I have been trying to find the closed form for ${}_2F_1\left(1, 2n+2; 2n+3; -\frac{1}{a} \right)$,
Background : I'm trying to solve the below integral,
$$I=\int_0^1 \frac{1}{x + a} \ln \left(\frac{1 + x}{1 - x} \right) \,dx, a\in \mathbb{C}$$
Here's how I am intending to solve,
$$\color{red}{\ln\left(\frac{1+x}{1-x}\right)=2\sum_{n\ge0}\frac{x^{2n+1}}{2n+1}}$$
$$I=2\sum_{n\ge0}\frac{1}{2n+1}\int_0^1 \frac{x^{2n+1}}{a+x} \,dx$$
$$\color{red}{\int_0^u \frac{x^{\mu - 1} \, dx}{(1 + \beta x)^\nu} = \frac{u^\mu}{\mu} \, {}_2F_1\left(\nu, \mu; 1 + \mu; -\beta u\right)}$$
$$I=\frac2a\sum_{n\ge0}\frac{{}_2F_1\left(1, 2n+2; 2n+3; -\frac{1}{a} \right)}{(2n+1)(2n+2)}$$
Now, I am not able to move forward.
It seems that the closed form I'm trying to look for is somewhat of the below form,
$${}_2F_1\left(a, b; b+1; z \right)=b z^{-b} \beta_z(b, 1 - a)$$
So, we can put in $a=1,b=2n+2, z=-\frac{1}{a}$, but it does not seem to be easy.
$${}_2F_1\left(1, 2n+2; 2n+3; {-\frac{1}{a}} \right)=(2n+2) \left(\frac{-1}{a}\right)^{-(2n+2)} \beta_{-\frac{1}{a}}(2n+2, 0)$$
$$I=\frac2 {a^3}\sum_{n\ge0}\frac{\beta_{-\frac{1}{a}}(2n+2, 0)}{a^{2n}(2n+1)}$$
