For $t>0$, we have $v^2+bt\le\max(2v^2,2bt)$, whence 
\begin{equation}
	P(X\ge t)\le \max[1\wedge Q_1(t),1\wedge Q_2(t)]\le[1\wedge Q_1(t)]+[1\wedge Q_2(t)], 
\end{equation}
where $x\wedge y:=\min(x,y)$, 
\begin{equation}
	Q_1(t):=C\exp\Big(-\frac{t^2}{4v^2}\Big),\quad 
	Q_2(t):=C\exp\Big(-\frac{t^2}{4bt}\Big).
\end{equation}
Next, for $t>0$ we have 
\begin{equation}
	1\le Q_1(t)\iff t\le t_1:=2v\sqrt{\ln C}, 
\end{equation}
\begin{equation}
	1\le Q_2(t)\iff t\le t_2:=4b\ln C.  
\end{equation}
So, 
\begin{equation}
	EX=\int_0^\infty P(X\ge t)\,dt\le I_1+I_2,
\end{equation}
where 
\begin{multline}
	I_1:=\int_0^\infty[1\wedge Q_1(t)]\,dt=t_1+\int_{t_1}^\infty Q_1(t)\,dt \\ 
	=t_1+Cv\sqrt 2\int_{\sqrt{2\ln C}}^\infty e^{-s^2/2}\,ds
	\le t_1+v\sqrt\pi
	=2v\big(\sqrt{\ln C}+\sqrt\pi/2\big)  
\end{multline}
and 
\begin{multline}
	I_2:=\int_0^\infty[1\wedge Q_2(t)]\,dt=t_2+\int_{t_2}^\infty Q_2(t)\,dt \\ 
	=t_2+C\int_{t_2}^\infty \exp\Big(-\frac{t}{4b}\Big)=t_2+4b=4b(1+\ln C); 
\end{multline}
the inequality in the penultimate display follows because $\int_t^\infty e^{-s^2/2}\,ds\le\sqrt{\pi/2}\,e^{-t^2/2}$ for $t\ge0$. 

Thus, 
\begin{equation}
	EX\le 2v\big(\sqrt{\ln C}+\sqrt\pi/2\big)+4b(1+\ln C),
\end{equation}
which is a bit better than the bound requested in the OP, where we see $\sqrt\pi$ in place of $\sqrt\pi/2$.