I use the Feynman parameterisation formula in the general form (valid for $ \text{Re} \alpha_{j} > 0 $ for all $j$):
$$\frac{1}{A_{1}^{\alpha_{1}}\cdots A_{n}^{\alpha_{n}}} = \frac{\Gamma(\alpha_{1}+\dots+\alpha_{n})}{\Gamma(\alpha_{1})\cdots\Gamma(\alpha_{n})}\int_{0}^{1}du_{1}\cdots\int_{0}^{1}du_{n}\frac{\delta(1-\sum_{k=1}^{n}u_{k})\;u_{1}^{\alpha_{1}-1}\cdots u_{n}^{\alpha_{n}-1}}{\left(\sum_{k=1}^{n}u_{k}A_{k}\right)^{\sum_{k=1}^{n}\alpha_{k}}} $$ In our case we take $n=3$, $\alpha_1=3/2$, $\alpha_2=\alpha_3=1/2$ to find $$F(w)=\frac{w_i}{\| w\|^3\|w-a\|\|w-b\|}=\frac{3w_i}{2\pi}\int_{0}^{1}du_{1}\int_{0}^{1}du_{2}\int_{0}^{1}du_{3}\frac{\delta(1-u_1-u_2-u_3)\;u_{1}^{1/2}u_{2}^{-1/2}u_{3}^{-1/2}}{\left(u_1\|w\|^2+u_2\|w-a\|^2+u_3\|w-b\|^2\right)^{5/2}}.$$ The integral of $F(w)$ over the 3-dimensional vector $w$ can be evaluated by first rewriting $$u_1\|w\|^2+u_2\|w-a\|^2+u_3\|w-b\|^2=(u_1+u_2+u_3)\bigl(\|w-v\|^2+d\bigr),$$ $$v=\frac{u_2 a+u_3b}{u_1+u_2+u_3},\;\;d=\frac{a^2 u_1 u_2+a^2 u_2 u_3-2 a b u_2 u_3+b^2 u_1 u_3+b^2 u_2 u_3}{(u_1+u_2+u_3)^2}.$$ The integral over $w$ then follows from the formula $$\int d^3 w \frac{w_i}{(\|w-v\|^2+d)^{5/2}}=\frac{4\pi v_i}{3d}.$$ So we arrive at $$\int d^3 F(w)=2\int_{0}^{1}du_{1}\int_{0}^{1}du_{2}\int_{0}^{1}du_{3}\frac{(u_2 a_i+u_3 b_i)\delta(1-u_1-u_2-u_3)\;u_{1}^{1/2}u_{2}^{-1/2}u_{3}^{-1/2}}{a^2 u_1 u_2+a^2 u_2 u_3-2 a b u_2 u_3+b^2 u_1 u_3+b^2 u_2 u_3}$$ $$\qquad=2\int_{0}^{1}du_{1}\int_{0}^{1}du_{3}\frac{(u_2 a_i+u_3 b_i)(1-u_2-u_3)^{1/2}u_{2}^{-1/2}u_{3}^{-1/2}}{\|a\|^2 u_2(1-u_2)+\|b\|^2(1-u_3)u_3-2(a\cdot b)u_2u_3}.$$
The paramaterization quoted by the OP is a bit different, not symmetric in $a$ and $b$, I presume it is equivalent.