Integration of $\int^{1}_{-1} \frac {1}{3} \sinh^{-1} \left( \frac {3\sqrt 3}{2} (1-t^2) \right) dt$ Recently I came across with respect to this  post of mine hyperbolic solution to the cubic equation for one real root given by
$$
t=-2\sqrt \frac {p}{3} \sinh \left( \frac {1}{3} \sinh^{-1} \left( \frac {3q}{2p} \sqrt \frac {3}{p} \right) \right)
$$
Intuitively I sought to find the related definitely integral,
$$
I=\int^{1}_{-1} \frac {1}{3} \sinh^{-1} \left( \frac {3\sqrt 3}{2} (1-t^2) \right) dt
$$
Unfortunately, there was no closed form solution. However, the Integral is amazingly near $\sqrt 2$.
$$
I=0.8285267994716327, \frac {I}{2} +1=1.4142633998
$$
To investigated more, I tried a heuristic expansion of the integral into Egyptian fractions. Although it gets problematic after the 4th term,
The first four terms are,
$$
\frac {I}{2} +1 = 1+ \frac {1}{2} - \frac {1}{12}-\frac {1}{416}
$$
Here the denominators can be given by,
$$
a_n = \sum_{k=0}^{n} { }^nC_k (2^n - 2^kq)^{n-k}q^k , q=\sqrt 2
$$
(Likewise, the denominators in the expansion for $\sqrt 2$ are related to Pell numbers, which makes me believe that my integral too is somewhat related to the numbers $a_n$.) Therefore, I am finding either a closed form or possibly a fast converging infinite series solution to the integral, just any of these. Thanks for any help.
The indefinite integral
For $t=\sin z$ and applying integration by parts, I get another, somewhat simpler, indefinite integral,
$$
\frac {\sin z}{3} \sinh^{-1} \left( \frac {3\sqrt 3}{2} \cos^2 z \right) + 2\sqrt 3 \int \frac {\sin^2 z \cos z dz}{\sqrt {27\cos^4 z + 4}}
$$
Then again I am stuck. Moreover, this expression ensures that my definite integral is an improper one.
Update
A closed solution in terms of incomplete elliptic integrals with complex arguments is, as given by a user in the comments section,
$$
\frac {4}{9} (9+2\sqrt 3 i) \left[ F \left( \sin^{-1} \sqrt {\frac {3}{31}(9+2\sqrt 3 i)} ; \frac {1}{31} (23-12\sqrt 3 i) \right)-
E \left( \sin^{-1} \sqrt {\frac {3}{31}(9+2\sqrt 3 i)} ; \frac {1}{31} (23-12\sqrt 3 i) \right) \right]
$$
However, I am still wondering how to transform this into a real number, especially the $a_n$ connection of the integral is fascinating my mind.

We have from symmetry that $$I=\frac23\int_0^1\sinh^{-1}\left[\frac{3\sqrt3}2(1-x^2)\right]dx$$
So we define 
$$f(a)=\int_0^1\sinh^{-1}[a(1-x^2)]dx$$
Then we recall that 
$$\sinh^{-1}(x)=x\,_2F_1\left(\frac12,\frac12;\frac32;-x^2\right)=\sum_{n\geq0}(-1)^n\frac{(1/2)_n^2}{(3/2)_n}\frac{x^{2n+1}}{n!}$$
so 
$$\sinh^{-1}[a(1-x^2)]=a(1-x^2)\,_2F_1\left(\frac12,\frac12;\frac32;-a^2(1-x^2)^2\right)\\
=\sum_{n\geq0}(-1)^n\frac{a^{2n+1}}{n!}\frac{(1/2)_n^2}{(3/2)_n}(1-x^2)^{2n+1}$$
so 
$$f(a)=\sum_{n\geq0}(-1)^n\frac{a^{2n+1}}{n!}\frac{(1/2)_n^2}{(3/2)_n}\int_0^1(1-x^2)^{2n+1}dx$$
For this integral, we use $x=\sin(t)$:
$$j_n=\int_0^1(1-x^2)^{2n+1}dx=\int_0^{\pi/2}\cos(t)^{4n+3}dt$$
I leave it as a challenge to you to show that $$\int_0^{\pi/2}\sin(t)^a\cos(t)^bdt=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$
choosing $b=4n+3$, $a=0$ we have 
$$j_n=\frac{\Gamma(1/2)\Gamma(2n+2)}{2\Gamma(2n+5/2)}$$
Then defining 
$$t_n=\frac{(1/2)_n^2}{(3/2)_n}j_n$$
we have $$\frac{t_{n+1}}{t_n}=\frac{(n+\frac12)^2(n+1)}{(n+\frac74)(n+\frac54)}$$
Which gives $$f(a)=a\,_3F_2\left(\frac12,\frac12,1;\frac74,\frac54;-a^2\right)$$
And since $I=\frac23f(3\sqrt3/2)$ we have (assuming I've made no mistakes),
$$I=\sqrt3\,_3F_2\left(\frac12,\frac12,1;\frac74,\frac54;-\frac{27}4\right)$$
