Looking for a bound on a contour integral I'm looking for an estimate of this contour integral
$$\left|\,\int_{\gamma}\,\left(\,1+\log^2(u)\,\right)^a\,du\,\right|$$
where parammeter $a\in\mathbb{C}$ and the path $\gamma$ is the circle with center $b$ and radius $b$, that is, $\gamma=\{\,u \,;\,|u-b|=b\,\}$ with $ 0.5\leq b<0.8.$
Since its calculation is difficult, it would be enough to find a bound. With the classical parameterization
$$\left|\,\int_{\gamma}\,\left(\,1+\log^2(u)\,\right)^a\,du\,\right|\leq\,b\,\int_0^{2\pi}\left|\,\left(\,1+\log^2(b+be^{i\,x})\,\right)^{\,a}\,\right|\,dx$$
Any help will be wellcome.
Edit_1:
We can split
$$\int_0^{2\pi}\left|\,\left(\,1+\log^2(b+be^{i\,x})\,\right)^{\,a}\,\right|\,dx=\int_0^{\pi}\left|\,\left(\,1+\log^2(b+be^{i\,x})\,\right)^{\,a}\,\right|\,dx+\int_{\pi}^{2\pi}\left|\,\left(\,1+\log^2(b+be^{i\,x})\,\right)^{\,a}\,\right|\,dx$$
and after a chenge of variable $x-\pi\to y $ we obtain two similar integrals
$$\int_0^{\pi}\left|\,\left(\,1+\log^2(b+be^{i\,x})\,\right)^{\,a}\,\right|\,dx+\int_0^{\pi}\left|\,\left(\,1+\log^2(b-be^{i\,x})\,\right)^{\,a}\,\right|\,dx$$
The question was solved by Szeto below.
Edit_2: Now, it's possible to find $\int_0^{2\pi}\left|\,\left(\,1+\log^2(b+be^{i\,x})\,\right)^{\,a}\,\right|\,dx$ ?
May be with a change of variable like $b+be^{i\,x}\to e^y$ ?
 A: Assuming the contour does not enclose any branch cuts, then

Claim: $$I:=\text{PV}\,\int_{\gamma}\,\left(\,1+\log^2(u)\,\right)^a\,du=0$$

Proof:
Consider $\Gamma(\epsilon)$, which is the boundary of $\{z~|~|z-b|=b\}\setminus\{z~|~|z|=\epsilon\}$.
Let $\Gamma(\epsilon)=\Gamma_1(\epsilon)\cup\Gamma_2(\epsilon)$, where $\Gamma_1(\epsilon)$ is the circular contour with a bit removed at the origin, and $\Gamma_2(\epsilon)$ is the tiny detour around origin.
By definition, $\displaystyle{I=\lim_{\epsilon\to 0^+}\int_{\Gamma_1(\epsilon)}\left(\,1+\log^2(u)\,\right)^a\,du}$
1.
Recognize that your contour does not enclose any branch points (the branch point of taking to the power of $a$ is at $e^{\pm i}$, which requires $b>\frac1{2\cos 1}\approx 0.9$ to enclose it) , nor singularities. 
By Cauchy's theorem,
$$\oint_{\Gamma(\epsilon)}\left(\,1+\log^2(u)\,\right)^a\,du=0$$ for every $\epsilon>0$.
2.
Let $a=p+iq$.
We observe that, for some $c\in[-\frac\pi2,\frac\pi2]$, 
$$|(1+\log^2(\epsilon e^{ic}))^a|=\left((1-c^2+\log^2\epsilon)^2+4c^2\log^2\epsilon\right)^{p/2}e^{-q\theta}$$ where $|\theta|$ is bounded by some constant $M$.
Secondly,
$$
\begin{align}
\left|\lim_{\epsilon\to0^+}\int_{\Gamma_2(\epsilon)}\left(\,1+\log^2(u)\,\right)^a\,du\right| 
&\le \lim_{\epsilon\to 0^+}\int_{\pi/2}^{-\pi/2}|\left(\,1+\log^2(\epsilon e^{it})\,\right)^a i\epsilon e^{it}|dt \\
&=\lim_{\epsilon\to 0^+}\int_{\pi/2}^{-\pi/2}|\left(\,1+\log^2(\epsilon e^{it})\,\right)^a |\epsilon dt \\
&=\lim_{\epsilon\to 0^+}\pi\cdot\epsilon |(1+\log^2(\epsilon e^{ic}))^a| \qquad{(1)} \\
&\le\lim_{\epsilon\to 0^+}\pi\cdot\epsilon \left((1-c^2+\log^2\epsilon)^2+4c^2\log^2\epsilon\right)^{p/2}e^{|q|M} \\
&=0
\end{align}
$$
$(1)$: By mean value theorem for integral, $c\in[-\frac\pi2,\frac\pi2]$.
Therefore,
$$\begin{align}
I
&=\lim_{\epsilon\to 0^+}\int_{\Gamma_1(\epsilon)}\left(\,1+\log^2(u)\,\right)^a\,du \\
&=\lim_{\epsilon\to0^+}\oint_{\Gamma(\epsilon)}\left(\,1+\log^2(u)\,\right)^a\,du-\lim_{\epsilon\to0^+}\int_{\Gamma_2(\epsilon)}\left(\,1+\log^2(u)\,\right)^a\,du \\
&=0-0 \\
&=0
\end{align}
$$
Q.E.D.
