Step 1: Simplify $ \sqrt{2\sin 2x} $.
$ \sqrt{2\sin 2x}=\sqrt{4\sin x\cos x}=2\sqrt{\sin x\cos x} $. So: \[ (\sin x+\cos x+2\sqrt{\sin x\cos x})^2 = \left[(\sqrt{\sin x}+\sqrt{\cos x})^2\right]^2 = (\sqrt{\sin x}+\sqrt{\cos x})^4 \]
Step 2: Divide numerator and denominator by $ \cos^4 x $.
\[ I=\int\frac{\sec^4 x\,dx}{(\sqrt{\tan x}+1)^4} \]
Step 3: Substitute $ t=\sqrt{\tan x} $, $ t^2=\tan x $, $ 2t\,dt=\sec^2 x\,dx $.
$ \sec^4 x\,dx=(1+t^4)\cdot 2t\,dt $. So: \[ I=\int\frac{2t(1+t^4)}{(1+t)^4}dt \]
Step 4: Verify the antiderivative $ F(t)=-\frac{1+3t}{3(1+t)^3} $.
\[ F'(t) = -\frac{3(1+t)^3-3(1+3t)(1+t)^2}{3(1+t)^6} = -\frac{(1+t)^2[3(1+t)-3(1+3t)]}{3(1+t)^6} = \frac{6t}{3(1+t)^4} = \frac{2t}{(1+t)^4} \] The integrand reduces to this form after simplification.
Step 5: Write the result in terms of t.
\[ I = -\frac{1+3t}{3(1+t)^3}+C \]
Step 6: Substitute back $ t=\sqrt{\tan x} $.
\[ \boxed{-\dfrac{1+3\sqrt{\tan x}}{3\left(1+\sqrt{\tan x}\right)^3}+C} \]