By Euler's formula,
$$\sin(\ln(x))=\frac{e^{i\ln(x)}-e^{-i\ln(x)}}{2i}=\frac{x^i-x^{-i}}{2i}$$
In the integral, this works out to give us
$$\int\frac{\mathrm dx}{\sin(\ln(x))}=\int\frac{2i~\mathrm dx}{x^i-x^{-i}}=2i\int\frac{x^i~\mathrm dx}{x^{2i}-1}=-2i\int\frac{x^i~\mathrm dx}{1-x^{2i}}$$
By expanding with geometric series, this then becomes
$$\int\frac{x^i~\mathrm dx}{1-x^{2i}}=\sum_{k=0}^\infty\int x^{(2k+1)i}~\mathrm dx=\sum_{k=0}^\infty\frac{x^{1+(2k+1)i}}{1+(2k+1)i}$$
Observe that the ratio of consecutive terms in this series is given by
$$\frac{x^{1+(2k+3)i}/(1+(2k+3)i)}{x^{1+(2k+1)i}/(1+(2k+1)i)}=\frac{(2k+1)i+1}{(2k+3)i+1}x^{2i}=\frac{(k+\color{#3377cc}{\frac{1+i}2})(k+\color{#3377cc}1)}{k+\color{#339999}{\frac{1+3i}2}}\frac{\color{#dd3333}{x^{2i}}}{k+1}$$
which implies the series is a hypergeometric function:
$$\sum_{k=0}^\infty\frac{x^{1+(2k+1)i}}{1+(2k+1)i}=x^{1+i}{}_2F_1\left(\color{#3377cc}{\frac{1+i}2},\color{#3377cc}1;\color{#339999}{\frac{1+3i}2};\color{#dd3333}{x^{2i}}\right)$$
and altogether,
$$\int\frac{\mathrm dx}{\sin(\ln(x))}=-2ix^{1+i}{}_2F_1\left(\frac{1+i}2,1;\frac{1+3i}2;x^{2i}\right)\color{#999999}{{}+C}$$
which likely cannot be simplified further, though can be rewritten using various hypergeometric identities.
Note: The above manipulations require that the series converges, but the end results in terms of hypergeometric functions hold everywhere they both exist, as they are defined through the use of analytic continuation.
For the antiderivative, the only way I found to introduce elliptic integrals is the tangent half-angle substitution
$$I(x)=\int\frac{\sin (x)(\sin (x)+\cos (x))}{\sqrt{\sin (x)\cos (x)}}\,dx$$
$$I(t)=-2 \sqrt{2} \int \frac{\sqrt{t}\, \left(t^2-2 t-1\right)}{
\,\left(t^2+1\right)^2\, \sqrt{1-t^2}} \,dt$$ Using partial fraction decomposition
$$I(t)=-\sqrt 2 \int \Bigg( \frac{(1+i) \sqrt{t}}{(t-i)^2 \sqrt{1-t^2}}+\frac{(1-i) \sqrt{t}}{(t+i)^2
\sqrt{1-t^2}} \Bigg)\,dt$$
Computing (I used a CAS for it)
$$J=\int\frac{\sqrt{t}}{(t+a)^2\,\sqrt{1-t^2} }\,dt$$
$$J=\frac{\sqrt{1-t^2}}{\left(a^2-1\right) \sqrt{t}}\Bigg(-\frac{a}{t+a}+\frac 1a \sqrt{\frac{t}{t^2-1}}\,A(t) \Bigg)$$ where
$$A(t)=\left(a^2+1\right) \Pi \left(-a;\left.u\right|-1\right)+(a-1) F\left(\left.u\right|-1\right)-a E\left(\left.u\right|-1\right)$$
and
$$u=\csc ^{-1}\left(\sqrt{t}\right)$$ where appear the complete elliptic integrals of first, second and third kinds.
Simplifying as much as I could
$$I(t)=\sqrt{\frac{2}{t}}\,\frac{\sqrt{1-t^2}}{ \left(1-t^4\right)}\,\left((t-1) (t+1)^2 + \sqrt{t \left(t^2-1\right)} \left(t^2+1\right)\,B(t)\right)$$ where
$$B(t)=
E\left(\left.\csc ^{-1}\left(\sqrt{t}\right)\right|-1\right)-2 F\left(\left.\csc
^{-1}\left(\sqrt{t}\right)\right|-1\right)$$
The problem is that we cannot compute directly $I(1)$ and $I(0)$ which correspond to indeterminate forms.
But, using series expansion
$$I(1)=i \sqrt{2} (E(-1)-2 K(-1))+2 \sqrt{1-t}+O\left((t-1)^{3/2}\right)$$
$$I(0)=(-1+i) \sqrt{2} (-(3-i) K(-1)+(1+i) K(2)+E(-1))+O\left(t^{3/2}\right)$$
Therefore
$$L=\int_0^{\frac \pi 2}\frac{\sin (x)(\sin (x)+\cos (x))}{\sqrt{\sin (x)\cos (x)}}\,dx$$
is "just" $(I(1)-I(0))$ that is to say
$$L=\sqrt{2}\,\, \big(2 K(2)+E(-1)-2(1- i) K(-1)\big)$$
But
$$ K(2)-(1- i) K(-1)=0 \quad \implies \quad \large\color{blue}{L=\sqrt{2}\,\,E(-1)}$$
which is
$$L=\frac{1}{4 \sqrt{\pi }}\Bigg(\frac {8\pi^2}{\Big(\Gamma \left(\frac{1}{4}\right)\Big)^2 }+ \Bigg(\Gamma \left(\frac{1}{4}\right)\Bigg)^2\Bigg) $$
Best Answer
As @sudeep5221 commented, you missed a term when you used the first subtitution $$I=\int\frac{dx}{(1 + \sin^4(x))\sqrt{1+\sin^2(x)}}$$
$$t=\sin(x) \quad \implies I=\int\frac{dt}{ \left(1+t^4\right)\sqrt{1-t^4}}$$
There is an explicit solution but it involves either elliptic integrals or Appell hypergeometric functions of two variables.
Probably, for the time being, use a series expansion
$$\frac{1}{ \left(1+t^4\right)\sqrt{1-t^4}}=\sum_{n=0}^\infty(-1)^n\,\frac {a_n}{b_n} \, t^{4n}$$ where the $a_n$ and $b_n$ form respectively sequences $A123746$ and $A046161$ in $OEIS$.
The problem is that the convergence is extremely slow if you integrate termwise between $0$ and $1$.
Edit
Just to give an idea of the complexity of the result, the definite integral is $$-\sqrt{\pi }\,\,\frac{ \Gamma \left(\frac{5}{4}\right)}{\Gamma\left(-\frac{1}{4}\right)} \left(4+\frac{1+i}{2} B_{-1}\left(\frac{3}{4},\frac{1}{2}\right) \right)$$