What does this imaginary number mean for time and velocity? As some have pointed out in the chat, perhaps the question that I should have asked is, am I really integrating for velocity? My integration might be misleading in that it integrates for something but probably not velocity. Velocity could not be in the same direction as acceleration unless the object is already at terminal velocity - in which case acceleration = 0 and it therefore has no direction. A paradox, which perhaps explains for my imaginary numbers. I believe that I have understood my conceptual misunderstanding.
In regards to 2D kinematics, I essentially did partial fractions and integrated for time as a function of velocity, and suddenly, I have to deal with an imaginary number! My specific problem is not so much about my working out - it's about why I am obtaining imaginary numbers? Is this inherent of the coefficient of drag and lift equations? I can't see why it would be.
I initially had multiple acceleration functions in terms of velocity, using the equations for coefficient of drag, lift and gravity.
\begin{align}
ma_x &= F_L\sin\theta-F_D\cos\theta \\
ma_y&=mg-F_L\cos\theta-F_D\sin\theta
\end{align}
with
$$
F=C\frac{pv^2}{2}A.
$$
I broke this down into horizontal ($h$) and vertical components ($l$). My third line comes from expanding and simplify the equations of coefficients of drag, lift and gravity. 
 $$ acceleration{^2} = h^2 +l^2$$ 
 $$  \frac{dv}{dt} = \sqrt{h^2 +l^2 }$$ 
$$\sqrt{h^2 +l^2 } = \sqrt{av^4 -bv^2 + \frac{c}{a} }$$
$v$ is velocity, $t$ is time, and all other letters are pre-determined constants.  Rearrange equation to produce integrals and isolate dt 
$$\begin{align}\int \left ( \frac{\mathrm dv}{\sqrt{av^4 -bv^2 + \frac{c}{a}}} \right ) &= \int \mathrm dt \\
\end{align} \\
\frac{\mathrm dv}{\sqrt{a[\left(v^2 - \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a}]}} \
$$
In order to keep things easier to read, substitute a value for $D$.
$$D = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$$ $$
\\
\frac{\mathrm dv}{\sqrt{a[\left(v^2 - \frac{b}{2a}\right)^2 - D]}} \\
\frac{1}{\sqrt{a}} \frac{\mathrm dv}{\sqrt{(v^2 - \frac{b}{2a})^{2} - D}} 
$$
Pretend for a moment that constant $1/\sqrt a$ is not there, to make it easier to write.
$$
\frac{\mathrm dv}{((v^2 - \frac{b}{2a}) + D^{1/2})^{1/2}\cdot((v^2 - \frac{b}{2a}) - D^{1/2})^{1/2}} 
$$
Partial fractions.
$$
\int  \frac{A}{((v^2 - \frac{b}{2a}) + D^{1/2})^{1/2}} + \frac{B}{((v^2 - \frac{b}{2a}) - D^{1/2})^{1/2}} \mathrm dv = t
$$
Equate coefficients.
$$
A\sqrt{-2\sqrt{D}} = 1.
$$
This is what makes us have the imaginary number.
$$
A = \frac{1}{\sqrt{-2\sqrt{D}}} $$ and $$ B = \frac{1}{({4D)^{1/4}}}.
$$
I did everything else on Wolfram Alpha, including the integration. The result is,
$$
\frac{\log\left(\sqrt{2}\sqrt{\frac{-2a\sqrt{D}+2av^2-b}{a}}+2v\right)}{\sqrt{2}\sqrt[4]{D}}+\frac{\log\left(\sqrt{2}\sqrt{\frac{2a\sqrt{D}+2av^2-b}{a}}+2v\right)}{\sqrt{2}\sqrt{-\sqrt D}}=t
$$

Complex velocity doesn't make sense in physics so you have to choose the parameters $a,b,c$ so you don't get an imaginary velocity.
\begin{align*}
 & \sqrt{a v^4- b v^2+\frac{c}{a}}\quad\Rightarrow\quad a v^4- b v^2+\frac{c}{a} \ge 0\\
  &v^2 \mapsto x\quad\Rightarrow\\ &g_1=a x^2- b x+\frac{c}{a} \ge 0\\
  &g_1=(x-\tau_1)\,(x-\tau_2)\ge 0\quad\text{with:}\\
  &\tau_1=-\frac{b}{2\,a}+\frac{1}{2\,a}\,\sqrt{b^2-4\,c}\\
   &\tau_2=-\frac{b}{2\,a}-\frac{1}{2\,a}\,\sqrt{b^2-4\,c}\\
   &\quad \Rightarrow\quad \\&b^2-4\,c \ge 0\quad b\ge 2\,\sqrt{c}
   \,\quad c \ge 0\\
   &\text{with:}\quad v^2=x\quad\Rightarrow\quad x > 0&\\\quad \Rightarrow\\
   &g_1\ge 0\quad \,\Rightarrow\quad \\\\&x - \tau_1\ge 0 \quad\text{and}\quad  x-\tau_2\ge 0 \\&\text{or}\\
   &x- \tau_1\le 0 \quad\text{and}\quad  x- \tau_2\le 0
\end{align*}
Consequences:
\begin{align*}
  &c \ge 0\\
  &b \ge 2\sqrt{c}\\
  &a > 0\,,\text{$x$ must be positive !!}\\\\
  & \tau_1 \le x \le \inf
\end{align*}
Example:
\begin{align*}
  &c=3\,,b=2\sqrt{3}+3\,,a=2\\
  &\Rightarrow\\
  &\tau_1=2.98\,,\tau_2=0.25\\
  &x > \tau_1=4\,\Rightarrow\quad g_1=7.6 > 0\\
  &v=\sqrt{x}=2\,,\checkmark
\end{align*}
