Verifying an antiderivative found in any integral table If $a > 0$, and $0 < b < c$.
\begin{equation*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
= \frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert .
\end{equation*}
(This is the antiderivative given in any calculus book.) With calculations that I made, I showed that
\begin{align*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
&= \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert
,
\end{align*}
and since the sum of an antiderivative of $1/[b + c\sin(ax)]$ and a constant is another antiderivate of $1/[b + c\sin(ax)]$, and since
\begin{equation*}
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b}
\qquad \text{and} \qquad
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b}
\end{equation*}
are reciprocals of each other,
\begin{align*}
&\int \frac{1}{b + c\sin(ax)} \, {\mathit dx} \\
&\qquad \qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert \\
&\qquad \qquad \qquad\qquad - \frac{1}{a\sqrt{c^{2} - b^{2}}} \ln\left( \frac{c + \sqrt{c^{2} - b^{2}}}{b} \right) \\
&\qquad \qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\sin(ax) + \dfrac{c + \sqrt{c^{2} - b^{2}}}{b} + \dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert \\
&\qquad\qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
\right\vert .
\end{align*}
Furthermore, we have the following trigonometric identity:
\begin{align*}
&\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)} \\
&\qquad\qquad =\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
} .
\end{align*}
So, the antiderivative given in any calculus book is the same function as the second antiderivative that I obtained:
\begin{align*}
&\frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert \\
&\qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
\right\vert .
\end{align*}
Here are my questions.  Is it evident to anyone that the function on the right side of the trigonometric identity simplifies to the function on the left side? (It is surprising that the two sides are equal. The numerators and denominators on each side are "almost" the same: the numerator and denominator on the right side have two more terms than those on the left side.) If it is not evident, can someone give me calculations, starting from integration using the technique of trigonometric substitution, that show that the antiderivative of $1/[b + c\sin(ax)]$ is the function that is found in the integral tables of any calculus book - calculations that avoid all the algebraic manipulations?

@Tunk-Fey makes a good point and the two expressions are not necessarily equal as they can differ by a constant. However, surprisingly, the two sides are equal in this case. Just ''rationalize'' the numerator of the expression inside the logarithm of the right side. To be more precise, perform the following computation:
\begin{equation}
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}\times 1,
\end{equation}
where we write 1 as
\begin{equation}
1=
\frac{
b + c\sin(ax) - \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
{
b + c\sin(ax) - \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
.
\end{equation}
You will then find that the denominator becomes:
$$
2\,b \left( 1+\cos \left( ax \right)  \right)  \left( c\sin \left( ax
 \right) +b \right)
$$
and the numerator becomes:
$$
2\,b \left( 1+\cos \left(a x \right)  \right)  \left( \sqrt {c^2-b^2 }\cos \left(a x \right) +b\sin \left( ax
 \right) +c \right)
$$
If you divide these two expressions, you do get the argument of the logarithmic function on the left side. 
