It seems your voice has been heard! All of the integrals you list are now again evaluated, in MMA 12.0.0 (for Mac). Note, however, that the new outputs may not be different from those originally generated by 11.2.0. I don't have the latter running to do a comprehensive check, but 12.0.0 does give results identical to 11.2.0 for the examples shown in the posts by gwr and Michael E2.
Here is a table comparing the respective (unsimplified) leaf counts of the integrals in Nasser's table, in MMA 12.0.0 and Rubi 4.16.0.2 (running within MMA 12.0.0), respectively.
Rubi can evaluate all of these except for integral no. 12:
Int[(c + d*Sec[e + f*x])^(3/2)/Sqrt[a + b*Sec[e + f*x]],x]
The MMA results for the remaining 22 integrals are, on average, substantially (160–fold, by leaf count) larger than those from Rubi. Evaluating all 22 integrals together in a single cell on a fresh kernel takes 19 times longer in MMA than Rubi.
[Equipment: Mid-2014 MacBook Pro, 2.8 GHz Core i7 (4980HQ Haswell/Crystalwell), MacOS 10.13.6 (High Sierra).]
One other notable distinction between MMA and Rubi (referring to Michael E2's post) is that, when the integrand has non-integrable singularities, Rubi typically (though not always) returns antiderivatives whose discontinuities "match" (i.e., occur at the same points in the domain) those of the integrand, while MMA (especially for more complicated antiderivatives) often does not. Furthermore, I've found that, for intervals over which an integrand is real-valued, Rubi's integrals are much more likely to be real-valued than MMA's (unless the integral is relatively simple, in which case both will be real-valued). For instance, here is the integrand from Michael E2's example (integrand no. 1 from Nasser's list):
expr = Log[x^2 + Sqrt[1 - x^2]];
intRUBI = Int[expr, x];
Plot[{expr, intRUBI}, {x, -2, 2}, PlotRange -> All, PlotLegends -> {"integrand", "Rubi integral"}]
Here's a related example, taken from a WRI blog, of an integrand that is continuous on the reals, but has simple poles elsewhere in the complex plane (https://blog.wolfram.com/2008/01/19/mathematica-and-the-fundamental-theorem-of-calculus/):
expr = 1/(5 + 4 Sin[x]);
intRUBI = Int[expr, x];
intMMA = Integrate[expr, x];
Plot[{expr, intRUBI, intMMA}, {x, -10, 10}, PlotRange -> All, PlotLegends -> {"integrand", "Rubi integral", "MMA integral"}]//Quiet
Note the blog is from 2008, and doesn't mention Rubi, which may not have yet been publicly released; it just happens that the "alternative" antiderivative to which the author compares the MMA result is the one that Rubi generates. Let's call the integrand $\mathcal{h}(z)$, and its MMA and Rubi antiderivatives $\mathcal{H}_1(z)$ and $\mathcal{H}_2(z)$, respectively.
An implicit message of the blog seems to be that $\mathcal{H}_1(z)$'s discontinuities on the reals do not make it an inferior result to $\mathcal{H}_2(z)$, since if the integrand has simple poles, discontinuities somewhere in the antiderivative are unavoidable: "Moreover, if a meromorphic integrand $\mathcal{h}(z)$ has simple poles in the complex plane, it is impossible to choose an antiderivative $\mathcal{H}(z)$ continuous along every imaginable path in the complex plane–because of branch cuts in $\mathcal{H}(z)$."
Specifically, the author explains that while $\mathcal{H}_1(z)$ may have discontinuities in the reals that aren't present in $\mathcal{H}_2(z)$, $\mathcal{H}_2(z)$ has discontinuities elsewhere in the complex plane that aren't present in $\mathcal{H}_1(z)$. For instance, he shows that $\mathcal{H}_2(z)$ has a discontinuity at $z = \frac{3}{2} + i \ln(2)$, while $\mathcal{H}_1(z)$ does not. I.e., it's a wash.
However, this notion (that $\mathcal{H}_1(z)$ and $\mathcal{H}_2(z)$ are equivalent merely because they both have discontinuities) doesn't make sense to me, since it ignores the importance of where the discontinuities occur. Here there is a choice between an antiderivative that is continous along the reals and one that is not, and I think the former is typically a more convenient choice. Yet, even when an antiderivative that is continous on the reals exists (which is generally the case if the integrand is continous on the reals*), Mathematica often instead provides one that is not.
*With the possible exceptions, as Daniel Lichblau points out, of "complex-analytic antiderivatives, or a different notion of integral e.g as might exist in the world of generalized functions".
