Closed form for $\int_0^1\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\log\left(\frac{1+2x}{x\,(1-x)}\right)\,dx$
$$Q=\frac{\Gamma\left(\frac56\right)}{\Gamma\left(\frac13\right)}\cdot\frac{\sqrt{27}+\sqrt3\,\ln\left(16+8\,\sqrt3\right)-2\pi}{\sqrt[3]{76+44\,\sqrt3}}\cdot\sqrt\pi$$
Proof:
Looking through Table of Integrals, Series, and Products, $7^{th}$ Edition, I.S. Gradshteyn, I.M. Ryzhik, I noticed that the formula 3.255 could be taken as a starting point: $$\int_0^1\frac{x^{\mu+\frac12}(1-x)^{\mu-\frac12}}{\left(c+2\,b\,x-a\,x^2\right)^{\mu+1}}dx=\frac{\sqrt\pi}{\left(a+\left(\sqrt{c+2\,b-a}+\sqrt{c}\right)^2\right)^{\mu+\frac12}\sqrt{c+2\,b-a}}\cdot\frac{\Gamma\left(\mu+\frac12\right)}{\Gamma(\mu+1)}$$ Fixing the parameters $a=0,\ b=1,\ c=1$, we can make an observation that the integrand in $Q$ can be represented as a derivative of the integrand in 3.255: $$\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\ln\left(\frac{1+2\,x}{x\,(1-x)}\right)=\partial_\mu\left(\frac{x^{\mu+\frac12}(1-x)^{\mu-\frac12}}{(1+2\,x)^{\mu+1}}\right)_{\mu=\frac13}$$ Now we need to calculate the derivative of the right-hand side of 3.255: $$Q=\partial_\mu\left(\frac{\sqrt\pi}{\left(\sqrt3+1\right)^{2\mu+1}\sqrt3}\cdot\frac{\Gamma\left(\mu+\frac12\right)}{\Gamma(\mu+1)}\right)_{\mu=\frac13}\\=\sqrt{\frac\pi3}\left(\frac{\psi\left(\mu+\frac12\right)-\psi(\mu+1)-2\ln\left(\sqrt3+1\right)}{\left(\sqrt3+1\right)^{2\mu+1}}\cdot\frac{\Gamma\left(\mu+\frac12\right)}{\Gamma(\mu+1)}\right)_{\mu=\frac13}\\=\sqrt{\frac\pi3}\cdot\frac{\psi\left(\frac56\right)-\psi\left(\frac43\right)-2\ln\left(\sqrt3+1\right)}{\left(\sqrt3+1\right)^{\frac53}}\cdot\frac{\Gamma\left(\frac56\right)}{\Gamma\left(\frac43\right)}$$ Here $\psi(z)=\partial_z\ln\Gamma(z)$ is the digamma function. Using Gauss digamma theorem we can expand the values of the digamma function that appear in this formula: $$\psi\left(\frac56\right)=\frac{\pi\,\sqrt3}2-\frac{3\ln3}2-2\ln2-\gamma$$ $$\psi\left(\frac43\right)=3-\frac{\pi\,\sqrt3}6-\frac{3\ln3}2-\gamma$$ Plugging these values back to the previous formula, we get the desired result.
Addendum (By editor): The formula $3.255$ in G&R that Vladimir quoted, can be proved by 6 substitutions, that is: $$x\to 1-t, t\to \frac1u, u\to v+1, v\to w^2, w\to y \sqrt[4]{\frac a{a+b-c}}$$ and the final one is the a V. Moll's variant of Cauchy-Schlomilch transform (i.e. dealing with integrals involving $\left(\frac{x^2}{x^4+2ax^2+1}\right)^c$), which can be find in Theorem $4.1$ of:
Amdeberhan, T. , et al. "The Cauchy-Schlomilch transformation." Mathematics (2010).
This justifies the correctness of $3.255$.
The same integral without the log term is given by $$ Q(a,b,c) = \int_0^1 \frac{x^a}{(1-x)^b(1+2x)^c}\,dx = \frac{\Gamma(1+a)\Gamma(1-b)}{\Gamma(2+a-b)} F\left(\begin{array}{c}1+a,c\\2+a-b\end{array}\middle|-2\right), $$ because this integral is one of the common integral representations of the hypergeometric function, such as DLMF 15.6.1.
At the point $(a,b,c)=(\frac56,\frac16,\frac43)$, the integral you want is given by a sum of partial derivatives of $Q$ at that point: $$ (-\partial_a+\partial_b-\partial_c)Q(a,b,c). $$ Derivatives of the hypergeometric function with respect to parameters generally aren't expected to have a closed form, except in terms of functions such as Appell functions or Kampe de Feriet functions if those can be considered closed form.