}}{1+t^2}\,dt=\color{red}{\frac{\pi^2}{24}}\tag{3}$$
and this is not difficult since both
$$\int_{0}^{1}\frac{dt}{1+t^2}(1-t^2)^{\frac{2m+1}{2}},\qquad \int_{0}^{\frac{1}{\sqrt{2}}}\frac{(1-2t^2)^{\frac{2m+1}{2}}}{1+t^2}\,dt $$
can be computed through the residue theorem or other techniques. For instance:
$$\int_{0}^{1}\frac{(1-t)^{\frac{2m+1}{2}}}{t^{\frac{1}{2}}(1+t)}\,dt = \sum_{n\geq 0}(-1)^n \int_{0}^{1}(1-t)^{\frac{2m+1}{2}} t^{n-\frac{1}{2}}\,dt=\sum_{n\geq 0}(-1)^n\frac{\Gamma\left(m+\frac{3}{2}\right)\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(m+n+2)}$$
or just:
$$\int_{0}^{1}\frac{\sqrt{\frac{1-t^2}{2}}}{(1+t^2)\left(1+\frac{1-t^2}{2}u^2\right)}\,dt = \frac{\pi}{2(1+u^2)}\left(1-\frac{1}{\sqrt{2+u^2}}\right)\tag{4}$$
from which:
$$\int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\frac{\pi}{2}\int_{0}^{1}\frac{du}{1+u^2}\left(1-\frac{1}{\sqrt{2+u^2}}\right) =\color{red}{\frac{\pi^2}{24}} $$
as wanted, since:
$$ \int \frac{du}{(1+u^2)\sqrt{2+u^2}}=\arctan\frac{u}{\sqrt{2+u^2}}.$$
Best Answer
With $K$ as the Complete Elliptic Integral of the First Kind with Paramater $k$, $$K:=K(k)=\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2\sin^2t}}dt$$
Also using Landen's Transformation:
$$K\left(\frac{2\sqrt{x}}{1+x}\right)=(1+x)K(x)$$
Here is a Collection of Proofs for the Transformation.
This transformation is also related to a larger family of Hypergeometric Transformations (Quadratic Transformation of Gauss).
The Integral in question is then reduced to:
$$I=2\int_0^1kKdk$$
$$=-2\int_0^{\pi/2}\csc^2(t)\left[\int_0^1\frac{d}{dk}\sqrt{1-k^2\sin^2t}\ dk\right]dt$$ $$=-2\int_0^{\pi/2}\frac{\cos t-1}{\sin^2 t}dt$$
Therefore, $$I=2$$