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 $(1)?$

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\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}
