I aim to determine the Laplace transform of $\log^2(1+t)$. Apparently, Mathematica gives me a closed formula $\log(2/s^3 + 2/s^2 + 1/s)$. Any idea how to obtain this formula?
I though run into a hypergeometric function: $ \begin{aligned}\mathcal{L} \left\{ \log^2(1+\cdot) \right\}(s)&=\int_{0}^{\infty} e^{-st}\log^2 (1+t) \, dt =e^s\int_{1}^{\infty} e^{-st}\log^2 (t) \, dt \\ &= e^s \left( \mathcal{L} \left\{ \log^2 \right\}(s) - \int_{0}^{1} e^{-st}\log^2 (t) \, dt \right)\\ &= \frac{e^s}{s} \left( \frac{\pi^2}{6} + \gamma^2 - \log^2(s) e^{-s} - 2 \log(s) E_1(s) - \underbrace{ \int_0^s e^{-u} \log^2(u) du }_{F(1, 1, 1; 2, 2, 2; -s)}\right) \end{aligned} $