I'm looking for a closed-form expression for the value of this integral:

$$I=int_0^pi frac{sin(x)}{sqrt{x^3+x+1}} dx$$

The graph of the integrand looks like this:

$hskip 2.4 in$

Numerically, the area is $0.875044...$ for which the Inverse Symbolic Calculator doesn't produce anything promising. My CAS finds neither an antiderivative nor a closed form for the definite integral, and my own manipulations haven't really got me anywhere either.

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,
$$a(n)= text {Min} left(x; π[x]-πleft[frac {x}{2}right]=nright)$$
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.
$$I = 2int frac {cosh J(t)}{cosh 3J(t)} sin left(-2 sqrt {frac {1}{3}} sinh J(t)right) dt$$
where,
$$J(t):=frac {1}{3} sinh^{-1} left[frac {3sqrt 3}{2} (1-t^2)right]$$
So , I am giving a series form solution. Consider,
$$F(x):=int frac {sin x}{sqrt {1+x(1+x^2)}} dx……(1)$$
Now, for $x<1$,
$$[1+x(1+x^2)]^{-frac {1}{2}} = -sum_{k=0}^∞ C^{k-frac {1}{2}}_{-frac {1}{2}}x^k(1+x^2)^k ……(2)$$
Plugging $(2)$ into $(1)$ we get,
$$F(x) = -sum_{k=0}^∞ C^{k-frac {1}{2}}_{-frac {1}{2}} G(k,x) ……(3),$$
where ,
$$G(k,x):= int x^k sin x (1+x^2)^k dx ……(4)$$
But,
$$(1+x^2)^k= sum_{r=0}^k C^k_r x^{2r} ……(5)$$
Plugging $(5)$ into $(4)$ we get,
$$G(k,x) = sum_{r=0}^k C^k_r H(r,k,x) ……(6),$$
where,
$$H(r,k,x) := int x^{k+2r} sin x dx$$
$$= - frac {Gamma (k+2r+1, ix) + (-1)^{k+2r}Gamma (k+2r+1, -ix)}{2(-1)^{frac {5}{2} (k+2r)}} …… (7)$$
Hence,
$$F(x)=sum_{k=0}^∞ C^{k-frac {1}{2}}_{-frac {1}{2}} sum_{r=0}^k C^k_r frac {Gamma (k+2r+1, ix) + (-1)^{k+2r}Gamma (k+2r+1, -ix)}{2(-1)^{frac {5}{2} (k+2r)}}……(8)$$
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,
$$F(x)=-sum_{k=0}^∞ C^{k-frac {1}{2}}_{-frac {1}{2}} sum_{r=0}^k C^k_r frac {x^{k+2r+2}}{k+2r+2} {}_1text {F}_2 left(frac {k+2r+4}{2},frac {3}{2} ; frac {k+2r+2}{2} ; -frac {x^2}{4}right)$$

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:
$$int frac {sin x}{sqrt {x^3}} dx = sqrt {8π} text {C} left(sqrt {frac {2x}{π}}right) - frac {2sin x}{sqrt x} + c$$
The integrand is almost the same, except for the extra linear term $x+1$ under square root.