Solving Riccati equation $v'=av^2+bv+c$ I am trying to solve Riccati equation 
$$
v(t)'=av(t)^2+bv(t)+c
$$
where $a$,$b$,$c$ are real-valued constants. Wolfram|Alpha gives the solution (here), but I am not able to reproduce the result.
Thank you in advance.
 A: $$v(t)'=av(t)^2+bv(t)+c\Longleftrightarrow$$
$$\frac{dv(t)}{dt}=av(t)^2+bv(t)+c\Longleftrightarrow$$
$$\frac{\frac{dv(t)}{dt}}{av(t)^2+bv(t)+c}=1\Longleftrightarrow$$
$$\int\left(\frac{\frac{dv(t)}{dt}}{av(t)^2+bv(t)+c}\right)dt=\int 1dt\Longleftrightarrow$$
$$\frac{2\tan^{-1}\left(\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\right)}{\sqrt{-b^2+4ac}}=t+k_1\Longleftrightarrow$$
$$\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)=2\tan^{-1}\left(\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\right)\Longleftrightarrow$$
$$\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}=\tan^{-1}\left(\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\right)\Longleftrightarrow$$
$$\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)=\tan\left(\tan^{-1}\left(\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\right)\right)\Longleftrightarrow$$
$$\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)=\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\Longleftrightarrow$$
$$\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)\left(\sqrt{-b^2+4ac}\right)=b+2av(t)\Longleftrightarrow$$
$$\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)\left(\sqrt{-b^2+4ac}\right)-b=2av(t)\Longleftrightarrow$$
$$\frac{\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)\left(\sqrt{-b^2+4ac}\right)-b}{2a}=v(t)\Longleftrightarrow$$
$$v(t)=\frac{1}{2a}\left(\sqrt{4ac-b^2}\tan\left(\frac{1}{2}\left(k_1\sqrt{4ac-b^2}+t\sqrt{4ac-b^2}\right)\right)-b\right)$$
