How can we prove that $8\int_{0}^{\infty}{\ln x\over x}\left(e^{-x}-{1\over \sqrt[4]{1+8x}}\right)\mathrm dx=-32C+4\gamma^2-5\pi^2?$ An integral exhibits $3$ interesting constants.

$$8\int_{0}^{\infty}{\ln x\over x}\left(e^{-x}-{1\over \sqrt[4]{1+8x}}\right)\mathrm dx=-32\color{red}C+4\color{blue}\gamma^2-5\color{green}\pi^2\tag1$$

I am only interested in $(1)$, because rarely Catalan's constant and Euler-Masheroni's constant they appear together!
Making an attempt:
It is too difficult here to make an attempt, apart from differentiating under the integral
$$I(a)=8\int_{0}^{\infty}{\ln x\over x^{a}}\left(e^{-x}-{1\over \sqrt[4]{1+8x}}\right)\mathrm dx\tag2$$
$$I{'}(a)=8\int_{0}^{\infty}{1\over x^{a}}\left(e^{-x}-{1\over \sqrt[4]{1+8x}}\right)\mathrm dx\tag3$$
I guess $(3)$ diverges
How may we prove $(1)?$
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&8\int_{0}^{\infty}{\ln\pars{x} \over x}
\pars{\expo{-x} - {1 \over \root[4]{1 + 8x}}}\,\dd x
\,\,\,\stackrel{\mrm{IBP}}{=}\,\,\,
-4\int_{0}^{\infty}\ln^{2}\pars{x}
\bracks{-\expo{-x} + {2 \over \pars{1 + 8x}^{5/4}}}\,\dd x
\\[5mm] & =
4\ \overbrace{\pars{\gamma^{2} + {\pi^{2} \over 6}}}^{\ds{\Gamma''\pars{1}}} - 
8\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{1 + 8x}^{5/4}}\,\dd x =
4\gamma^{2} + {2 \over 3}\,\pi^{2} -
\left.8\,\partiald[2]{}{\mu}\int_{0}^{\infty}{x^{\mu} \over \pars{1 + 8x}^{5/4}}\,\dd x\,\right\vert_{\ \mu\ =\ 0}
\end{align}

\begin{align}
&\int_{0}^{\infty}{x^{\mu} \over \pars{1 + 8x}^{5/4}}\,\dd x =
\int_{0}^{\infty}x^{\mu}\bracks{{1 \over \Gamma\pars{5/4}}\int_{0}^{\infty}t^{1/4}\expo{-\pars{1 + 8x}t}\,\dd t}\,\dd x
\\[5mm] = &\
{1 \over \Gamma\pars{5/4}}
\int_{0}^{\infty}t^{1/4}\expo{-t}\int_{0}^{\infty}x^{\mu}\expo{-8tx}
\,\dd x\,\dd t =
{1 \over \Gamma\pars{5/4}}\int_{0}^{\infty}t^{1/4}\expo{-t}
\,{\Gamma\pars{\mu + 1} \over \pars{8t}^{\mu + 1}}\,\dd t
\\[5mm] = &\
{8^{-\mu - 1}\,\Gamma\pars{\mu + 1} \over \Gamma\pars{5/4}}
\int_{0}^{\infty}t^{-3/4 - \mu}\expo{-t}\,\dd t =
{1 \over \Gamma\pars{5/4}}\,8^{-\mu - 1}\,\Gamma\pars{\mu + 1}
\Gamma\pars{{1 \over 4} - \mu}
\end{align}

$$
\mbox{and}\ 
\left.\partiald[2]{}{\mu}\bracks{{1 \over \Gamma\pars{5/4}}\,
8^{-\mu - 1}\,\Gamma\pars{\mu + 1}\Gamma\pars{{1 \over 4} - \mu}}
\right\vert_{\ \mu\ =\ 0} = 4C + {17 \over 24}\,\pi^{2}
$$

\begin{align}
&8\int_{0}^{\infty}{\ln\pars{x} \over x}
\pars{\expo{-x} - {1 \over \root[4]{1 + 8x}}}\,\dd x =
4\gamma^{2} + {2 \over 3}\,\pi^{2} -
8\pars{4C + {17 \over 24}\,\pi^{2}}
\\[5mm] = &
\bbx{\ds{-32C + 4\gamma^{2} - 5\pi^{2}}}
\end{align}
A: For any $a\in(0,1)$ we have
$$ \int_{0}^{+\infty}\left(e^{-x}-\frac{1}{x+1}\right)\frac{dx}{x^a} = -\frac{\pi}{\sin(\pi a)}+\Gamma(1-a)\tag{1} $$
due to the Laplace transform and the integral definition of the $\Gamma$ function. By differentiating with respect to $a$ we get:
$$ \int_{0}^{+\infty}\left(e^{-x}-\frac{1}{x+1}\right)\frac{\log x}{x^a}\,dx = \psi(1-a)\Gamma(1-a)-\frac{\pi^2\cos(\pi a)}{\sin^2(\pi a)}\tag{2} $$
and by considering the limit as $a\to 1^-$ we get
$$ \int_{0}^{+\infty}\left(e^{-x}-\frac{1}{x+1}\right)\frac{\log x}{x}\,dx =\frac{6\gamma^2-\pi^2}{12}.\tag{3}$$
On the other hand
$$\begin{eqnarray*} \int \frac{1}{x}\left(\frac{1}{x+1}-\frac{1}{(1+8x)^{1/4}}\right)\,dx &=& \log(x)-\log(x+1)+2\arctan((1+8x)^{1/4})\\&+&\log(1-(1+8x)^{1/4})-\log(1+(1+8x)^{1/4})\end{eqnarray*}\tag{4} $$
hence the remaining part can be tackled by integration by parts, or by exploiting Mariusz Iwaniuk's comment.
