Eilenberg-Maclane space $K(G\rtimes H, 1)$ for a semi-direct product.

$K(G \rtimes H, 1)$ fits into a fibration sequence

$$K(G, 1) \to K(G \rtimes H, 1) \to K(H, 1).$$

So for example one can access the homology and cohomology using the Serre spectral sequence. See this answer for some context.

Here are some details on how to get the fibration in Qiaochu Yuan's answer:$\newcommand{\Z}{\mathbb{Z}}$

Let $H \curvearrowright E_H$ act freely on a weakly contractible space $E_H$, and let the exact sequence of groups $1 \to G \to G \rtimes H \xrightarrow{\phi} H \to 1$ be given. Associated to the map $\phi : G \rtimes H \to H$, there is an action $G \curvearrowright E_H$. The induced map $\Phi: E_H/G \to E_H/H$ given by $Ge \to He$ is well defined because $\phi(G)\subset H$.

Now make the additional assumption that the groups are discrete so that $Z G=BG$, $Z G \rtimes H=B G \rtimes H $, $Z H=BH$ : After all $\pi_1 B(G \rtimes H)=G \rtimes H$ and same for $G, H$. $\pi_k(B(\text{any of the groups }))=0$ for $k>2$(since they are homotopy groups of contractible spaces).

Turn the map $\mathbb{Z}G \rtimes H \to \mathbb{Z}H $ into a fibration with fiber $F$ via a space homotopy equivalent to a $\mathbb{Z} G \rtimes H$. This will be another model for $\mathbb{Z}G \rtimes H$. The homotopy long exact sequence for the fibration $F \hookrightarrow \mathbb{Z} G \rtimes H \to\mathbb{Z} H$ implies that $\pi_k(F)=0$ for $k>1$. Since the induced map $\pi_1 (B(G \rtimes H)) \to \pi_1(BH)$ is exactly $\phi$, and is therefore surjective, we get the top row and the horizontal isomorphisms of the following commutative diagram $\require{AMScd} \begin{CD} 1 @>>>\pi_1(F) @>>> \pi_1(\mathbb{Z}G \rtimes H) @>\phi>> \pi_1(\mathbb{Z}H) @>>> 1\\ @|@. @| @| @|\\ 1 @>>>\pi_1(\mathbb{Z} G ) @>\psi >> \pi_1(\mathbb{Z} G \rtimes H) @>>> \pi_1(\mathbb{Z} H) @>>> 1 \end{CD}$

We can define a map from $\pi_1( \Z G) \to \pi_1(F)$ by chasing the diagram because $\phi \circ (\text{ equals sign }) \circ \psi=0$. Since we defined the map $\pi_1( \Z G) \to \pi_1(F)$ by chasing the diagram, the following diagram commutes:

$ \begin{CD} 1 @>>>\pi_1(F) @>>> \pi_1(\mathbb{Z}G \rtimes H) @>\phi>> \pi_1(\mathbb{Z}H) @>>> 1\\ @|@AAA @| @| @|\\ 1 @>>>\pi_1(\mathbb{Z} G ) @>\psi >> \pi_1(\mathbb{Z} G \rtimes H) @>>> \pi_1(\mathbb{Z} H) @>>> 1 \end{CD}$ By the five lemma, the fiber $F$ is $ZG$ up to weak homotopy equivalence.

Summarizing, we get the fibration in Qiaochu Yuan's answer $\Z G \hookrightarrow \Z G\rtimes H \to \Z H$.

I have not shown that the fiber will be a CW complex. Notice that I didn't assume that the map on $\pi_1$ was induced topologically. But because I showed that it has the same homotopy groups as an eilenberg mclane space, the postnikov invariants vanish and I have shown that there is indeed a weak homotopy equivalence between the fiber and $\Z G$. But it is possible to show that the fiber is a CW complex in which case I can say that it has the homotopy type of a $\Z G$.

I'll put something on the cohomology calculation in a little bit. There are way too many ...by a spectral sequence argument.... and to the best of my knowledge no calculations of this type done on all of mathstackexchange.