The closed form of $\lim_{x\to\frac{4}{3}}\frac{\partial}{\partial x}\left[\,_2{\rm{F}}_1\left(\frac{1}{3},1;x;-1\right)\right]$

(Update: Closed form found)

Problem. Show that the limit $L$ defined below has the following closed form value: $$L:=\lim_{x\to\frac43}\frac{\partial}{\partial x}\left[{_2F_1}{\left(\frac13,1;x;-1\right)}\right]\\ =\color{blue}{\frac{\pi}{\sqrt{3}}+\ln{(2)}+\frac{\pi^2}{6}-\frac{\ln^2{(2)}}{3}-\frac{\pi\ln{\left(\frac{27}{4}\right)}}{6\sqrt{3}}-\frac{1}{4}\Psi^{(1)}{\left(\frac13\right)}-\Re{\left[\operatorname{Li}_{2}{\left(\frac{\sqrt{3}}{2}e^{\frac{i\pi}{6}}\right)}\right]}}\\ \approx 0.097647124907961878899227966043695113852476374.$$

Hint. Here's a plan of attack for evaluating $L$ that is pretty straightforward. Represent the hypergeometric function as an integral, and then systematically calculate the derivative followed by the limit; the result is a reduction of $L$ to a sum of two definite integrals of elementary functions. We shall use Euler's integral representation, valid for $\Re{(c)}>\Re{(b)}>0\land|\arg{(1-z)}|<\pi$:

$$\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)}=\int_{0}^{1}t^{b-1}(1-t)^{c-b-1}(1-z\,t)^{-a}\,\mathrm{d}t.$$

Let $a=\frac13$, $b=1$, $c=x$, and $z=-1$. We then have $|\arg{(1-z)}|\bigg{|}_{z=-1}=|\arg{(2)}|=0<\pi$, and we also have $\Re{(b)}\bigg{|}_{b=1}=\Re{(1)}=1>0$. So for $\Re{(x)}>1$, we have:

$${_2F_1}{\left(\frac13,1;x;-1\right)}=\frac{1}{\operatorname{B}{\left(1,x-1\right)}}\int_{0}^{1}(1-t)^{x-2}(1+t)^{-\frac13}\,\mathrm{d}t\\ =(x-1)\int_{0}^{1}\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\,\mathrm{d}t.$$

Using the product rule to take the derivative and using Leibniz's integral rule to differentiate the integral under the integral sign gives us:

$$\begin{align} \frac{\partial}{\partial x}\left[{_2F_1}{\left(\frac13,1;x;-1\right)}\right] &=\frac{\partial}{\partial x}\left[(x-1)\int_{0}^{1}\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\,\mathrm{d}t\right]\\ &=\int_{0}^{1}\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\,\mathrm{d}t+(x-1)\frac{\partial}{\partial x}\int_{0}^{1}\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\,\mathrm{d}t+(x-1)\int_{0}^{1}\frac{\partial}{\partial x}\left[\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\right]\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\,\mathrm{d}t+(x-1)\int_{0}^{1}\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\ln{\left(1-t\right)}\,\mathrm{d}t. \end{align}$$

Finally, we arrive at a representation for $L$ as a sum of two integrals by taking the limit as $x\to\frac43$, which can be found straightforwardly by substituting $\frac43$ everywhere for $x$:

$$\begin{align} L &=\lim_{x\to\frac43}\frac{\partial}{\partial x}\left[{_2F_1}{\left(\frac13,1;x;-1\right)}\right]\\ &=\lim_{x\to\frac43}\left[\int_{0}^{1}\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\,\mathrm{d}t+(x-1)\int_{0}^{1}\frac{(1-t)^{x-2}}{\sqrt[3]{1+t}}\ln{\left(1-t\right)}\,\mathrm{d}t\right]\\ &=\int_{0}^{1}\frac{(1-t)^{-\frac23}}{\sqrt[3]{1+t}}\,\mathrm{d}t+\frac13\int_{0}^{1}\frac{(1-t)^{-\frac23}}{\sqrt[3]{1+t}}\ln{\left(1-t\right)}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{(1-t)^{\frac23}\,\sqrt[3]{1+t}}+\frac13\int_{0}^{1}\frac{\ln{\left(1-t\right)}}{(1-t)^{\frac23}\,\sqrt[3]{1+t}}\,\mathrm{d}t\\ &=:I+J, \end{align}$$

where $I$ and $J$ have introduced to denote the two integrals in the second-to-last line above for convenient reference.

The integral $I$ actually has a very simple closed form:

$$I=\frac{\pi}{\sqrt{3}}+\ln{(2)}\approx 2.5069465448.$$

To derive this value, start by making the sequence of substitutions $\frac{t}{1-t}=u$, $\sqrt[3]{1+2u}=v$, and $\frac{1}{v}=w$:

$$\begin{align} I &=\int_{0}^{1}\frac{\mathrm{d}t}{(1-t)^{\frac23}\,\sqrt[3]{1+t}}\\ &=\int_{0}^{\infty}\frac{\mathrm{d}u}{(1+u)^2\left(\frac{1}{1+u}\right)^{\frac23}\,\sqrt[3]{\frac{1+2u}{1+u}}}\\ &=\int_{0}^{\infty}\frac{\mathrm{d}u}{(1+u)\,\sqrt[3]{1+2u}}\\ &=\int_{1}^{\infty}\frac{\frac32v^2\,\mathrm{d}v}{\frac12(1+v^3)\,v}\\ &=3\int_{1}^{\infty}\frac{v\,\mathrm{d}v}{1+v^3}\\ &=3\int_{0}^{1}\frac{\mathrm{d}w}{1+w^3}\\ &=3\left[\frac{1}{\sqrt{3}}\arctan{\left(\frac{2w-1}{\sqrt{3}}\right)}+\frac16\ln{\left[\frac{(1+w)^3}{1+w^3}\right]}\right]_{0}^{1}\\ &=\left[\sqrt{3}\arctan{\left(\frac{2w-1}{\sqrt{3}}\right)}+\frac12\ln{\left[\frac{(1+w)^3}{1+w^3}\right]}\right]_{0}^{1}\\ &=\frac{\pi}{2\sqrt{3}}+\ln{(2)}-\left(-\frac{\pi}{2\sqrt{3}}\right)\\ &=\frac{\pi}{\sqrt{3}}+\ln{(2)}. \end{align}$$

Thus, the problem of finding a closed form for $L$ reduces to that of finding a closed form evaluation of a single integral, $\int_{0}^{1}\frac{\ln{\left(1-t\right)}}{(1-t)^{\frac23}\,\sqrt[3]{1+t}}\,\mathrm{d}t$. Given the similarity of this last integral to others which have been successfully evaluated on this site, I have confidence a closed form may be found for it.

( Edit: Reduction of $J$ to the integral $\int_{0}^{1}\frac{\ln{\left(1+w^3\right)}}{1+w^3}\,\mathrm{d}w$.)

The same sequence of substitutions which we used to evaluate $I$, $\frac{t}{1-t}=u$, $\sqrt[3]{1+2u}=v$, and $\frac{1}{v}=w$, transform the integral $J$ to an integral of a product of a rational function with the logarithm of a rational function:

$$\begin{align} J &=\frac13\int_{0}^{1}\frac{\ln{\left(1-t\right)}}{(1-t)^{\frac23}\,\sqrt[3]{1+t}}\,\mathrm{d}t\\ &=\frac13\int_{0}^{\infty}\frac{\ln{\left(\frac{1}{1+u}\right)}}{\left(\frac{1}{1+u}\right)^{\frac23}\,\sqrt[3]{\frac{1+2u}{1+u}}}\cdot\frac{\mathrm{d}u}{(1+u)^2}\\ &=-\frac13\int_{0}^{\infty}\frac{\ln{\left(1+u\right)}}{(1+u)\,\sqrt[3]{1+2u}}\,\mathrm{d}u\\ &=-\frac13\int_{1}^{\infty}\frac{\ln{\left(\frac{1+v^3}{2}\right)}}{\frac12(1+v^3)\,v}\,\left(\frac32v^2\right)\,\mathrm{d}v\\ &=-\int_{1}^{\infty}\frac{v\ln{\left(\frac{1+v^3}{2}\right)}}{1+v^3}\,\mathrm{d}v\\ &=-\int_{1}^{0}\frac{1}{w}\cdot\frac{w^3\ln{\left(\frac{1+w^3}{2w^3}\right)}}{1+w^3}\cdot\frac{(-1)\,\mathrm{d}w}{w^2}\\ &=-\int_{0}^{1}\frac{\ln{\left(\frac{1+w^3}{2w^3}\right)}}{1+w^3}\,\mathrm{d}w\\ &=\ln{(2)}\int_{0}^{1}\frac{\mathrm{d}w}{1+w^3}+3\int_{0}^{1}\frac{\ln{\left(w\right)}}{1+w^3}\,\mathrm{d}w-\int_{0}^{1}\frac{\ln{\left(1+w^3\right)}}{1+w^3}\,\mathrm{d}w. \end{align}$$

The first integral of the last line above is of course just,

$$\ln{(2)}\int_{0}^{1}\frac{\mathrm{d}w}{1+w^3}=\frac{\ln{(2)}}{3}I=\frac{\pi\,\ln{(2)}}{3\sqrt{3}}+\frac{\ln^2{(2)}}{3}.$$

The value of the second integral can be written as a difference of values of the trigamma function:

$$\begin{align} 3\int_{0}^{1}\frac{\ln{\left(w\right)}}{1+w^3}\,\mathrm{d}w &=\frac13\int_{0}^{1}\frac{z^{-2/3}\ln{\left(z\right)}}{1+z}\,\mathrm{d}z\\ &=\frac13\left[\int_{0}^{1}\frac{z^{a-1}\ln{\left(z\right)}}{1+z}\,\mathrm{d}z\right]_{a=\frac13}\\ &=\frac13\left[\int_{0}^{1}\frac{\partial}{\partial a}\left(\frac{z^{a-1}}{1+z}\right)\,\mathrm{d}z\right]_{a=\frac13}\\ &=\frac13\left[\frac{d}{da}\int_{0}^{1}\left(\frac{z^{a-1}}{1+z}\right)\,\mathrm{d}z\right]_{a=\frac13}\\ &=\frac13\left[\frac{d}{da}\left(\frac{\Psi{\left(\frac{a+1}{2}\right)}-\Psi{\left(\frac{a}{2}\right)}}{2}\right)\right]_{a=\frac13}\\ &=\frac16\left[\frac{d}{da}\left(\Psi{\left(\frac{a+1}{2}\right)}-\Psi{\left(\frac{a}{2}\right)}\right)\right]_{a=\frac13}\\ &=\frac16\left[\frac12\left(\Psi^{(1)}{\left(\frac{a+1}{2}\right)}-\Psi^{(1)}{\left(\frac{a}{2}\right)}\right)\right]_{a=\frac13}\\ &=\left[\frac{1}{12}\left(\Psi^{(1)}{\left(\frac{a+1}{2}\right)}-\Psi^{(1)}{\left(\frac{a}{2}\right)}\right)\right]_{a=\frac13}\\ &=\frac{1}{12}\left(\Psi^{(1)}{\left(\frac23\right)}-\Psi^{(1)}{\left(\frac16\right)}\right)\\ &=\frac{2\pi^2}{9}-\frac12\Psi^{(1)}{\left(\frac13\right)}. \end{align}$$

This leaves leaves us with a representation of $J$ (and of $L$) as a sum of constants and one unevaluated integral, which we'll denote $J_0:=-\int_{0}^{1}\frac{\ln{\left(1+w^3\right)}}{1+w^3}\,\mathrm{d}w$:

$$J=\frac{\pi\,\ln{(2)}}{3\sqrt{3}}+\frac{\ln^2{(2)}}{3}+\frac{2\pi^2}{9}-\frac12\Psi^{(1)}{\left(\frac13\right)}-\int_{0}^{1}\frac{\ln{\left(1+w^3\right)}}{1+w^3}\,\mathrm{d}w\\ \implies L=\frac{\pi}{\sqrt{3}}+\ln{(2)}+\frac{2\pi^2}{9}+\frac{\ln^2{(2)}}{3}+\frac{\pi\,\ln{(2)}}{3\sqrt{3}}-\frac12\Psi^{(1)}{\left(\frac13\right)}+J_0.$$

The integration of $J_0$ is not as easy as its deceptively simple integrand might lead one to believe. Using properties of logarithms and partial fractions though, we can split up $J_0$ into a sum of integrals of the form $\int\frac{P_1(w)\ln{P_3(w)}}{P_2(w)}\,\mathrm{d}w$, where $P_{1,2,3}(w)$ are polynomial functions of $w$ of degree no higher than $2$:

$$\begin{align} -\int_{0}^{1}\frac{\ln{\left(1+w^3\right)}}{1+w^3}\,\mathrm{d}w &=-\frac13\int_{0}^{1}\frac{\ln{\left(1+w^3\right)}}{1+w}\,\mathrm{d}w-\frac13\int_{0}^{1}\frac{(2-w)\ln{\left(1+w^3\right)}}{1-w+w^2}\,\mathrm{d}w\\ &=-\frac13\int_{0}^{1}\frac{\ln{\left(1+w\right)}}{1+w}\,\mathrm{d}w-\frac13\int_{0}^{1}\frac{\ln{\left(1-w+w^2\right)}}{1+w}\,\mathrm{d}w\\ &~~~~~ -\int_{0}^{1}\frac{(2-w)\ln{\left(1+w\right)}}{3(1-w+w^2)}\,\mathrm{d}w-\int_{0}^{1}\frac{(2-w)\ln{\left(1-w+w^2\right)}}{3(1-w+w^2)}\,\mathrm{d}w\\ &=-\frac{\ln^2{(2)}}{6}-\frac13\int_{0}^{1}\frac{\ln{\left(1-w+w^2\right)}}{1+w}\,\mathrm{d}w\\ &~~~~~ -\int_{0}^{1}\frac{(2-w)\ln{\left(1+w\right)}}{3(1-w+w^2)}\,\mathrm{d}w-\int_{0}^{1}\frac{(2-w)\ln{\left(1-w+w^2\right)}}{3(1-w+w^2)}\,\mathrm{d}w\\ &=-\frac{\ln^2{(2)}}{6}+\frac13\int_{0}^{1}\frac{(2w-1)\ln{\left(1+w\right)}}{1-w+w^2}\,\mathrm{d}w\\ &~~~~~ -\int_{0}^{1}\frac{(2-w)\ln{\left(1+w\right)}}{3(1-w+w^2)}\,\mathrm{d}w-\int_{0}^{1}\frac{(2-w)\ln{\left(1-w+w^2\right)}}{3(1-w+w^2)}\,\mathrm{d}w\\ &=-\frac{\ln^2{(2)}}{6}+\int_{0}^{1}\frac{(w-1)\ln{\left(1+w\right)}}{1-w+w^2}\,\mathrm{d}w\\ &~~~~~ +\int_{0}^{1}\frac{(2w-1)\ln{\left(1-w+w^2\right)}}{6(1-w+w^2)}\,\mathrm{d}w-\int_{0}^{1}\frac{\ln{\left(1-w+w^2\right)}}{2(1-w+w^2)}\,\mathrm{d}w\\ &=-\frac{\ln^2{(2)}}{6}+\int_{0}^{1}\frac{(w-1)\ln{\left(1+w\right)}}{1-w+w^2}\,\mathrm{d}w\\ &~~~~~ +0-\frac12\int_{0}^{1}\frac{\ln{\left(1-w+w^2\right)}}{1-w+w^2}\,\mathrm{d}w\\ &=-\frac{\ln^2{(2)}}{6}+\int_{0}^{1}\frac{(w-1)\ln{\left(1+w\right)}}{1-w+w^2}\,\mathrm{d}w-\int_{0}^{1}\frac{\ln{\left(1-w+w^2\right)}}{2(1-w+w^2)}\,\mathrm{d}w\\ &=-\frac{\ln^2{(2)}}{6}+J_1+J_2.\\ \end{align}$$

$$\implies L=\frac{\pi}{\sqrt{3}}+\ln{(2)}+\frac{2\pi^2}{9}+\frac{\ln^2{(2)}}{6}+\frac{\pi\,\ln{(2)}}{3\sqrt{3}}-\frac12\Psi^{(1)}{\left(\frac13\right)}+J_1+J_2.$$

Substituting $w=\frac{1-x}{1+x}$ in the integrals $J_1$ and $J_2$, we get:

$$\begin{align} J_1 &=\int_{0}^{1}\frac{(w-1)\ln{\left(1+w\right)}}{1-w+w^2}\,\mathrm{d}w\\ &=-\int_{1}^{0}\frac{2x(1+x)\ln{\left(\frac{2}{1+x}\right)}}{1+3x^2}\cdot\frac{-2\,\mathrm{d}x}{(1+x)^2}\\ &=\int_{0}^{1}\frac{4x\ln{\left(\frac{1+x}{2}\right)}}{(1+x)(1+3x^2)}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{4x\ln{\left(1+x\right)}}{(1+x)(1+3x^2)}\,\mathrm{d}x-\int_{0}^{1}\frac{4\ln{(2)}\,x\,\mathrm{d}x}{(1+x)(1+3x^2)}\\ &=\int_{0}^{1}\frac{(3x+1)\ln{\left(1+x\right)}}{1+3x^2}\,\mathrm{d}x-\int_{0}^{1}\frac{\ln{\left(1+x\right)}}{1+x}\,\mathrm{d}x-4\ln{(2)}\int_{0}^{1}\frac{x\,\mathrm{d}x}{(1+x)(1+3x^2)}\\ &=\int_{0}^{1}\frac{(3x+1)\ln{\left(1+x\right)}}{1+3x^2}\,\mathrm{d}x-\frac{\ln^2{(2)}}{2}-\frac{\pi\ln{(2)}}{3\sqrt{3}}\\ &=-\frac{\ln^2{(2)}}{2}-\frac{\pi\ln{(2)}}{3\sqrt{3}}+\int_{0}^{1}\frac{(3x+1)\ln{\left(1+x\right)}}{1+3x^2}\,\mathrm{d}x,\\ \end{align}$$

and,

$$\begin{align} J_2 &=-\frac12\int_{0}^{1}\frac{\ln{\left(1-w+w^2\right)}}{1-w+w^2}\,\mathrm{d}w\\ &=-\frac12\int_{1}^{0}\frac{(1+x)^2\ln{\left(\frac{1+3x^2}{(1+x)^2}\right)}}{1+3x^2}\cdot\frac{-2\,\mathrm{d}x}{(1+x)^2}\\ &=\int_{0}^{1}\frac{\ln{\left(\frac{(1+x)^2}{1+3x^2}\right)}}{1+3x^2}\,\mathrm{d}x\\ &=2\int_{0}^{1}\frac{\ln{\left(1+x\right)}}{1+3x^2}\,\mathrm{d}x-\int_{0}^{1}\frac{\ln{\left(1+3x^2\right)}}{1+3x^2}\,\mathrm{d}x.\\ \end{align}$$

Thus, adding $J_1$ and $J_2$ yields:

$$\begin{align} J_1+J_2 &=-\frac{\ln^2{(2)}}{2}-\frac{\pi\ln{(2)}}{3\sqrt{3}}+3\int_{0}^{1}\frac{(x+1)\ln{\left(1+x\right)}}{1+3x^2}\,\mathrm{d}x-\int_{0}^{1}\frac{\ln{\left(1+3x^2\right)}}{1+3x^2}\,\mathrm{d}x\\ &=-\frac{\ln^2{(2)}}{2}-\frac{\pi\ln{(2)}}{3\sqrt{3}}+3\int_{0}^{1}\frac{(x+1)\ln{\left(1+x\right)}}{1+3x^2}\,\mathrm{d}x-\frac{1}{\sqrt{3}}\int_{0}^{\sqrt{3}}\frac{\ln{\left(1+z^2\right)}}{1+z^2}\,\mathrm{d}z\\ &=:-\frac{\ln^2{(2)}}{2}-\frac{\pi\ln{(2)}}{3\sqrt{3}}+K_1+K_2.\\ \end{align}$$

$$\implies L=\frac{\pi}{\sqrt{3}}+\ln{(2)}+\frac{2\pi^2}{9}-\frac{\ln^2{(2)}}{3}-\frac12\Psi^{(1)}{\left(\frac13\right)}+K_1+K_2.$$

The integral $K_2$ can readily be evaluated using the trigonometric substitution $z=\tan{\frac{\varphi}{2}}$, as well as series techniques to convert the resulting log-trig integral into a sum of dilogarithmic values:

$$\begin{align} K_2 &=-\frac{1}{\sqrt{3}}\int_{0}^{\sqrt{3}}\frac{\ln{\left(1+z^2\right)}}{1+z^2}\,\mathrm{d}z\\ &=\frac{1}{\sqrt{3}}\int_{0}^{\frac{2\pi}{3}}\ln{\left(\cos{\frac{\varphi}{2}}\right)}\,\mathrm{d}\varphi\\ &=\frac{1}{\sqrt{3}}\int_{0}^{\frac{2\pi}{3}}\ln{\left(\frac{2\cos{\frac{\varphi}{2}}}{2}\right)}\,\mathrm{d}\varphi\\ &=\frac{1}{\sqrt{3}}\int_{0}^{\frac{2\pi}{3}}\ln{\left(2\cos{\frac{\varphi}{2}}\right)}\,\mathrm{d}\varphi-\frac{1}{\sqrt{3}}\int_{0}^{\frac{2\pi}{3}}\ln{(2)}\,\mathrm{d}\varphi\\ &=-\frac{2\pi\ln{(2)}}{3\sqrt{3}}+\frac{1}{\sqrt{3}}\int_{0}^{\frac{2\pi}{3}}\ln{\left(2\cos{\frac{\varphi}{2}}\right)}\,\mathrm{d}\varphi\\ &=-\frac{2\pi\ln{(2)}}{3\sqrt{3}}+\frac{1}{\sqrt{3}}\int_{0}^{\frac{2\pi}{3}}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{\cos{k\varphi}}{k}\,\mathrm{d}\varphi\\ &=-\frac{2\pi\ln{(2)}}{3\sqrt{3}}+\frac{1}{\sqrt{3}}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{1}{k}\int_{0}^{\frac{2\pi}{3}}\cos{k\varphi}\,\mathrm{d}\varphi\\ &=-\frac{2\pi\ln{(2)}}{3\sqrt{3}}+\frac{1}{\sqrt{3}}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{\sin{\left(\frac{2\pi k}{3}\right)}}{k^2}\\ &=-\frac{2\pi\ln{(2)}}{3\sqrt{3}}-\frac{1}{\sqrt{3}}\sum_{k=1}^{\infty}\frac{(-1)^{k}\sin{\left(\frac{2\pi k}{3}\right)}}{k^2}\\ &=-\frac{2\pi\ln{(2)}}{3\sqrt{3}}-\frac{1}{\sqrt{3}}\sum_{k=1}^{\infty}\frac{\cos{\left(\pi k\right)}\sin{\left(\frac{2\pi k}{3}\right)}}{k^2}\\ &=-\frac{2\pi\ln{(2)}}{3\sqrt{3}}-\frac{1}{\sqrt{3}}\sum_{k=1}^{\infty}\frac{\sin{\left(\frac{5\pi k}{3}\right)}+\sin{\left(\frac{-\pi k}{3}\right)}}{2k^2}\\ &=-\frac{2\pi\ln{(2)}}{3\sqrt{3}}+\frac{1}{\sqrt{3}}\sum_{k=1}^{\infty}\frac{\sin{\left(\frac{\pi k}{3}\right)}}{k^2}\\ &=-\frac{2\pi\ln{(2)}}{3\sqrt{3}}+\frac{1}{\sqrt{3}}\Im{\left[\operatorname{Li}_{2}{\left(e^{\frac{i\pi}{3}}\right)}\right]}\\ &=-\frac{\pi^2}{9}-\frac{2\pi\ln{(2)}}{3\sqrt{3}}+\frac{1}{6}\Psi^{(1)}{\left(\frac13\right)}.\\ \end{align}$$

Edit #2

It was suggested that I simplify the last term of the final value involving the real part of a dilogarithm, $\Re{\left[\operatorname{Li}_{2}{\left(\frac{\sqrt{3}}{2}e^{\frac{i\pi}{6}}\right)}\right]}$, by re-expressing it as an ordinary dilogarithmic. Note that $\frac{\sqrt{3}}{2}e^{\frac{i\pi}{6}}=\frac34+i\frac{\sqrt{3}}{4}$. As I explained in my response here, this term is related to the dilogarithm with argument $-\frac13$:

$$\begin{align} \Re{\left[\operatorname{Li}_{2}{\left(\frac34+i\frac{\sqrt{3}}{4}\right)}\right]} &=\frac{7\pi^2}{72}+\frac{\ln^2{(3)}}{8}-\frac{\ln^2{\left(2\right)}}{2}+\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}.\\ \end{align}$$

Thus,

$$\begin{align} L &=\frac{\pi}{\sqrt{3}}+\ln{(2)}+\frac{\pi^2}{6}-\frac{\ln^2{(2)}}{3}-\frac{\pi\ln{\left(\frac{27}{4}\right)}}{6\sqrt{3}}-\frac{1}{4}\Psi^{(1)}{\left(\frac13\right)}-\Re{\left[\operatorname{Li}_{2}{\left(\frac{\sqrt{3}}{2}e^{\frac{i\pi}{6}}\right)}\right]}\\ &=\frac{\pi}{\sqrt{3}}+\ln{(2)}+\frac{5\pi^2}{72}+\frac{\ln^2{(2)}}{6}-\frac{\ln^2{(3)}}{8}-\frac{\pi\ln{\left(\frac{27}{4}\right)}}{6\sqrt{3}}-\frac{1}{4}\Psi^{(1)}{\left(\frac13\right)}-\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}.\\ \end{align}$$

Note that while the new expression is arguably simpler, it has the disadvantage of being less compact. Ultimately, I think which form one prefers is just a matter of taste.

Using formula $3.197(3)$ page $317$ from Table of Integrals, Series, and Products Seventh Edition by I.S. Gradshteyn and I.M. Ryzhik, we have

\begin{equation} \int_0^1 t^{\lambda-1} (1-t)^{\mu-1} (1-\beta t)^{-\nu}\ dt={\rm{B}}(\lambda,\mu) \,_2{\rm{F}}_1\left(\nu,\lambda;\lambda+\mu;\beta\right) \end{equation} for $\Re\, \lambda>0, \Re\, \mu>0, |\beta|<1$.

Therefore, by setting $\nu=\frac{1}{3},\,\lambda=1,\,$ and $\,\mu=x-1,$ as $\beta\to -1^+$ we have \begin{equation}\int_0^1 (1-t)^{x-2} (1+ t)^{-\frac{1}{3}}\ dt={\rm{B}}\left(1,x-1\right) \,_2{\rm{F}}_1\left(\frac{1}{3},1;x;-1\right)\tag1 \end{equation} Differentiating $(1)$ w.r.t. $x$ and taking the limit as $x\to\frac{4}{3}$, we obtain \begin{align*} \lim_{x\to\frac{4}{3}}\partial_x\int_0^1 (1-t)^{x-2} (1+ t)^{-\frac{1}{3}}\ dt&=-\sqrt{3}\pi-3\ln2+3\lim_{x\to\frac{4}{3}}\partial_x\,_2{\rm{F}}_1\left(\frac{1}{3},1;x;-1\right)\\ \int_0^1 \frac{\ln(1-t)}{(1-t)^{\frac{2}{3}} (1+ t)^{\frac{1}{3}}}\, dt&=-\sqrt{3}\pi-3\ln2+3\lim_{x\to\frac{4}{3}}\partial_x\,_2{\rm{F}}_1\left(\frac{1}{3},1;x;-1\right)\\ \lim_{x\to\frac{4}{3}}\partial_x\,_2{\rm{F}}_1\left(\frac{1}{3},1;x;-1\right)&=\frac{1}{3}\int_0^1 \frac{\ln(1-t)}{(1-t)^{\frac{2}{3}} (1+ t)^{\frac{1}{3}}}\, dt+\frac{\pi}{\sqrt{3}}+\ln2\\ &=I+\frac{\pi}{\sqrt{3}}+\ln2 \end{align*} Let's us try a magic substitution $2y=1+t$ to evaluate $I$. \begin{align*} I&=\frac{2}{3}\int_{\frac{1}{2}}^1 \frac{\ln(2-2y)}{(2-2y)^{\frac{2}{3}} (2y)^{\frac{1}{3}}}\, dy\\ &=\frac{1}{3}\left[\int_{\frac{1}{2}}^1 \frac{\ln2}{(1-y)^{\frac{2}{3}} y^{\frac{1}{3}}}\, dy+\int_{\frac{1}{2}}^1 \frac{\ln(1-y)}{(1-y)^{\frac{2}{3}} y^{\frac{1}{3}}}\, dy\right]\\ &=\frac{1}{3}\left[\int_0^{\frac{1}{2}} \frac{\ln2}{t^{\frac{2}{3}} (1-t)^{\frac{1}{3}}}\, dt+\int_0^{\frac{1}{2}} \frac{\ln t}{t^{\frac{2}{3}} (1-t)^{\frac{1}{3}}}\, dt\right]\quad\Rightarrow\quad t=1-y\\ &=\frac{1}{3}\left[{\rm{B}}_\frac{1}{2}\left(\frac{1}{3},\frac{2}{3}\right)\ln2+\lim_{x\to\frac{1}{3}}\partial_x{\rm{B}}_\frac{1}{2}\left(x,\frac{2}{3}\right)\right]\\ &=\frac{1}{3}\left[{\rm{B}}_\frac{1}{2}\left(\frac{1}{3},\frac{2}{3}\right)\ln2-\frac{9}{\sqrt[3]{2}}\,_3{\rm{F}}_2\left(\frac{1}{3},\frac{1}{3},\frac{1}{3};\frac{4}{3},\frac{4}{3};\frac{1}{2}\right)-{\rm{B}}_\frac{1}{2}\left(\frac{1}{3},\frac{2}{3}\right)\ln2\right]\\ &=-\frac{3}{\sqrt[3]{2}}\,_3{\rm{F}}_2\left(\frac{1}{3},\frac{1}{3},\frac{1}{3};\frac{4}{3},\frac{4}{3};\frac{1}{2}\right) \end{align*} where ${\rm{B}}_z(\cdot)$ is incomplete beta function and its derivation is obtained by Wolfram Alpha. Thus

\begin{align*} \lim_{x\to\frac{4}{3}}\partial_x\,_2{\rm{F}}_1\left(\frac{1}{3},1;x;-1\right)&=\frac{\pi}{\sqrt{3}}+\ln2-\frac{3}{\sqrt[3]{2}}\,_3{\rm{F}}_2\left(\frac{1}{3},\frac{1}{3},\frac{1}{3};\frac{4}{3},\frac{4}{3};\frac{1}{2}\right)\\ &\approx0.09764712490796187889922796604369511385247637400 \end{align*}

Sorry, I'm unable to get its closed-form without using hypergeometric function.

\begin{align} \small\frac{\partial}{\partial x}\left.{}_2\mathrm{F}_1\left(\left.\begin{matrix}\frac{1}{3},1\\x\end{matrix}\right|-1\right)\right|_{x=\frac{4}{3}} &\small=\frac{\partial}{\partial x}\left.(x-1)\int^1_0\frac{(1-t)^{x-2}}{(1+t)^\frac{1}{3}}dt\right|_{x=\frac{4}{3}}\tag{1a}\\ &\small=\int^1_0\frac{dt}{(1-t)^{\frac{2}{3}}(1+t)^{\frac{1}{3}}}+\frac{1}{3}\int^1_0\frac{\ln(1-t)}{(1-t)^{\frac{2}{3}}(1+t)^{\frac{1}{3}}}dt\tag{1b}\\ &\small=\left(1+\frac{1}{3}\ln{2}\right)\int^1_0\frac{t^{-\frac{2}{3}}}{1+t}dt+\frac{1}{3}\int^1_0\frac{t^{-\frac{2}{3}}\ln{t}}{1+t}dt-\frac{1}{3}\int^1_0\frac{t^{-\frac{2}{3}}\ln(1+t)}{1+t}dt\tag{1c}\\ &\small=\frac{\pi}{\sqrt{3}}+\ln{2}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{\ln^2{2}}{3}+\frac{2\pi^2}{9}-\frac{1}{2}\psi_1\left(\frac{1}{3}\right)-\int^1_0\frac{\ln(1+t^3)}{1+t^3}dt\tag{1d} \end{align} Explanation:

$(\text{1a})$: Used the integral representation of ${}_2\mathrm{F}_1$.
$(\text{1b})$: Applied the product rule.
$(\text{1c})$: Applied the substitution $t\mapsto\frac{1-t}{1+t}$.
$(\text{1d})$: Used the fact that $\int^1_0\frac{t^a}{1+t}dt=\frac{1}{2}\left(\psi_0\left(\frac{a}{2}+1\right)-\psi_0\left(\frac{a+1}{2}\right)\right)$, substituted $t\mapsto t^3$.

\begin{align} \small{\int^1_0\frac{\ln(1+t^3)}{1+t^3}dt} =&\ \small{4\int^1_{-1}\frac{\ln\left(\frac{1}{8}(x+3)(x^2+3)\right)}{(x+3)(x^2+3)}dx\tag{2a}}\\ =&\ \small{-\frac{\pi\ln{2}}{\sqrt{3}}-\frac{\ln^2{2}}{2}+\int^1_{-1}\frac{\ln(x+3)}{x^2+3}dx-\frac{1}{3}\int^1_{-1}\frac{x\ln(x+3)}{x^2+3}dx+\frac{1}{3}\int^1_{-1}\frac{\ln(x^2+3)}{x+3}dx}\\ &\small{+2\int^1_0\frac{\ln(x^2+3)}{x^2+3}dx\tag{2b}}\\ =&\ \small{-\frac{\pi\ln{2}}{\sqrt{3}}-\frac{5\ln^2{2}}{6}+\int^1_{-1}\frac{\ln(x+3)}{x^2+3}dx+\frac{1}{2}\int^1_{-1}\frac{\ln(x^2+3)}{x+3}dx+2\int^1_0\frac{\ln(x^2+3)}{x^2+3}dx}\\ &\tag{2c} \end{align} Explanation:

$(2\text{a})$ Used the substitution $x=2t-1$.
$(2\text{b})$ Laws of logarithms, Partial Fractions.
$(2\text{c})$ Integrated by parts.

\begin{align} \small{2\int^1_0\frac{\ln(x^2+3)}{x^2+3}dx} &\small=\frac{2}{\sqrt{3}}\int^\frac{\pi}{6}_0\left(\ln{3}-2\ln(\cos{x})\right).dx\tag{3a}\\ &\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}-\frac{4}{\sqrt{3}}\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\int^\frac{\pi}{6}_0\cos(2nx)dx\tag{3b}\\ &\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}-\frac{2}{\sqrt{3}}\sum^\infty_{n=1}\frac{(-1)^{n-1}\sin(\pi n/3)}{n^2}\\ &\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}-\sum^\infty_{n=0}\frac{1}{(6n+1)^2}+\sum^\infty_{n=0}\frac{1}{(6n+2)^2}-\sum^\infty_{n=0}\frac{1}{(6n+4)^2}+\sum^\infty_{n=0}\frac{1}{(6n+5)^2}\\ &\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}+\frac{1}{36}\left(-\psi_1\left(\frac{1}{6}\right)+\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)+\psi_1\left(\frac{5}{6}\right)\right)\tag{3c}\\ &\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}+\frac{4\pi^2}{27}-\frac{2}{9}\psi_1\left(\frac{1}{3}\right)\tag{3d}\\ \end{align} Explanation:

$(3\text{a})$ Used the substitution $x\mapsto\sqrt{3}\tan{x}$.
$(3\text{b})$ Used the Fourier series of $\ln(\cos{x})$.
$(3\text{c})$ Recognised the series representation of $\psi_1$.
$(3\text{d})$ Used the duplication & reflection formulae.

\begin{align} \small\frac{1}{2}\int^1_{-1}\frac{\ln(x^2+3)}{x+3}dx &\small=\Re\int^1_{-1}\frac{\ln(x+i\sqrt{3})}{x+3}dx\\ &\small=\Re\int^\frac{i}{\sqrt{3}}_{-\frac{1}{2}+\frac{i}{2\sqrt{3}}}\frac{\ln\left((3-i\sqrt{3})x\right)}{1+x}dx\tag{4a}\\ &\small=\Re\Bigg{[}\ln(1+x)\ln\left((3-i\sqrt{3})x\right)+\mathrm{Li}_2(-x)\Bigg{]}^\frac{i}{\sqrt{3}}_{-\frac{1}{2}+\frac{i}{2\sqrt{3}}}\tag{4b}\\ &\small=\ln^2{2}+\frac{\pi^2}{18}+\Re\left[\mathrm{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\mathrm{Li}_2\left(\frac{1}{2}-\frac{i}{2\sqrt{3}}\right)\right]\\ &\small=\ln^2{2}+\frac{\ln^2{3}}{8}-\frac{\pi^2}{72}+\frac{1}{4}\mathrm{Li}_2\left(-\frac{1}{3}\right)\tag{4c} \end{align}

Explanation:

$(4\text{a})$ Used the substitution $x\mapsto(3-i\sqrt{3})x-i\sqrt{3}$.
$(4\text{b})$ Integrated by parts.
$(4\text{c})$ Used dilogarithm duplication & reflection formulae.

\begin{align} \small\int^1_{-1}\frac{\ln(x+3)}{x^2+3}dx &\small=\frac{1}{\sqrt{3}}\int^\frac{\pi}{6}_{-\frac{\pi}{6}}\ln\left(\frac{2\sqrt{3}\sin\left(x+\frac{\pi}{3}\right)}{\cos{x}}\right)dx\tag{5a}\\ &\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{2\pi^2}{27}-\frac{1}{9}\psi_1\left(\frac{1}{3}\right)+\frac{1}{\sqrt{3}}\int^\frac{\pi}{2}_\frac{\pi}{6}\ln(2\sin{x})dx\tag{5b}\\ &\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{2\pi^2}{27}-\frac{1}{9}\psi_1\left(\frac{1}{3}\right)-\frac{1}{\sqrt{3}}\sum^\infty_{n=1}\frac{1}{n}\int^\frac{\pi}{2}_\frac{\pi}{6}\cos(2nx)dx\tag{5c}\\ &\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{2\pi^2}{27}-\frac{1}{9}\psi_1\left(\frac{1}{3}\right)+\frac{1}{2\sqrt{3}}\sum^\infty_{n=1}\frac{\sin(\pi n/3)}{n^2}\\ &\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{2\pi^2}{27}-\frac{1}{9}\psi_1\left(\frac{1}{3}\right)+\frac{1}{144}\left(12\psi_1\left(\frac{1}{3}\right)-8\pi^2\right)\tag{5d}\\ &\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{\pi^2}{54}-\frac{1}{36}\psi_1\left(\frac{1}{3}\right)\\ \end{align} Explanation:

$(5\text{a})$ Used the substitution $x\mapsto\sqrt{3}\tan{x}$.
$(5\text{b})$ Used $3\text{d}$, substituted $x\mapsto x-\frac{\pi}{3}$.
$(5\text{c})$ Used the Fourier seires of $\ln(2\sin{x})$.
$(5\text{d})$ Same explanation as $3\text{b}$ to $3\text{d}$.

Putting all of this together, $$\small\int^1_0\frac{\ln(1+t^3)}{1+t^3}dt=\frac{\pi\ln{3}}{2\sqrt{3}}+\frac{\ln^2{2}}{6}+\frac{\ln^2{3}}{8}+\frac{11\pi^2}{72}-\frac{1}{4}\psi_1\left(\frac{1}{3}\right)+\frac{1}{4}\mathrm{Li}_2\left(-\frac{1}{3}\right)$$ and \begin{align} \small\frac{\partial}{\partial x}\left.{}_2\mathrm{F}_1\left(\left.\begin{matrix}\frac{1}{3},1\\x\end{matrix}\right|-1\right)\right|_{x=\frac{4}{3}} &\small\color{red}{=\frac{\pi}{\sqrt{3}}+\ln{2}+\frac{\pi\ln{2}}{3\sqrt{3}}-\frac{\pi\ln{3}}{2\sqrt{3}}+\frac{\ln^2{2}}{6}-\frac{\ln^2{3}}{8}+\frac{5\pi^2}{72}-\frac{1}{4}\psi_1\left(\frac{1}{3}\right)-\frac{1}{4}\mathrm{Li}_2\left(-\frac{1}{3}\right)} \end{align}