Recently I came across with respect to this post of mine hyperbolic solution to the cubic equation for one real root given by
$$
t=-2sqrt frac p3 sinh left( frac 13 sinh^-1 left( frac 3q2p sqrt frac 3p right) right)
$$
Intuitively I sought to find the related definitely integral,
$$
I=int^1_-1 frac 13 sinh^-1 left( frac 3sqrt 32 (1-t^2) right) dt
$$
Unfortunately, there was no closed form solution. However, the Integral is amazingly near $sqrt 2$.
$$
I=0.8285267994716327, frac I2 +1=1.4142633998
$$
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,
$$
frac I2 +1 = 1+ frac 12 - frac 112-frac 1416
$$
Here the denominators can be given by,
$$
a_n = sum_k=0^n ^nC_k (2^n - 2^kq)^n-kq^k , q=sqrt 2
$$
(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,
$$
frac sin z3 sinh^-1 left( frac 3sqrt 32 cos^2 z right) + 2sqrt 3 int frac sin^2 z cos z dzsqrt 27cos^4 z + 4
$$
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,
$$
frac 49 (9+2sqrt 3 i) left[ F left( sin^-1 sqrt frac 331(9+2sqrt 3 i) ; frac 131 (23-12sqrt 3 i) right)-
E left( sin^-1 sqrt frac 331(9+2sqrt 3 i) ; frac 131 (23-12sqrt 3 i) right) right]
$$
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 $$I=frac23int_0^1sinh^-1left[frac3sqrt32(1-x^2)right]dx$$
So we define
$$f(a)=int_0^1sinh^-1[a(1-x^2)]dx$$
Then we recall that
$$sinh^-1(x)=x,_2F_1left(frac12,frac12;frac32;-x^2right)=sum_ngeq0(-1)^nfrac(1/2)_n^2(3/2)_nfracx^2n+1n!$$
so
$$sinh^-1[a(1-x^2)]=a(1-x^2),_2F_1left(frac12,frac12;frac32;-a^2(1-x^2)^2right)\
=sum_ngeq0(-1)^nfraca^2n+1n!frac(1/2)_n^2(3/2)_n(1-x^2)^2n+1$$
so
$$f(a)=sum_ngeq0(-1)^nfraca^2n+1n!frac(1/2)_n^2(3/2)_nint_0^1(1-x^2)^2n+1dx$$
For this integral, we use $x=sin(t)$:
$$j_n=int_0^1(1-x^2)^2n+1dx=int_0^pi/2cos(t)^4n+3dt$$
I leave it as a challenge to you to show that $$int_0^pi/2sin(t)^acos(t)^bdt=fracGamma(fraca+12)Gamma(fracb+12)2Gamma(fraca+b2+1)$$
choosing $b=4n+3$, $a=0$ we have
$$j_n=fracGamma(1/2)Gamma(2n+2)2Gamma(2n+5/2)$$
Then defining
$$t_n=frac(1/2)_n^2(3/2)_nj_n$$
we have $$fract_n+1t_n=frac(n+frac12)^2(n+1)(n+frac74)(n+frac54)$$
Which gives $$f(a)=a,_3F_2left(frac12,frac12,1;frac74,frac54;-a^2right)$$
And since $I=frac23f(3sqrt3/2)$ we have (assuming I've made no mistakes),
$$I=sqrt3,_3F_2left(frac12,frac12,1;frac74,frac54;-frac274right)$$