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
Here's how an electrician tries to solve this problem
First, we introduce the following notation
$$X(x)=X=ax^2+bx+c$$
Then we can present the integral in the form
$$I_n=\int\frac{x^ndx}{\sqrt{X}}$$
Now, let's differentiate $x^{n-1}\sqrt{X}$ with respect to $x$
$$\frac{\mathrm{d} }{\mathrm{d} x}(x^{n-1}\sqrt{X})=an\frac{x^n}{\sqrt{X}}+b(n-\frac{1}{2})\frac{x^{n-1}}{\sqrt{X}}+c(n-1)\frac{x^{n-2}}{\sqrt{X}}$$
Here we took into account that $X=ax^2+bx+c$
Integrating the resulting equation with respect to x we can write
$$x^{n-1}\sqrt{X}=anI_n+b(n-\frac{1}{2})I_{n-1}+c(n-1)I_{n-2}$$
Set here $n=1$ $$\sqrt{X}=aI_1+\frac{b}{2}I_0$$
We're almost done.
The only thing left is to calculate
$$I_0=\int\frac{dx}{\sqrt{ax^2+bx+c}}$$
After that we can calculate the integrals $I_1,I_2...$ successively using the recurrence equation.
Further development
Let's introduce a generating function
$$I(s)=I=\sum_{n=0}^{\infty}I_ns^n$$
Now, multiply both sides of the recurrence relation above by $s^n$ and sum over $n\geqslant 1$.
(I'll skip the long but simple algebraic transformations)
We get the following differential equation $$(cs^2+bs+a)\frac{\mathrm{d} I}{\mathrm{d} s}+(cs+\frac{b}{2})I=\frac{\sqrt{X}}{1-xs}$$
Let's remember that $X=ax^2+bx+c$
All the family of the given integral $I_n$ is hidden in this differential equation.
But how to drag them all out, that's the problem!