Let's start with this Pfaff transformation for $\,a=\frac 14,b=\frac34,c=\frac23$ :
$$\tag{1}_2F_1\left(a,b\;;c\;;z\right)=(1-z)^{-a}\;_2F_1\left(a,c-b\;;c\;;\frac z{z-1}\right)$$
The 'Darboux evaluation' $(42)$ of Vidunas' "Transformations of algebraic Gauss hypergeometric functions" is :
$$\tag{2}_2F_1\left(\frac14,-\frac 1{12};\,\frac23;\,\frac {x(x+4)^3}{4(2x-1)^3}\right)=(1-2x)^{-1/4}$$
Solving $\,\displaystyle\frac {x(x+4)^3}{4(2x-1)^3}=\frac z{z-1}\,$ gives :
$$\tag{3}z=\frac{x(x+4)^3}{(x^2-10x-2)^2} $$
that we will use as :
$$\tag{4}z-1=\frac {4(2x-1)^3}{(x^2-10x-2)^2}$$
while $(1)$ and $(2)$ return :
$$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[(z-1)(2x-1)\right]^{-1/4}$$
so that :
$$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[\frac{4\,(2x-1)^4}{(x^2-10x-2)^2}\right]^{-1/4}$$
and (up to a minus sign) :
$$\tag{5}_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)^2=-\frac{x^2-10x-2}{2\,(2x-1)^2}$$
and indeed the substitution of $(z-1)$ and $_2F_1()^2$ with $(4)$ and $(5)$ in your formula gives :
$$27\,(z-1)^2\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^8+18\,(z-1)\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^4-8\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^2=1$$
Many other formulae of this kind may be deduced using Vidunas' paper.
Consider the hypergeometric equation with parameters $(a,b,c)=\left(\frac16,\frac12,\frac13\right)$, and build from its two canonical solutions near $z=0$ the vector
$$\vec{y}(z)=\left(\begin{array}{c}
y_1 \\ y_2
\end{array}\right)=\left(\begin{array}{c}
_2F_1(a,b;c;z) \\ z^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;z)
\end{array}\right).\tag{1}$$
This is a single-valued vector function on $\mathbb{C}\backslash\{(-\infty,0]\cup[1,\infty)\}$. Its analytic continuation along a closed loop $\gamma$ gives rise to monodromy representation of $\pi_1(\mathbb{C}\backslash\{0,1\})$:
$$ \gamma\mapsto M_{[\gamma]},\qquad y(\gamma z)=M_{[\gamma]}y(z).$$
The monodromy group $G\subset GL(2,\mathbb{C})$ of the hypergeometric equation is generated by two matrices corresponding to simple loops around $0$ and $1$. In the case we are interested in these matrices are explicitly given by
$$M_0=\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{-2\pi i /3}\end{array}\right),\qquad
M_1=C\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{2\pi i /3}\end{array}\right)C^{-1},\tag{2}$$
where the connection matrix $C=\left(\begin{array}{cc} 1 & 2^{\frac43} \\ -2^{\frac83} & 8\end{array}\right)$. If $G$ is finite, then $\vec{y}(z)$ has a finite number of branches, and moreover (Schwarz 1872), is algebraic.
It is not difficult to check that the monodromy group $G$ generated by $M_0$, $M_1$ from (2) is indeed finite. In particular, note that
$$M_0^3=M_1^3=I,\qquad M_1^2=-M_0M_1M_0, $$ $$M_1M_0M_1=-M_0^2,\qquad M_1M_0^2M_1=M_0^2M_1M_0^2.$$
It turns out that $G$ has order $24$ and is isomorphic to the binary tetrahedral group:
$$G\cong 2T=\langle s,t\,|\,(st)^2=s^3=t^3\rangle, $$
where the generators can be identified as $s=M_0M_1M_0M_1M_0$, $t=M_1M_0M_1M_0M_1$.
Corollary: The hypergeometric functions in (1) are algebraic.
Algebraic solutions of the hypergeometric equations are classified by the so-called Schwarz table, and have been studied by many mathematicians, see e.g. the bibliography in this paper by R.Vidunas. Their explicit construction is somewhat involved but relatively straightforward - at least when the corresponding algebraic curve has genus $0$ (the genus can be determined independently from the Riemann-Hurwitz formula).
In our case the task simplifies even more as our parameter values can be obtained from the genus $0$ tetrahedral formula (2.4) of the above mentioned paper by a combination of a linear trasformation (sending $\frac56$ to $\frac43-\frac56=\frac12$) and differentiation (transforming $\frac43$ into $\frac13$). The result is
$$_2F_1\left(\frac16,\frac12;\frac13;-\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)=\frac{\sqrt{1-r^2}}{2r+1}.$$
Corollary: The antiderivative $\displaystyle\int \mathcal{R}\left(x,y(x)\right)dx$, where $y(x)={}_2F_1\left(\frac16,\frac12;\frac13;-x\right)$ and $\mathcal{R}(x,y)$ is rational in both arguments, can be expressed in terms of elementary functions.
Example:
The transformation $r\mapsto x(r)=\frac{r(r+2)^3}{(r+1)(1-r)^3}$ bijectively maps $(0,1)$ to $(0,\infty)$, and therefore the initial integral becomes
\begin{align}\mathcal{I}&=\int_0^1 \left(\frac{\sqrt{1-r^2}}{2r+1}\right)^{12}\left(\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)'dr=\\&=
2\int_0^1 \frac{(1+r)^4(1-r)^2(r+2)^2}{(2r+1)^{10}} dr=\\&=\frac{80\,663}{153\,090}.
\end{align}
Best Answer
After first multiplying and dividing the integrand by 2, substitute $x=\frac{t}{2}$:
$$I=\int_{0}^{\infty}\frac{2\ln{(x)}}{\sqrt{2x}\,\sqrt{2x+2}\,\sqrt{2x+1}}\mathrm{d}x=\int_{0}^{\infty}\frac{\ln{\left(\frac{t}{2}\right)}}{\sqrt{t}\,\sqrt{t+2}\,\sqrt{t+1}}\mathrm{d}t.$$
Next, substituting $t=\frac{1}{u}$ yields:
$$\begin{align} I &=-\int_{0}^{\infty}\frac{\ln{(2u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\ &=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-\int_{0}^{\infty}\frac{\ln{(u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\ &=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-I\\ \implies I&=-\frac{\ln{(2)}}{2}\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}. \end{align}$$
Making the sequence of substitutions $x=\frac{u-1}{2}$, then $u=\frac{1}{t}$, and finally $t=\sqrt{w}$, puts this integral into the form of a beta function:
$$\begin{align} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}} &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\ &=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\ &=\frac12\int_{0}^{1}\frac{\mathrm{d}w}{w^{3/4}\,\sqrt{1-w}}\\ &=\frac12\operatorname{B}{\left(\frac14,\frac12\right)}\\ &=\frac12\frac{\Gamma{\left(\frac12\right)}\Gamma{\left(\frac14\right)}}{\Gamma{\left(\frac34\right)}}\\ &=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}} \end{align}$$
Hence,
$$I=-\frac{\ln{(2)}}{2}\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}=-\frac{\pi^{3/2}\,\ln{(2)}}{2^{3/2}\,\Gamma^2{\left(\frac34\right)}}.~~~\blacksquare$$
Possible Alternative: You could also derive the answer from the complete elliptic integral of the first kind instead of from the beta function by making the substitution $t=z^2$ instead of $t=\sqrt{w}$.
$$\begin{align} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}} &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\ &=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\ &=2\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{1-z^4}}\\ &=2\,K{(-1)}\\ &=\frac{\Gamma^2{\left(\frac14\right)}}{2\sqrt{2\pi}}\\ &=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}. \end{align}$$