In the second case for the setting where Do > Di > Dv and te < 0 the conditional expression under the remaining integral works into your hands stating
-Inf (Dv + te)^2 (-2 Di + Dv + te)/(2 (Dv Inf + (Inf-Ip) te)) for D0>= Dv + te && Di > Dv + te && Dv + te > 0
0 elsewhere
because you say te < 0 the upper integration limitD0 +Di -Dv for te must be smaller than 0
there is a contradiction: If Di > Dv and D0 > Dv then D0 + Di - Dv > 0 so te > 0 (not te < 0) on the upper limit .... okay, let's send this back to you and continue with the integration limits:
The lower limit is fullfilled: D0 > 0 && Di > 0 && 0 >= 0 because D0> Dv in the assumptions and if te < 0 at least on the lower limit one has Dv > 0.
The upper limit D0 >= D0 + Di && Di > D0 + Di && D0 + Di > 0 is not fullfilled because Di > 0; one has to determine where the integrand is 0 and where not. To get out of this mess one could plot it
With[{Dv = 1, Di = 4.5, D0 = 8},
Plot[If[D0 >= Dv + te && Di > Dv + te && Dv + te > 0,
1, -1], {te, -Dv, D0 + Di + Dv}]
]
fortunately there is only one switch from 1 to -1 at te = Di - Dv leaving us behind with the integral
In[30]:= Clear[Dv, Di, D0]
Inf phi tau Integrate[-((Inf (Dv + te)^2 (-2 Di + Dv + te))/(
2 (Dv Inf + (Inf - Ip) te))), {te, -Dv, Di - Dv},
Assumptions -> Dv > 0]/(D0 N)
Out[31]= ConditionalExpression[-((
Inf^2 phi tau (-Di (Inf - Ip) (4 Di^2 (Inf - Ip)^2 - 6 Dv^2 Ip^2 +
(-Inf + Ip)) +
6 Dv^2 Ip^2 (2 Di (Inf - Ip) + Dv Ip) Log[(Dv Ip)/(
Di Inf - Di Ip + Dv Ip)]))/(12 D0 (Inf - Ip)^4 N)),
Di Ip^2 < Di Inf Ip || Di (Inf - Ip) (Di (Inf - Ip) + Dv Ip) < 0]
the integrals are done. Please check for errors.
