Definite integral involving hypergeometric functions I would like to know if the following definite integral of the product of elementary hypergeometric functions is known in closed form
$$
\int_0^1\,_2F_1\left(-n,1+n;1;x^3\right) \left[3 \, _2F_1\left(1-n,2+n;1;x^3\right)-\, _2F_1\left(1-n,2+n;2;x^3\right)\right]\, x \, \mathrm{d}x,
$$
with $n\geq1$.
 A: Let us denote the integral by $I_n$:
$$
I_n \equiv
\int_0^1\,_2F_1\left(-n,1+n;1;x^3\right) \left[3 \, _2F_1\left(1-n,2+n;1;x^3\right)-\, _2F_1\left(1-n,2+n;2;x^3\right)\right]\, x \, \mathrm{d}x\,.
$$
We first expand both hypergeometric functions on the right of the integral using the series representation around $x=0$. This yields
$$
I_n=\int_0^1 \sum_{j=0}^{+\infty}\frac{3 j x^{3 j+1} \Gamma (j+1) \, _2F_1\left(-n,n+1;1;x^3\right) (1-n)_j (n+2)_j}{j! (1)_j (2)_j} \mathrm{d}x=\sum_{j=0}^{+\infty}\int_0^1 \frac{3 j x^{3 j+1} \Gamma (j+1) \, _2F_1\left(-n,n+1;1;x^3\right) (1-n)_j (n+2)_j}{j! (1)_j (2)_j} \mathrm{d}x
$$
The resulting integral can then be expressed in terms of $_3 F_2$ generalised hypergeometric functions:
$$
I_n = \sum_{j=0}^{+\infty} \frac{3 j \Gamma (j+1) (1-n)_j (n+2)_j \, _3F_2\left(j+\frac{2}{3},-n,n+1;1,j+\frac{5}{3};1\right)}{(3 j+2) j! (1)_j (2)_j}\,.
$$
Now one can use the Pfaff–Saalschütz Balanced Sum (see for instance https://dlmf.nist.gov/16.4) to express the $_3F_2$ at unit value in terms of more Pochhammer symbols. In particular, we get
$$
I_n = -\sum_{j=0}^{+\infty} \frac{\Gamma \left(j+\frac{2}{3}\right) (1-n)_j \Gamma \left(-j+n+\frac{1}{3}\right) \Gamma (j+n+2)}{\Gamma \left(-j-\frac{2}{3}\right) \Gamma (j+1) \Gamma (j+2) \Gamma (n+2) \Gamma
   \left(j+n+\frac{5}{3}\right)}
$$
which one recognises as being related to a $_4 F_3$ generalised hypergeometric function evaluated at unit argument. After some simplifications, one finds
$$
I_n=\frac{\Gamma \left(\frac{5}{3}\right) \Gamma \left(n+\frac{1}{3}\right)}{\Gamma \left(\frac{1}{3}\right) \Gamma
   \left(n+\frac{5}{3}\right)} \, _4F_3\left(\frac{2}{3},\frac{5}{3},1-n,n+2;2,\frac{2}{3}-n,n+\frac{5}{3};1\right)\,.
$$
