The wikipedia page says that the equation for impulse-based collision response is:
\begin{equation} j_r = \frac{ -(1 + e) v_r \cdot n } { {m_1}^{-1} + {m_2}^{-1} + ({I_1}^{-1}(r_1 \times n) \times r_1 + {I_2}^{-1}(r_2 \times n) \times r_2 ) \cdot n}\end{equation}
What is the full working out for this equation?
To derive the equation:
\begin{equation} \tag{1} j_r = \frac{ -(1 + e) v_r \cdot n } { {m_1}^{-1} + {m_2}^{-1} + ({I_1}^{-1}(r_1 \times n) \times r_1 + {I_2}^{-1}(r_2 \times n) \times r_2 ) \cdot n}\end{equation}
we know that for linear velocity, \begin{equation}\tag{2a} v_1' = v_1 - \frac{j_r}{m_1} n \end{equation} \begin{equation}\tag{2b} v_2' = v_2 + \frac{j_r}{m_2} n \end{equation}
and for angular velocity, \begin{equation}\tag{3a} {\omega}_1' = {\omega}_1 - \frac{j_r}{I_1} {r_1 \times n} \end{equation} \begin{equation}\tag{3b} {\omega}_2' = {\omega}_2 + \frac{j_r}{I_2} {r_2 \times n} \end{equation}
at the contact point, inital velocity is, \begin{equation}\tag{4a} v_{pi} = v_i + {\omega}_i \times r_i \end{equation}
and final velocity is, \begin{equation}\tag{4b} v_{pi}' = v_i' + {\omega}_i' \times r_i \end{equation} similarly for relative velocity, \begin{equation}\tag{5a} v_r = v_{p2} - v_{p1} \end{equation} \begin{equation}\tag{5b} v_r' = v_{p2}' - v_{p1}' \end{equation}
Newton's law of restitution, \begin{equation}\tag{6a} v_r' \cdot n = -e v_r \cdot n \end{equation}
substitute (5b) into (6a), \begin{equation}\tag{6b} (v_{p2}' - v_{p1}') \cdot n = -e v_r \cdot n \end{equation}
substitute (4b) into (6b),
\begin{equation}\tag{6c} \Big((v_2' + {\omega}_2' \times r_2) - (v_1' + {\omega}_1' \times r_1)\Big) \cdot n = -e v_r \cdot n \end{equation}
substitute (2a) and (2b) into (6c),
\begin{equation}\tag{6d} \bigg(\Big((v_2 + \frac{j_r}{m_2} n) + {\omega}_2' \times r_2) - ((v_1 - \frac{j_r}{m_1} n) + {\omega}_1' \times r_1\Big)\bigg) \cdot n = -e v_r \cdot n \end{equation}
expanding and simplifying left side,
\begin{equation}\tag{7a} LHS = (v_2 + \frac{j_r}{m_2} n + {\omega}_2' \times r_2 - v_1 + \frac{j_r}{m_1} n - {\omega}_1' \times r_1) \cdot n \end{equation} \begin{equation}\tag{7b} \hspace{5em} = \Big(v_2 - v_1 + j_r({m_2}^{-1} + {m_1}^{-1})n + {\omega}_2' \times r_2 - {\omega}_1' \times r_1\Big) \cdot n \end{equation}
substitute (3a) and (3b) into (7b),
\begin{equation}\tag{8a} = \bigg(v_2 - v_1 + j_r({m_2}^{-1} + {m_1}^{-1})n + {\Big(\omega}_2 + \frac{j_r}{I_2} (r_2 \times n)\Big)\times r_2 - \Big({\omega}_1 - \frac{j_r}{I_1} (r_1 \times n) \Big) \times r_1\bigg) \cdot n \end{equation} expanding and simplifying left side,
\begin{equation}\tag{8b} = \Big(v_2 - v_1 + j_r({m_2}^{-1} + {m_1}^{-1})n + {\omega}_2 \times r_2 + j_r{I_2}^{-1} (r_2 \times n) \times r_2 - {\omega}_1 \times r_1 + j_r{I_1}^{-1} (r_1 \times n) \times r_1\Big) \cdot n \end{equation} \begin{equation}\tag{8c} = \bigg(v_2 + {\omega}_2 \times r_2 - (v_1 + {\omega}_1 \times r_1) + j_r\Big(({m_2}^{-1} + {m_1}^{-1})n + {I_2}^{-1} (r_2 \times n) \times r_2 + {I_1}^{-1} (r_1 \times n) \times r_1\Big) \bigg) \cdot n \end{equation}
\begin{equation}\tag{8d} = \bigg(v_r + j_r\Big(({m_2}^{-1} + {m_1}^{-1})n + {I_2}^{-1} (r_2 \times n) \times r_2 + {I_1}^{-1} (r_1 \times n) \times r_1\Big) \bigg) \cdot n \end{equation}
\begin{equation}\tag{8e} = v_r \cdot n + \bigg(j_r\Big(({m_2}^{-1} + {m_1}^{-1})n + {I_2}^{-1} (r_2 \times n) \times r_2 + {I_1}^{-1} (r_1 \times n) \times r_1\Big) \bigg) \cdot n \end{equation}
and equating with RHS,
\begin{equation}\tag{9a} j_r \bigg(\Big(({m_2}^{-1} + {m_1}^{-1})n + {I_2}^{-1} (r_2 \times n) \times r_2 + {I_1}^{-1} (r_1 \times n) \times r_1\Big) \bigg) \cdot n = -e v_r \cdot n - v_r \cdot n \end{equation}
\begin{equation}\tag{9b} j_r = \frac{-(1 + e)v_r \cdot n}{\Big(({m_2}^{-1} + {m_1}^{-1})n + {I_2}^{-1} (r_2 \times n) \times r_2 + {I_1}^{-1} (r_1 \times n) \times r_1\Big) \cdot n} \end{equation}
which simplifies to,
\begin{equation}\tag{10} j_r = \frac{-(1 + e)v_r \cdot n}{ {m_1}^{-1} + {m_2}^{-1} + \Big({I_1}^{-1} (r_1 \times n) \times r_1 + {I_2}^{-1} (r_2 \times n) \times r_2\Big) \cdot n} \end{equation}