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.
$$I(a, b) =\int_0^\frac{\pi}{2} \frac{1}{\sqrt{a^2 \sin^2 t + b^2 \cos^2 t}}dt$$
$$ = \int_0^\frac{\pi}{2} \frac{1}{\sqrt{b^2-(b^2-a^2)\sin^2 t}}dt=\frac1b \int_0^\frac{\pi}{2} \frac{1}{\sqrt{1-\left(1-\frac{a^2}{b^2}\right)\sin^2 t}}dt=\frac1b K\left(1-\frac{a^2}{b^2}\right)$$ Or this can be written as: $I(a, b)=\frac{\pi}{2M(a,b)},$ where $M(a,b)$ is the arithmetic-geometric mean of $a,b$ defined by $M(a,b):=\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n$ where
$$a_0=a, \ b_0=b, \ a_{n+1}=\frac{a_n+b_n}{2}, \ b_{n+1}=\sqrt{a_nb_n}$$
Best Answer
Hint:
$$\sin2\theta=(\sin\theta+\cos\theta)^2-1$$
and $\sin\theta+\cos\theta=\sqrt2\sin\left(\dfrac\pi4+\theta\right)$
Use Integral of $\csc(x)$