Solving a given complex integral I am trying to solve a problem that involves solving the integral
$$\int\frac{1}{\sqrt{y^2 + a^2}} \left(\frac{\sqrt{y^2 + a^2}}{k} - 1\right)^pdy$$
Where $$p=1-\frac{1}{1+n}, n>1$$$, $n$ is an integer.
A simpler representation 
$$\int\left(\frac{a\cosh(t)}{k} - 1\right)^pdt$$
where $y=\sinh(t)$.
I tried solving the above  integral using Mathematica, but with no help. Does anyone know how to solve the above integral. An approximate answer may help.
 A: I reached an equivalent differential equation, see equation (09) below, and  I'll try to solve it. May be I converted the problem to one of equal difficulty, so creating a "hole in  the water": the indefinite integral we try to determine to be the solution of this differential equation (vicious cycle).
\begin{equation}
\int\frac{1}{\sqrt{y^2 + a^2}} \left(\frac{\sqrt{y^2 + a^2}}{k} - 1\right)^pdy
\tag{01} 
\end{equation}
\begin{equation}
p=1-\frac{1}{1+n}, n>1
\tag{02} 
\end{equation} 
Change of variable
\begin{equation}
z = \sqrt{y^2 + a^2}
\tag{03} 
\end{equation}
then
\begin{equation*}
z^2 = y^2 + a^2\Longrightarrow 2zdz=2ydy\Longrightarrow dy=\dfrac{zdz}{y}
\end{equation*}
so
\begin{equation}
dy=\dfrac{zdz}{\sqrt{z^2 - a^2}}
\tag{04} 
\end{equation}
and (01) yields
\begin{equation}
\int\frac{1}{\sqrt{z^2 - a^2}} \biggl(\dfrac{z}{k} - 1\biggr)^pdz = F(z)
\tag{05} 
\end{equation}
that is we must have
\begin{equation}
 dF(z)=\dfrac{1}{\sqrt{z^2 - a^2}} \biggl(\dfrac{z}{k} - 1\biggr)^pdz 
\tag{06} 
\end{equation}
Because of the power term in the right hand side let suppose the following expression for $\:F(z)\:$
\begin{equation}
 F(z)= \biggl(\dfrac{z}{k} - 1\biggr)^{p+1} G(z)+ C_{1}\;,  \quad C_{1}=\text{constant}
\tag{07}
\end{equation}
where $\:G(z)\:$ unknown  function to be determined.
Then
\begin{equation*}
 dF(z)= \biggl(\dfrac{z}{k} - 1\biggr)^{p+1}dG + \dfrac{p+1}{k}\biggl(\dfrac{z}{k} - 1\biggr)^{p}Gdz 
\end{equation*}
or
\begin{equation}
 dF(z)= \left[\biggl(\dfrac{z}{k} - 1\biggr)dG + \dfrac{p+1}{k}Gdz\right]\biggl(\dfrac{z}{k} - 1\biggr)^{p}
\tag{08}   
\end{equation}
Comparing equation (08) with (06) the unknown function $\:G(z)\:$ must satisfy the following differential equation
\begin{equation*}
\biggl(\dfrac{z}{k} - 1\biggr)dG + \biggl(\dfrac{p+1}{k}\biggr)Gdz-\dfrac{1}{\sqrt{z^2 - a^2}}dz=0  
\end{equation*}
or
\begin{equation}
\bbox[#FFFF88,12px]{\biggl(\dfrac{z}{k} - 1\biggr)\dfrac{dG}{dz} + \biggl(\dfrac{p+1}{k}\biggr)G-\dfrac{1}{\sqrt{z^2 - a^2}}=0}
\tag{09}   
\end{equation}
To any specific solution of "inhomogeneous" equation (09) we might add the solution $\:G_{0}(z)\:$ of the homogeneous differential equation
\begin{equation}
\biggl(\dfrac{z}{k} - 1\biggr)\dfrac{dG_{0}}{dz} + \biggl(\dfrac{p+1}{k}\biggr)G_{0}=0
\tag{10}   
\end{equation}
which is
\begin{equation}
G_{0}(z)=\dfrac{C_{2}}{\biggl(\dfrac{z}{k} - 1\biggr)^{p+1}}\;,  \quad C_{2}=\text{constant}
\tag{11}   
\end{equation}
but contributes nothing essential since it changes the constant $\:C_{1}\:$ in the solution (07) to
$\:C_{1}^{\prime}=C_{1}+C_{2}$.  
UNFORTUNATELY my effort failed till now. If you try to solve the linear 1st order differential equation (09) you'll follow exactly the inverse steps and end up to equation (01) for its solution. I leave my answer here (not deleted) in order to show the equivalence of (01) and (09).  
(Finale with help from WolframAlpha : )
The solution to the differential equation (09) is :  
\begin{equation}
\bbox[#FFFF88,12px]{
G(z)=\dfrac{k\sqrt{\dfrac{z-k}{k-\left| a\right|}+1}\;\sqrt{\dfrac{z+k}{k+\left| a\right|}}\;F_{1}\biggl(p+1;\frac{1}{2};\frac{1}{2};p+2;\dfrac{k-z}{k+\left| a\right|};\dfrac{z-k}{\left| a\right|-k}\biggr)}{\left(p+1\right)\sqrt{z^2 - a^2}}+G_{0}(z)}   
\end{equation}
.......................................................................................................................................................(12)
where $\:F_{1}\:$ the  Appell Hypergeometric Function of two variables, as given in : "A Course of Modern Analysis" E.T.Whittaker-G.N.Watson, 4th Edition 1927, Cambridge University Press (Example 22, page 300):
\begin{equation}
F_{1}\left(\alpha\; ; \beta, \beta^{\prime}\; ; \gamma\; ; x,y \right) =\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\;\dfrac{\alpha_{m+n}\beta_{m}\beta_{n}^{\prime}}{m!\;n!\; \gamma_{m+n}}\;x^{m}y^{n}  
\tag{13}   
\end{equation}
and $\:\rho_{\ell}\:$ the so called Pochhammer symbol :
\begin{equation}
\rho_{\ell} = \rho\left(\rho+1\right)\left(\rho+2\right)\cdots\left(\rho+\ell-1\right)=\dfrac{\Gamma\left(\rho+\ell
\right)}{\Gamma\left(\rho\right)}
\tag{14}   
\end{equation}
(in present case $\:\rho = \alpha,\beta,\beta^{\prime},\gamma \: \text{and}\: \ell=m,n,m+n\:$)
Note that this function obeys the following system of differential equations :                                                                            
\begin{align}
D\left(x,y,\alpha,\beta,\gamma\right)F_{1}\left(x,y\right)&=0
\tag{15a}\\
D\left(y,x,\alpha,\beta^{\prime},\gamma\right)F_{1}\left(x,y\right)&=0
\tag{15b}
\end{align}
where for the differential operators $\:D\:$
\begin{equation}
\begin{split}
&D\left(x,y,\alpha,\beta,\gamma\right)\equiv\\
&x\left(1-x\right)\dfrac{\partial^{2}}{\partial x^{2}}+y\left(1-x\right)\dfrac{\partial^{2}}{\partial x \partial y}+\left[\gamma-\left(\alpha+\beta+1\right)x\right]\dfrac{\partial}{\partial x}-\beta y \dfrac{\partial}{\partial y}-\alpha\beta\\
\end{split} 
\tag{16a}   
\end{equation}
\begin{equation}
\begin{split}
&D\left(y,x,\alpha,\beta^{\prime},\gamma\right)\equiv\\
&y\left(1-y\right)\dfrac{\partial^{2}}{\partial y^{2}}+x\left(1-y\right)\dfrac{\partial^{2}}{\partial y \partial x}+\left[\gamma-\left(\alpha+\beta^{\prime}+1\right)y\right]\dfrac{\partial}{\partial y}-\beta^{\prime} x \dfrac{\partial}{\partial x}-\alpha\beta^{\prime}\\
\end{split} 
\tag{16b}   
\end{equation}
Equation (16a) yields (16b) after the transposition $\:x \longleftrightarrow  y \:$
and the replacement $\:\beta \longrightarrow \beta^{\prime}\:$.
Combining equations (03),(05),(07),(12),(13),(14) we have an expression for the indefinite integral (01). But I don't think that such a complex expression is useful in any case.
 
