In case of coupling of the Einstein's Field Equations to fermionic matter torsion is generated. However for a consistent set of EFEs it is required that the Einstein tensor
$$G_{ik} = R_{ik} -\frac{1}{2}g_{ik} R\quad \text{fulfills}\quad G_{i\, ;k}^k=0$$
which yields from the contraction of the second Bianchi identity. However, under non-zero torsion the second Bianchi identity changes its form to ($R^i_{jkl}$ are the components of the curvature tensor and $T^s_{km}$ the components of the torsion tensor):
$$\sum_{cyclic\, klm} R^i_{jkl;m} = \sum_{cyclic\, klm} T^s_{km} R^i_{jsl} \quad\quad\text{from N.Straumann: GR & relativistic Astrophysics}$$
With zero torsion the rhs of this equation would be zero and by using the definition of the Ricci tensor $R_{jl} = g^{ik}R_{ijkl}$ and the "usual" contracted Bianchi identity $G_{i\, ;k}^k=0$ would follow rather easily: by use of the definition and the symmetry of the curvature tensor we can write:
$$R_{j\,;m}^m = g^{ml} R_{jl;m} = g^{ml}g^{ik} R_{ijkl;m} = g^{ml}g^{ik} R_{klij;m}$$
followed by -- using the symmetry properties of the curvature tensor and the definition of the curvature scalar $R =g^{ik}R_{ik}$:
$$ R_{j\, ;m}^m = -g^{ml}g^{ik}(R_{kljm;i} + R_{klmi;j}) = - g^{ik} R_{kj;i} + g^{ik}R_{ki;j} = -R_{j\,;m}^m + R_{;j}$$
from which $G_{i\,\,;k}^k=0$ immediately follows. However, with non-zero torsion this equation would get the form:
$$ R_{j;m}^m = -g^{ml}g^{ik}(R_{kljm;i} + R_{klmi;j}) + g^{ml}g^{ik}\sum_{cyclic\, ijm} T^s_{im} R_{klsj}$$
EDIT:
I will expand the cyclic sum a bit further:
$$\text{cyclic sum}=g^{ml}g^{ik} ( T^s_{im} R_{klsj} + T^s_{mj} R_{klsi} + T^s_{ji} R_{klsm}) = T^{s\,\,l}_i R^i_{lsj} +T^{sl}_j R^i_{lsi} -T^{s\,\,k}_{j} R^m_{ksm}$$
Moreover:
$$\text{cyclic sum} = T^s_{il} R^{il}_{sj} -T^s_{lj} R^l_s +T^s_{jk} R^k_s =T^s_{il} R^{il}_{sj} -T^s_{lj} R^l_s +T^s_{jl} R^l_s $$
So if the torsion tensor were symmetric in its 2 last indices the sum would indeed vanish, however, it does not, otherwise the torsion form $\Theta^s$
$$\Theta^s = \frac{1}{2}T^s_{il} dx^i \wedge dx^l $$
would vanish too, but its non-zero-ness was actually the starting point of my question.
So how non-zero torsion is with the integrability condition $G_{i\,\,;k}^k=0$ compatibel ? Does the contracted torsion term eventually disappear or do we have to add a new contribution from the torsion to the energy-momentum tensor ?
EDIT 2
For clarification I'd like to add the definition of cyclic sum used in this post:
$$\sum_{cyclic\,\, klm} R^i_{klm}:=R^i_{klm} + R^i_{lmk} + R^i_{mkl}$$
If this sum is set to zero it corresponds to the first Bianchi-identity (here without torsion). This how the above cited source (N.Straumann) also notes the first Bianchi-identity. So there cannot be an ambiguity with respect to the definition of the cyclic sum.