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
$$I=\int_{-\infty}^{\infty} \frac{\log(x^2+1)}{x^4+x^2+1}\,dx=2\int_{0}^{\infty} \frac{\log(x^2+1)}{x^4+x^2+1}\,dx$$ $$J(a)=\int_{0}^{\infty} \frac{\log(ax^2+1)}{x^4+x^2+1}\,dx$$ $$ J'(a)=\int_{0}^{\infty}\frac{x^2}{\left(x^4+x^2+1\right) \left(a x^2+1\right)}$$ Use the roots of unity and write (shorter) $$\frac{x^2}{\left(x^4+x^2+1\right) \left(a x^2+1\right)}=\frac{x^2}{(x^2-r)(x^2-s) \left(a x^2+1\right)}=$$ $$-\frac{a}{(a r+1) (a s+1) \left(a x^2+1\right)}+\frac{r}{(a r+1) (r-s) \left(x^2-r\right)}-\frac{s}{(a s+1) (r-s) \left(x^2-s\right)}$$ Simple integrals; using the bounds $$J'(a)=\frac {\pi \sqrt 3} 6\,\, \frac{a-\sqrt{3a} +1}{a^2-a+1}$$ $$\int_0^1 \frac{a-\sqrt{3a} +1}{a^2-a+1}\,da=2\int_0^1 \frac {db}{ b^2+b\,\sqrt{3}+1}=\log \left(2+\sqrt{3}\right)-\frac{\pi }{2 \sqrt{3}}$$ Recombining $$I=\frac{\pi }{\sqrt{3}}\log \left(2+\sqrt{3}\right)-\frac{\pi ^2}{6}$$