Help for integration of $3$-form on the sphere $S^3$

Question

Let $\omega$ be the $3$-form in $S^3$ $$ \omega = 4 (- t \, dx \wedge dy \wedge dz + z \, dx \wedge dy \wedge dt - y \, dx \wedge dz \wedge dt + x \, dy \wedge dz \wedge dt) $$ on the sphere. I need to calculate $\int_{S^3} \omega$. What I have tried is to take the stereographic projection ${\varphi}^{- 1} : {\mathbb{R}}^3 \to S^3 \setminus \{(0 , 0 , 0 , - 1)\}$, $$ {\varphi}^{- 1}(u , v , w) = \left(x = \frac{2 u}{1 + u^2 + v^2 + w^2} , y = \frac{2 v}{1 + u^2 + v^2 + w^2} , z = \frac{2 w}{1 + u^2 + v^2 + w^2} , t = \frac{1 - (u^2 + v^2 + w^2)}{1 + u^2 + v^2 + w^2}\right) $$ and compute ${\left({\varphi}^{- 1}\right)}^*(\omega)$, but what I got was $$ - t {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dy) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) = - 8 \frac{{(r^2 - 1)}^2}{{(1 + r^2)}^5} \, du \wedge dv \wedge d w, $$ $$ z {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dy) \wedge {\left({\varphi}^{- 1}\right)}^*(dt) = \frac{2 w}{1 + r^2} \frac{16}{{(1 + r^2)}^5} (- (1 + r^2) u + 2 (u^2 + v^2)(u - w))\, du \wedge dv \wedge d w, $$ $$ - y {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) \wedge {\left({\varphi}^{- 1}\right)}^*(dt) = - \frac{2 v}{1 + r^2} \frac{16}{{(1 + r^2)}^5} (2 v (w^2 - v^2) + v (1 + r^2))\, du \wedge dv \wedge d w, $$ $$ x {\left({\varphi}^{- 1}\right)}^*(dy) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) \wedge {\left({\varphi}^{- 1}\right)}^*(dt) = \frac{2 u}{1 + r^2} \frac{16}{{(1 + r^2)}^5} (2 u (w^2 - u w) + u (1 + r^2)) \, du \wedge dv \wedge d w. $$ But now I am stuck because I did not get many simplifications here, I do not know pretty sure that my computations are all correct. If they were correct, should I take another chart to compute the integral? Any suggestion? Thanks