The Laplace transform of $\frac{\ln(1+at)}{1+t}$

Let the integral in question be \begin{align}\tag{1} I_{a} = \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+a t)}{1 + t} \, dt. \end{align} For the case of $a=1$ the following is obtained.

By utilizing \begin{align} \int_{0}^{\infty} e^{-s t} \, \ln^{2} t \, dt = \frac{1}{s} \, ( \zeta(2) + (\gamma + \ln s)^{2} ) \end{align} for the change $t \to t+1$ leads to \begin{align}\tag{2} \int_{1}^{\infty} e^{-s t} \, \ln^{2}(t+1) \, dt = \frac{e^{s}}{s} \, ( \zeta(2) + (\gamma + \ln s)^{2} ) . \end{align} The integral \begin{align}\tag{3} \int_{0}^{1} e^{-st} \, \ln^{2}(t+1) \, dt &= e^{s} \left\{ 4 g(-2s) - 2 g(-s) + \frac{e^{-2s}}{s} \, (e^{2s} -1) \, \ln^{2} 2 - \frac{2 \, \ln 2}{s} (\gamma + \ln(2s) + \Gamma(0,2s)) \right\}, \end{align} where $g(x) = {}_{3}F_{3}(1,1,1; 2,2,2; x)$. Now, it is readily seen that \begin{align} I_{1} &= \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+t)}{1+t} \, dt = \frac{s}{2} \, \int_{0}^{\infty} e^{-st} \, \ln^{2}(1+t) \, dt. \end{align} Making use of equations (2) and (3) the result \begin{align} \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+t)}{1+t} \, dt &= e^{s} \, \left\{ 2 s \, g(-2s) - s \, g(-s) + \frac{1}{2} \, (1 - e^{-2s}) \, \ln^{2} 2 - \ln 2 \, (\gamma + \ln(2s) + \Gamma(0,2s)) \right. \\ & \hspace{15mm} \left. + \frac{1}{2} \, ( \zeta(2) + (\gamma + \ln s)^{2}) \right\} \end{align} where $g(x) = {}_{3}F_{3}(1,1,1; 2,2,2; x)$.