What is the equation for the scale factor of the universe, a(t), for the best fit of data to the $\Lambda CDM$ Model of Cosmology?

Ideally I like a single formula or multiple formulas for different time ranges that would cover the time from the end of inflation through 100+ billion years after the big bang using the $\Lambda CDM$ Model. I know that from the end of inflation back to the time of the big bang would be much more speculative, but some wild estimate would be appreciated for that time range also!

The Friedmann–Lemaître–Robertson–Walker metric defines the scale factor, a(t), from the metric: $$-c^2d\tau^2 = -c^2dt^2+a(t)^2d\Sigma^2$$ where $d\Sigma^2$ ranges over the 3 dimensional space of the universe and does not depend on time. Usually the scale of the scale factor is set by defining $a(t_{now}) = a(13.78 B yr) \equiv 1$

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The scale factor can be derived from the Hubble parameter $$H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}.$$ The latest values of the parameters, obtained from the Planck mission, are $$H_0 = 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\\ \Omega_{R,0} = 9.24\times 10^{-5},\qquad\Omega_{M,0} = 0.315,\\ \Omega_{\Lambda,0} = 0.685,\qquad\Omega_{K,0} = 0.$$ From $$\dot{a} = \frac{\text{d}a}{\text{d}t}$$ we get $$\text{d}t = \frac{\text{d}a}{\dot{a}} = \frac{\text{d}a}{aH(a)} = \frac{a\,\text{d}a}{a^2H(a)},$$ so that \begin{align} t(a) &= \int_0^a \frac{a'\,\text{d}a'}{a'^2H(a')}\\ &= \frac{1}{H_0}\int_0^a \frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a' + \Omega_{K,0}\,a'^2 + \Omega_{\Lambda,0}\,a'^4}}, \end{align} which is the age of the universe as a function of $a$. By numerically inverting this relation, we get $a(t)$. For more information, see these posts:
Simply from the form your solution takes one can tell something is wrong here. According to this, there is only one value of $t$ for each $a$, which is not necessarily true for the properly chosen values of the various $\Omega_i$. Also, care to justify this a bit more : $dt = \frac{da}{\dot{a}}$ ? –  ticster Aug 5 '14 at 22:26
@ticster: if the universe doesn't recontract, then $a(t)$ is a one-to-one function. Current observation has ruled out the recollapsing models of the universe. So, it is completely valid that there is only one value of $t$ for each value of $a$ –  Jerry Schirmer Aug 5 '14 at 22:40
@JerrySchirmer No, it isn't. His solution is a general one and is clearly wrong for the values of the $\Omega$ where the universe recollapses. Those scenarios are implicitly part of his solution. It's a sanity check that his solution clearly fails. –  ticster Aug 5 '14 at 23:02
@JerrySchirmer It'd be like someone calculating the maximal height of a cannonball as a function of initial velocity, and got an answer that wasn't $0$ for initial velocity $0$, then saying "that's fine, the particular case I want to apply it to has $v_0 \neq 0$". Again, failing a very specific sanity check doesn't mean your solution might be correct for other, more general situations. That's why it's called a sanity check. –  ticster Aug 5 '14 at 23:07
@ticster: it's still true over the piecewise invertible part of the function. You'd get a singularity in the integral at the recollapse point ($\frac{da}{dt} =0 \rightarrow \frac{dt}{da} = \infty$), and then you'd know that you'd have to assemble a piecewise solution. It's no fancier than inverting a parabola to get a square root function in the recollapsing case. –  Jerry Schirmer Aug 6 '14 at 14:53