The effects of Op amp Bias current
By Kirchhoff Current Law applied to the x node in your circuit, we have
$$
I_f= I_{in}+I_b. \tag{1} \label{1}
$$
The bias current \$I_b\$ can be positive i.e. can flow from the inverting input of the OpAmp to the x node or it can be negative if it flows the other way around. Now, you want to calculate the absolute error caused on the rise rate of output voltage \$V_\mathrm{out}\$ by the presence of \$I_b\$: then, 
$$
\left|\Delta \frac{d V_\mathrm{out}}{d t}\right|_{I_b}=\left|\frac{d V_\mathrm{out}}{d t} - \frac{d V_\mathrm{out}|_{I_b=0}}{d t}\right|=\frac{|I_b|}{C_f} = 30\mathrm{mV/s} \tag{2} \label{2}
$$
as you already calculated.

A few notes


You say that the slew rate of \$V_\mathrm{out}\$ is \$150\mathrm{V/s}\$ and also \$I_{in}\approx 130\mathrm{nA}\$: this cannot be even if you assume that the bias current \$I_b\$ has its maximum value. Perhaps the cause of this strong discrepancy is the tolerance of the value of \$C_f\$: a 15% tolerance can cause a slew rate error far superior to the one caused by the bias current of the OpAmp.
The error formula you report refers to the case of an inverting amplifier, i.e. when \$I_f\$ flows across a feedback resistor: in that case the error on the output voltage is proportional to the bias current through the value of the resistor connected between output and inverting input. In your circuit, a simple inverting integrator, there is not such resistor thus there is not such error.


An addendum

The so called "static errors" in an OpAmp are not limited to the bias currents required by the transistors in its input stages. The imperfect symmetry of these device models as an intrinsic more or less small input voltage which should be applied in order to have a zero output: this is the offset voltage \$V_\mathrm{IO}\$. Said that, the equivalent circuit of the OpAmp A above can be shown to be the following one



We have already accounted for \$I_b\$ (precisely \$I_{b-}\$) explicitly in the contribution to the feedback branch current \$I_f\$ (equation \eqref{1}): does the offset voltage contribute to it? Yes, but in a subtler way: the true value of \$I_{in}\$ is expressible as
$$
I_{in}=\frac{V_{in}-V_{IO}}{R_{in}}
$$
For a TL084 OpAmp, the typical value of \$V_\mathrm{IO}\$ is \$3\mathrm{mV}\$ up to a maximum of \$15\mathrm{mV}\$ (datasheet here). In our example, \$V_{in}\$ is  \$\approx30\mathrm{mV}\$ thus, assuming a typical offset  \$V_\mathrm{IO}\approx-3\mathrm{mV}\$ (its sing varies randomly, so it can add or subtract from the input voltage) we have that 
$$
I_{in}=\frac{30\mathrm{mV}+3\mathrm{mV}}{230\mathrm{k}\Omega}=\frac{33\mathrm{mV}}{230\mathrm{k}\Omega}\cong143\mathrm{nA}
$$
The value obtained is very similar to the one the OP obtains by measuring the rise rate of the output voltage \$V_\mathrm{out}\$: this implies that the major contribution to the error on the output is due to \$V_\mathrm{IO}\$ and is precisely given by the following expression
$$
\left|\Delta \frac{d V_\mathrm{out}}{d t}\right|_{V_\mathrm{IO}}=\frac{1}{C_f}\left|I_{in}-\frac{V_{in}}{R_{in}}\right|=\frac{|V_\mathrm{IO}|}{R_{in}C_f} = 13\mathrm{V/s} \tag{3} \label{3}
$$
which is a far higher contribution than the one given by \$I_b\$ (equation \eqref{2}).

How to reduce the error caused by the offset voltage?

There are two possibilities


Increase ten times \$V_{in}\$ and decrease ten times \$C_f\$: in this way the rise rate is kept constant while the output error caused by \$V_\mathrm{IO}\$ is decreased to one tenth. However, this means reducing the dynamic range of the input, and sometimes this is not possible nor auspicable.
Change the OpAmp A with one with a lower \$V_\mathrm{IO}\$. Perhaps this is a more costly solution: however, OpAmps with a maximum \$V_\mathrm{IO}\$ \$&lt;0.3\mathrm{mV}\$ (like this one) are easily found and not so expensive.