Analyzing an op amp circuit (transfer function) that does not match the basic cases
You can determine the transfer function \$H(s)\$ of the circuit reasoning on the following circuit:



and thinking of \$V_1\$ and \$V_2\$ as two independent inputs. Since the circuit is linear superimposition applies, and the output (in the s-domain) of the circuit when \$V_2\$ is off is simply that of an inverting amplifier (\$R_3\$ shorts the non inverting input to ground, assuming an ideal op-amp):

\$ V_{out1} = - \dfrac{R_2}{R_1} V_1 \$

Analogously, when \$V_1\$ is off, the circuit acts as a non-inverting amplifier whose input is filtered by the series \$C-R_3\$. Thus applying the non-inverting amp gain formula and the voltage divider formula you get:

\$ V_{out2} = \left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}} V_2 \$

The full response is the sum of the two above:

\$ V_{out} = V_{out1} + V_{out2} = 
- \dfrac{R_2}{R_1} V_1 +
\left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}} V_2 \$

Your circuit is like the one I posted, but with \$V_1 = V_2\$, therefore the full response becomes:

\$ V_{out} = V_{in} \cdot \left[
- \dfrac{R_2}{R_1} + \left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}}
\right] \$

from which you get:

\$ H(s) = \dfrac{V_{out}}{V_{in}} = 
- \dfrac{R_2}{R_1} + \left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}} \$

This simplifies, after a bit of algebra, into:

\$H(s) = \dfrac{s - \frac{R_2}{R_1 R_3 C}}{s + \frac{1}{R_3 C}} \$

Which shows that the circuit acts as an active filter with a 1st order frequency response.

Such a topology is used, for example, to create all-pass filters if \$R_2 = R_1\$.

EDIT

The derivation of the final form of H(s) follows:

\$ H(s) 
= - \dfrac{R_2}{R_1} + \left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}}
= - \dfrac{R_2}{R_1} + \dfrac{R_1 + R_2}{R_1} \dfrac{R_3 C s}{R_3 C s + 1} = \$

\$ 
= - \dfrac{R_2}{R_1} + \dfrac{(R_1 + R_2)R_3 C s}{R_1(R_3 C s + 1)}
= \dfrac{-R_2(R_3 C s + 1) + (R_1 + R_2)R_3 C s}{R_1(R_3 C s + 1)}
\$

\$
= \dfrac{-R_2 R_3 C s - R_2 + R_1 R_3 C s + R_2 R_3 C s}{R_1(R_3 C s + 1)}
= \dfrac{- R_2 + R_1 R_3 C s }{R_1(R_3 C s + 1)}
= \dfrac{R_1 R_3 C s - R_2 }{R_1 R_3 C s + R_1}
\$

dividing numerator and denominator by \$R_1 R_3 C \$ we get:

\$ H(s) 
= \dfrac{s - \frac{R_2}{R_1 R_3 C}}{s + \frac{R_1}{R_1 R_3 C}}
= \dfrac{s - \frac{R_2}{R_1 R_3 C}}{s + \frac{1}{R_3 C}}
\$