How do you compute the successive action of two angular momentum operators on a multivector in geometric algebra?

I’m trying to compute the action of two angular momentum operators $J_i$ on some multivector $psi$ in geometric algebra as in page 290 of Doran & Lasenby (Geometric Algebra for Physicists).

Specifically the problem is to show that:

$J_iJ_ipsi = 3/4psi – 2textbf{x}wedgenablapsi + textbf{x}wedgenabla(textbf{x}wedgenablapsi)$

where $J_k psiequiv J_{Isigma_k}psi = left( (Isigma_k)cdot(textbf{x}wedgenabla) – frac{1}{2}Isigma_k right)psi Isigma_3 $

I have a partial solution so far:

$begin{align}
J_iJ_ipsi &= left( (Isigma_i)cdot(textbf{x}wedgenabla) -frac{1}{2}Isigma_iright)
left[ left((Isigma_i)cdot(textbf{x}wedgenabla) – frac{1}{2}Isigma_iright)psiright] Isigma_3Isigma_3
&= -langle langle Isigma_itextbf{x}wedgenabla rangle langle Isigma_itextbf{x}wedgenabla ranglerangle psi
&quad+(Isigma_i)cdot(textbf{x}wedgenabla)frac{1}{2}Isigma_ipsi
&quad+frac{1}{2}Isigma_i(Isigma_i)cdot(textbf{x}wedgenabla)psi
&quad-frac{1}{4}Isigma_i Isigma_ipsi
&= -langle Isigma_i Isigma_i textbf{x}wedgenabla textbf{x}wedgenabla rangle psi + Isigma_i(Isigma_i)cdot(textbf{x}wedgenabla)psi – frac{1}{4}Isigma_i Isigma_ipsi
&= frac{3}{4}psi + Isigma_i (Isigma_i)cdot (textbf{x}wedgenabla)psi + textbf{x}wedgenabla(textbf{x}wedgenablapsi)
end{align}$

The terms containing $(Isigma_i)cdot(textbf{x}wedgenabla)$ terms are equal and add up to $Isigma_i (Isigma_i)cdot (textbf{x}wedgenabla)psi$, but I don’t know how to manipulate it further.

Edit:

$Isigma_i (Isigma_i)cdot(textbf{x}wedgenabla) = langle Isigma_i (Isigma_i) cdot (textbf{x}wedgenabla)rangle_2 = langle Isigma_i Isigma_i textbf{x}wedgenabla rangle_2 = -textbf{x}wedgenabla$

Closer but still not correct.

Edit2: There must be an implied sum over $i$ which would explain the $3$ in $3/4psi$.

Also, the factor of 3 is avoided in the $textbf{x}wedgenabla(textbf{x}wedgenablapsi)$ term because the sum over $i$ serves to reconstitute the original $textbf{x}wedgenabla$ bivectors from the projections. The geometric product $(textbf{x}wedgenabla)(textbf{x}wedgenabla)$ only contains a scalar because the cross product is zero and the wedge product between spatial bivectors is zero.