The quadratic form of spin-3 Lagrangian in the momentum space is: \begin{eqnarray} && (Kn)^{\nu_1\nu_2\nu_3}_{\mu_1\mu_2\mu_3} = \delta_{(\mu_1}^{(\nu_1} \delta_{\mu_2}^{\nu_2} \delta_{\mu_3)}^{\nu_3)} + 6 g_{(\mu_1\mu_2} g^{(\nu_1\nu_2} \delta_{\mu_3)}^{\nu_3)} \Big( - \frac{1}{2} +\lambda\beta^2 \Big) \nonumber\\ && + 3 (-1+\lambda) n_{(\mu_1} n^{(\nu_1} \delta_{\mu_2}^{\nu_2} \delta_{\mu_3)}^{\nu_3)} + 6\Big(\frac{1}{2} +\lambda\beta\Big) \Big( n_{(\mu_1} n_{\mu_2} g^{(\nu_1\nu_2} \delta_{\mu_3)}^{\nu_3)} \nonumber\\ && + n^{(\nu_1} n^{\nu_2} g_{(\mu_1\mu_2} \delta_{\mu_3)}^{\nu_3)}\Big) + 6\Big(-\frac{1}{4} + \lambda\beta^2\Big) n_{(\mu_1} g_{\mu_2\mu_3)} n^{(\nu_1} g^{\nu_2\nu_3)} \end{eqnarray}
where $n_\mu$ is a momentum. The equation for the propagator is: \[ (Kn)^{\nu_1\nu_2\nu_3}_{\mu_1\mu_2\mu_3} {(Kn)^{-1}}^{\mu_1\mu_2\mu_3}_{\alpha_1\alpha_2\alpha_3} = \delta^{(\nu_1}_{(\alpha_1}\delta^{\nu_2}_{\alpha_2}\delta^{\nu_3)}_{(\alpha_3)}. \]
The following Redberry code produces propagator in four dimensions (here we used l for $\lambda$ and b for $\beta$):
//quadratic form
def K = '''
d^a_p*d^b_q*d^c_r
+ 6*g_pq*g^ab*d^c_r*(-1/2 + l*b**2)
+ 3*(-1 + l)*n_p*n^a*d^b_q*d^c_r
+ 6*(1/2 + l*b)*(n_p*n_q*g^ab*d^c_r + n^a*n^b*g_pq*d^c_r)
+ 6*(-1/4+l*b**2)*n_p*g_qr*n^a*g^bc
'''.t
//symmetrize upper indices
def upper = K.indices.upper.si
upper.symmetries.setSymmetric()
K = Symmetrize[upper] >> K
//symmetrize lower indices
def lower = K.indices.lower.si
lower.symmetries.setSymmetric()
K = Symmetrize[lower] >> K
//propagator
addSymmetries 'P^{abc}_{mnk}',
[[0, 1]].p, [[0, 1, 2]].p,
[[3, 4]].p, [[3, 4, 5]].p
// l.h.s of the equation
def lhs = 'K^abc_mnk*P^mnk_pqr'.t
lhs = "K^abc_pqr = $K".t >> lhs
//r.h.s. of the equation
def rhs = 'd^a_p*d^b_q*d^c_r'.t
rhs = (Symmetrize[upper] & Symmetrize[lower]) >> rhs
def options = [Transformations: 'd_n^n = 4'.t & 'n_a*n^a = 1'.t,
ExternalSolver : [
Solver: 'Mathematica',
Path : '/Applications/Mathematica.app/Contents/MacOS']]
def s = Reduce(["$lhs = $rhs".t], ['P^{abc}_{pqr}'], options)
println s
Here we assumed $n^2 = 1$, but the correct dimension can be easily recovered.
The produced result is rather huge. It can be written in the following form: \begin{eqnarray} && {(Kn)^{-1}}^{\alpha_1 \alpha_2 \alpha_3}_{\beta_1 \beta_2 \beta_3} = Sym[ c_0 n_{ \beta_1 } n_{ \beta_2 } n_{ \beta_3 } n^{ \alpha_1 } n^{ \alpha_2 } n^{ \alpha_3 } + c_1 \delta_{ \beta_1 }^{ \alpha_1 } n_{ \beta_2 } n_{ \beta_3 } n^{ \alpha_2 } n^{ \alpha_3 }+ c_2 \delta_{ \beta_1 }^{ \alpha_1 } \delta_{ \beta_2 }^{ \alpha_2 } n_{ \beta_3 } n^{ \alpha_3 }+\nonumber\\ && + c_3 \delta_{ \beta_1 }^{ \alpha_1 } \delta_{ \beta_2 }^{ \alpha_2 } \delta_{ \beta_3 }^{ \alpha_3 }+ c_4 g^{ \alpha_1 \alpha_2 } n_{ \beta_1 } n_{ \beta_2 } n_{ \beta_3 } n^{ \alpha_3 }+ c_5 g^{ \alpha_1 \alpha_2 } \delta_{ \beta_1 }^{ \alpha_3 } n_{ \beta_2 } n_{ \beta_3 }+ c_6 g_{ \beta_1 \beta_2 } n_{ \beta_3 } n^{ \alpha_1 } n^{ \alpha_2 } n^{ \alpha_3 }+\nonumber\\ && + c_7 g_{ \beta_1 \beta_2 } \delta_{ \beta_3 }^{ \alpha_1 } n^{ \alpha_2 } n^{ \alpha_3 }+ c_8 g_{ \beta_1 \beta_2 } g^{ \alpha_1 \alpha_2 } n_{ \beta_3 } n^{ \alpha_3 }+ c_9 g_{ \beta_1 \beta_2 } g^{ \alpha_1 \alpha_2 } \delta_{ \beta_3 }^{ \alpha_3 }], \end{eqnarray} Where $Sym$ is a symmetrisation over upper and lower indices (as in the quadratic form) and the coefficients are: \begin{gather*} c_0 = \,{\frac {20\,{b}^{2}+4\,b-1}{4 \left( b+1 \right) ^{2}}} + % 16\,{\frac { \left( 1+6\,b+6\,{b}^{2} \right) ^{2}}{ \left( 12\,l{b}^{ 2}-24\,{b}^{2}-36\,b+12\,lb-14+3\,l \right) \left( 7+12\,{b}^{2}+18\, b \right) }}+\\ % +\,{\frac { 3\left( 4\,{b}^{2}+10\,b+5 \right) \left( 2\,b+1 \right) ^{2}}{2l \left( b+1 \right) ^{2} \left( 7+12\,{b}^{2}+18\,b \right) }} \\ c_1 = -\frac{3}{4}\,{\frac { \left( 2\,b+1 \right) \left( 2\,lb+4\,b+6-3\,l \right) }{ \left( b+1 \right) ^{2}l}} \\ c_2=-3\,{\frac {-1+l}{l}} \\ c_3=1\\ c_4= -\frac{3}{4}\,{\frac {2\,b+1}{b+1}} -6\,{\frac { \left( 4\,b+3 \right) \left( 1+6\,b+6\,{b}^{2} \right) \left( 2\,b+1 \right) }{ \left( 12\,l{b}^{2}-24\,{b}^{2}-36\,b+12\,lb -14+3\,l \right) \left( 7+12\,{b}^{2}+18\,b \right) }}+\\ +6\,{\frac { \left( b+1 \right) \left( 2\,b+1 \right) }{l \left( 7+12 \,{b}^{2}+18\,b \right) }} \\ c_5=\frac{3}{4}\,{\frac {2\,b+1}{b+1}}\\ c_6=c_4\\ c_7=c_5\\ c_8= \frac{1}{4}\,{\frac {8-6\,l+3\,{l}^{2}}{l \left( l-2 \right) }}+ \frac{1}{2}\,{\frac {24\,lb+15\,l+2}{ \left( l-2 \right) \left( 12\,l{b}^{2}- 24\,{b}^{2}-36\,b+12\,lb-14+3\,l \right) l}} \\ c_9=-\frac{3}{4} \end{gather*}