Prove ${\large\int}_0^\infty\frac{\ln x}{\sqrt{x}\ \sqrt{x+1}\ \sqrt{2x+1}}dx\stackrel?=\frac{\pi^{3/2}\,\ln2}{2^{3/2}\Gamma^2\left(\tfrac34\right)}$ I discovered the following conjecture by evaluating the integral numerically and then using some inverse symbolic calculation methods to find a possible closed form:
$$\int_0^\infty\frac{\ln x}{\sqrt{x\vphantom{1}}\ \sqrt{x+1}\ \sqrt{2x+1}}dx\stackrel{\color{#808080}?}=\frac{\pi^{3/2}\,\ln2}{2^{3/2}\,\Gamma^2\left(\tfrac34\right)}.\tag1$$
The equality holds numerically with a precision of at least $1000$ decimal digits. But so far I was not able to find a proof of it.
Because the integral can be represented as a derivative of a hypergeometic function with respect to its parameter, the conjecture can be rewritten as
$$\frac{d}{da}{_2F_1}\left(a,\ \tfrac12;\ 1;\ \tfrac12\right)\Bigg|_{a=\frac12}\stackrel{\color{#808080}?}=\frac{\sqrt\pi\,\ln2}{2\,\Gamma^2\left(\tfrac34\right)}\tag2$$
or, using a series expansion of the hypergeometric function, as
$${\large\sum}_{n=0}^\infty\frac{H_{n-\frac12}\ \Gamma^2\left(n+\tfrac12\right)}{2^n\ \Gamma^2\left(n+1\right)}\stackrel{\color{#808080}?}=-\frac{3\,\pi^{3/2}\,\ln2}{2\,\Gamma^2\left(\tfrac34\right)}\tag3,$$
where $H_q$ is the generalized harmonic number, $H_q=\gamma+\psi_0\left(q+1\right).$
Could you suggest any ideas how to prove this?
 A: 
$$I:=\int_{0}^{\infty}\frac{\ln{(x)}}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\mathrm{d}x.$$

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 $t=\sqrt{w}$.
$$\begin{align}
\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\
&=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\
&=2\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{1-z^4}}\\
&=2\,K{(-1)}\\
&=\frac{\Gamma^2{\left(\frac14\right)}}{2\sqrt{2\pi}}\\
&=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}.
\end{align}$$
