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Following $\verb=@Alexandre Eremenko=$ above answer:
$\ds{\int_{0}^{\infty}\exp\pars{cx^{a} + Kx^{b}}\,\dd x
=\int_{\xi}^{\infty}
\exp\pars{c\bracks{\xi + x}^{a} + K\bracks{\xi + x}^{b}}\,\dd x}$ where $\xi$ satisfies $\ds{c\xi^{a} + K\xi^{b} = 0}$. The $\large{\rm OP}$ already calculated $\xi = \pars{K \over c}^{1/\pars{a  b}}$. Then,
\begin{align}
&\int_{0}^{\infty}\exp\pars{cx^{a} + Kx^{b}}\,\dd x \approx
\int_{\xi}^{\infty}\exp\pars{\bracks{ca\xi^{a  1} + Kb\xi^{b  1}}x}\,\dd x
\\[3mm]&=
{\exp\pars{ca\xi^{a}  Kb\xi^{b}}
\over ca\xi^{a  1} + Kb\xi^{b  1}}
=
{\exp\pars{ca\xi^{a}  Kb\xi^{b}} \over ca\xi^{a}  Kb\xi^{b}}\,\xi
\end{align}
Also, $\ds{\xi^{a} \sim K^{a/\pars{a  b}}\,,\quad}$
$\ds{\xi^{b} \sim K^{b/\pars{a  b}}\,,\quad}$
$\ds{K\xi^{b} \sim K^{a/\pars{a  b}}}$. Then,
\begin{align}
&ca\xi^{a}  Kb\xi^{b} = ca\bracks{\pars{K \over c}^{1/\pars{a  b}}}^{a}

Kb\bracks{\pars{K \over c}^{1/\pars{a  b}}}^{b}
\\[3mm]&=\bracks{c^{a/\pars{a  b}}\,a  c^{b/\pars{a  b}}\,b}K^{a/\pars{a  b}}
\end{align}
\begin{align}
\int_{0}^{\infty}\exp\pars{cx^{a} + Kx^{b}}&\approx
{\exp\pars{\bracks{c^{a/\pars{a  b}}\,a
 c^{b/\pars{a  b}}\,b}K^{a/\pars{a  b}}}
\over \bracks{c^{a/\pars{a  b}}\,a  c^{b/\pars{a  b}}\,b}K^{a/\pars{a  b}}}
\,c^{1/\pars{a  b}}K^{1/\pars{a  b}}
\end{align}
\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}\exp\pars{cx^{a} + Kx^{b}}}\\[3mm]& \approx
\color{#00f}{\large{1 \over c^{\pars{1  a}/\pars{a  b}}\,a  c^{\pars{1  b}/\pars{a  b}}\,b}\times}
\\[3mm]&\color{#00f}{\large%
\exp\pars{\bracks{c^{a/\pars{a  b}}\,a
 c^{b/\pars{a  b}}\,b}K^{a/\pars{a  b}}}
K^{\pars{1  a}/\pars{a  b}}}
\end{align}