Closed form for $\int_0^{\pi/2}\frac{\sqrt{1+\sin\phi}}{\sqrt{\sin2\phi}\,\sqrt{\sin\phi+\cos\phi}}d\phi$

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.

$$ \begin{align} \int_0^\frac{\pi}{2}\dfrac{\sqrt{1+\sin\phi}}{\sqrt{\sin2\phi}\sqrt{\sin\phi+\cos\phi}}d\phi &=\int_0^1\dfrac{2}{1+t^2}\dfrac{\sqrt{1+\dfrac{2t}{1+t^2}}}{\sqrt{\dfrac{4t(1-t^2)}{(1+t^2)^2}}\sqrt{\dfrac{2t}{1+t^2}+\dfrac{1-t^2}{1+t^2}}}dt\\ &=\int_0^1\dfrac{1+t}{\sqrt t\sqrt{1-t^2}\sqrt{1+2t-t^2}}dt\\ &=\int_0^1\dfrac{\sqrt{1+t}}{\sqrt t\sqrt{1-t}\sqrt{1+(\sqrt2+1)t}\sqrt{1-(\sqrt2-1)t}}dt\\ &=\pi F_D^{(3)}\left(\dfrac{1}{2},-\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},1;-1,-\sqrt2-1,\sqrt2-1\right). \end{align} $$

According to Lauricella hypergeometric series.