Diodes are fast switching for extra op amp input protection, capacitors are variable caps to compensate for parasitic capacitance in diodes(~20pF).
20pF? Is that for each diode, if so, it's huge. If it's for both diodes added together, then it's not so bad. What diodes are you using?
I'd recommend a Schottky diode, such as the BAT54, because it has a forward voltage of only 300mV which will limit the input voltage below the maximum specified on the op-amp data sheet. It has a capacitance of 10pF.
http://assets.nexperia.com/documents/data-sheet/BAT54_SER.pdfThe capacitance of both of the diodes and the op-amp's common mode input are effectively in parallel at AC. One of the capacitors (C1 in your schematic) is making matters worse, rather than better, because it adds to the parasitic capacitance of the diodes and op-amp.
Only one capacitor is required across R3 and it should be
1/
9 of the value of the parasitic capacitance of the op-amp + diodes, forming a 10:1 capacitive divider, in parallel with the resistive divider.
If the op-amp has a common mode capacitance of 6pF and the capacitance of both diodes added together is 20pF, the total parasitic capacitance is 26pF, so the compensation capacitor, in parallel with R1 should be 26pF/9pF = 2.889pF, with 3pF being the nearest standard value.
http://ww1.microchip.com/downloads/en/DeviceDoc/21685d.pdfYou could have also chosen better resistor values, see attached.

What bandwidth do you need?
With the component values shown above, the bandwidth, without the compensation capacitor will be 51kHz. Adding the 3pF compensation capacitor, across R3, will increase the upper cut-off frequency to above the op-amp's gain bandwidth product.
F
C = 1/(2pi*RC)
Where
R = R1|R2|R3 = (R1
-1+R2
-1+R3
-1)
-1C = parasitic capacitance = 26pF