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.
Best Answer
After substitution $t=\tan{x}$ use $$\frac{a^2+b^2t^2}{(a^4+b^4t^2)(1+t^2)}=\frac{1}{a^2+b^2}\left(\frac{1}{1+t^2}+\frac{a^2b^2}{a^4+b^4t^2}\right).$$