where In terms of the dimensionless time T = t t$_{pl}$ this reads as: where z$_{1}$ = α$_{1}$β$_{2}$ , z$_{2}$ = α$_{1}$γ$_{2}$ + β$_{1}$β$_{2}$ , z$_{3}$ = α$_{1}$δ$_{2}$ + β$_{1}$γ$_{2}$ + γ$_{1}$β$_{2}$ and z$_{4}$ = α$_{1}$ Σ$_{2}$ + β$_{1}$δ$_{2}$ + γ$_{1}$γ$_{2}$ + δ$_{1}$β$_{2}$ . Since at the end of inflation: where where K ′′ is a constant. It is notable that the parameter α in Hankel function H (1) α ( Q$_{q}$ T ) is 18 R$_{q}$$_{(0)}$ ( T ) = K $^{′′}$T $^{3}$H (1) 3 ( Q$_{q}$ T ) (57) Q$_{q}$ = qt$_{pl}$ a ≃ qt$_{pl}$ a$_{0}$ . (56) d $^{2}$R$_{q}$ dT 2 − 5 T dR$_{q}$ dT + Q 2 $_{q}$R$_{q}$ = a$_{0}$z$_{1}$T dR$_{q}$ dT (55) z$_{1}$T 2 EI ≃ 1 . 32 × 10 − $^{12}$, z$_{2}$ ≃ − 1 . 57 × 10 − $^{36}$, z$_{3}$T − 2 EI ≃ 1 . 19 × 10 − $^{24}$, z$_{4}$T − 4 EI ≃ − 1 . 42 × 10 − $^{48}$, we can ignore z$_{2}$ , z$_{3}$T − $^{2}$, z$_{4}$T − 4 and the constant √ 3Λ$_{0}$ t$_{pl}$ . As a result the IMSE in this case is given by: S ( T ) ≃ + a$_{0}$ T $^{[}$z$_{1}$ T 2 + z$_{2}$ + z$_{3}$T − 2 + z$_{4}$T − 4 $^{]}$+ $^{√}$3Λ$_{0}$ t$_{pl}$ − 5 T + O$^{(}$$^{(}$ t t$_{pl}$$^{)}$$^{−}$$^{5}$$^{)}$ . (54) α$_{1}$ = − Λ$_{0}$˜ ω 2˜ νm 2 pl , β$_{1}$ = − Λ$_{0}$ 2˜ νm 2 pl ( − 2Ω + ˜ ω $^{2}$) , β$_{2}$ = − 2 γ$_{1}$ = − Λ$_{0}$ 2˜ νm 2 pl ( − 3Ω˜ ω + ˜ ν ˜ ωm 2 pl Λ$_{0}$ ) , γ$_{2}$ = − 4( Ω ˜ ω − ˜ ω ) , Σ$_{2}$ = − 80Ω( Ω ˜ ω − ω ) δ$_{1}$ = − Λ$_{0}$ 2˜ νm 2 pl (2Ω 2 − 2Ω ˜ νm 2 pl Λ$_{0}$ + ˜ ω $^{2}$˜ νm 2 pl Λ$_{0}$ ) , δ$_{2}$ = 16(Ω + ( Ω ˜ ω − ˜ ω ) $^{2}$) . It has to be noted that the term 5 T comes from a d dt ln( a 2 √ 3 Λ ˙ Λ $_{2Λ}$) and has the most contribution in this model. In contrast, − a$_{0}$z$_{1}$T is a consequence of the running gravitational coupling. In a very similar way like the previous subsection the solution to (55) can be obtained as: