A Neater Expression
$$I=\int^π_0 \frac {\sin x}{\sqrt {x^3+x+1}} dx = \sum_0^∞ A_k \sum_0^k (-1)^r {}^kP_{2r} π^{k-2r}$$
Where,
$$(3+2k)A_k + (5+2k)A_{2+k} + 2(3+k)A_{3+k} = 0,$$
$$A_0=1, A_1=\frac {-1}{2}, A_2=\frac {3}{8}$$
A greedy approach
$$I=\int^π_0 \frac {\sin x}{\sqrt {x^3+x+1}} dx = 0.8750439062939084$$
Using the greedy Egyptian fraction algorithm,
$$x_{k+1} = x_k - \frac {1}{\lceil \frac {1}{x_k} \rceil}$$
where, $x_0 = I$, I got an expansion,
$$I = \frac {1}{2} + \frac {1}{3} + \frac {1}{2^3.3} +\frac {1}{2^3.3.13.73}+\frac {1}{2^2.13.113.397547} +……$$
I couldn't go farther , for my limited computational capacity (which is my laptop), however I indeed see one pattern : the prime factors in the denominators $(2,3,13,73,113,…)$ belong to the set of primes given by,
$$a(n)= \text {Min} \left(x; π[x]-π\left[\frac {x}{2}\right]=n\right)$$
I got it on OEIS(https://oeis.org/A080359). Yet it needs much more insight.
Original answer
A closed form would be extremely difficult to get. This appears to be a new function. Substituting $t$ for the denominator, we get a beautiful form of the integral, however potentially latent in the present context.
$$I = 2\int \frac {\cosh J(t)}{\cosh 3J(t)} \sin \left(-2 \sqrt {\frac {1}{3}} \sinh J(t)\right) dt$$
where,
$$J(t):=\frac {1}{3} \sinh^{-1} \left[\frac {3\sqrt 3}{2} (1-t^2)\right]$$
So , I am giving a series form solution. Consider,
$$F(x):=\int \frac {\sin x}{\sqrt {1+x(1+x^2)}} dx\tag{1}$$
Now, for $x<1$,
$$[1+x(1+x^2)]^{-\frac {1}{2}} = -\sum_{k=0}^∞ C^{k-\frac {1}{2}}_{-\frac {1}{2}}x^k(1+x^2)^k \tag{2}$$
Plugging $(2)$ into $(1)$ we get,
$$F(x) = -\sum_{k=0}^∞ C^{k-\frac {1}{2}}_{-\frac {1}{2}} G(k,x) \tag{3},$$
where ,
$$G(k,x):= \int x^k \sin x (1+x^2)^k dx \tag{4}$$
But,
$$(1+x^2)^k= \sum_{r=0}^k C^k_r x^{2r} \tag{5}$$
Plugging $(5)$ into $(4)$ we get,
$$G(k,x) = \sum_{r=0}^k C^k_r H(r,k,x) \tag{6},$$
where,
$$H(r,k,x) := \int x^{k+2r} \sin x dx$$
$$= - \frac {\Gamma (k+2r+1, ix) + (-1)^{k+2r}\Gamma (k+2r+1, -ix)}{2(-1)^{\frac {5}{2} (k+2r)}} \tag{7}$$
Hence,
$$F(x)=\sum_{k=0}^∞ C^{k-\frac {1}{2}}_{-\frac {1}{2}} \sum_{r=0}^k C^k_r \frac {\Gamma (k+2r+1, ix) + (-1)^{k+2r}\Gamma (k+2r+1, -ix)}{2(-1)^{\frac {5}{2} (k+2r)}}\tag{8}$$
On the same lines an integral exists for the case $x>1$, the only difference being in the binomial expansion for the denominator of the original integral . Call it $F'(x)$. Then,
$$\int_0^π \frac {\sin x}{\sqrt {1+x(1+x^2)}} dx = [F(1) -F(0)] +[F'(π) - F'(1)]$$
This solution is in terms of upper incomplete gamma functions with complex arguments. The notation $C^n_r$ stands for combinatorial coefficients.
Note-1
Alternatively, one could use hypergeometric functions to express the final result,
$$F(x)=-\sum_{k=0}^∞ C^{k-\frac {1}{2}}_{-\frac {1}{2}} \sum_{r=0}^k C^k_r \frac {x^{k+2r+2}}{k+2r+2} {}_1\text {F}_2 \left(\frac {k+2r+4}{2},\frac {3}{2} ; \frac {k+2r+2}{2} ; -\frac {x^2}{4}\right)$$
Note-2
There is still hope for a closed form of the indefinite integral in terms of the Fresnel integral $C(z)$, consider a related integral:
$$\int \frac {\sin x}{\sqrt {x^3}} dx = \sqrt {8π} \text {C} \left(\sqrt {\frac {2x}{π}}\right) - \frac {2\sin x}{\sqrt x} + c$$
The integrand is almost the same, except for the extra linear term $x+1$ under square root.
Best Answer
Integrate by parts to obtain a recursive formula as follows
\begin{align} I_n&=\int_0^{\pi} \sin^{n}x \ln(\sin x) dx\\ &= -\int_0^{\pi} \sin^{n-1}x \ln(\sin x)\> d(\cos x)\>\\ & =\int_0^{\pi}((n-1) \sin^{n-2}x \cos^2x\ln(\sin x)+ \sin^{n-2}x\cos^2x)dx\\ &= (n-1) (I_{n-2}-I_n)+ \frac1{n-1}\int_0^{\pi}\sin^{n}x\>dx \end{align} Thus $$I_n = \frac{n-1}n I_{n-2} +\frac1{n(n-1)} \int_0^{\pi}\sin^{n}x\>dx $$ with $I_0 = -\pi\ln2$ and $I_1= \ln2 -1$. (See here for evaluating $\int_0^{\pi/2}\sin^{n}x\>dx$.)