Calculate the integral : $ I=\int_{x_0}^x (1+s^2)^{7/2} ds $. i have difficulties to calculate the following integral 
$$
I=\displaystyle\int_{x_0}^x (1+s^2)^{7/2} ds
$$
i try to do this 
we put $s= \sinh(t)$ then $ds=\cosh(t)$. Then
$$
I= \displaystyle\int_{argsh(x_0)}^{argsh(x)} (1+\sinh^2(t))^{7/2} \cosh(t) dt= \displaystyle\int_{argsh(x_0)}^{argsh(x)} (\cosh(t)^8 dt
$$
We have that $\cosh(t)=\dfrac{e^t+e^{-t}}{2}$, then 
$$
I= \dfrac{1}{2^8} \displaystyle\int_{argsh(x_0)}^{argsh(x)} (e^t+e^{-t})^8 dt
$$
but i can found the result. How we calculate it please. There is an simple method?
 A: Here are two practical methods:
(1) Use hyperbolic double angle identities Use identity $\cosh^2 t = \frac{1}{2} (1 + \cosh 2t)$, to rewrite the integrand as
$$\cosh^8 t = \left[\frac{1}{2}(1 + \cosh 2t)\right]^4 = \frac{1}{2^4} \left(\cosh^4 2t + 3 \cosh^3 2 t + \cdots \right).$$
Then, the even powers can be handled in the same way and the $\cosh^3 2t$ term can be handled with the usual trick of writing $\cosh^3 u = (1 + \sinh^2 u) \cosh u$ and making the natural substitution.
In the end one gets an antiderivative that is a linear combination of expressions of the form $\cosh j t \sinh k t$. Now, we can simply back-substitute and so produce an antiderivative with expressions with $\cosh j \operatorname{arsinh s}$, but those expressions are not in their most familiar form. Instead, using the hyperbolic double-angle identities (including the previously mentioned one, rearranged, as well as $\sinh 2 x = 2 \sinh x \cosh x)$, we can rewrite the antiderivative as a linear combination of terms of the form $\sinh^{\ell} t \cosh^m t$ and rewrite these in terms of $s$ using the identity $\cosh \operatorname{arsinh s} = \sqrt{1 + s^2}$ to produce an antiderivative of the form
$$\int (1 + s^2)^{7 / 2} ds = A \operatorname{arsinh s} + P(s, \sqrt{1 + s^2})$$ for some constant $A$ and polynomial $P$ (odd in the first argument, since the original integrand is even).
(2) Derive and apply a reduction formula For any integral of the form $\int (1 + s^2)^m ds$, we can apply integration by parts (with $u = (1 + s^2)^m$, $dv = ds$) to produce (for $m \neq 0$) an identity
$$\int (1 + s^2)^m = s (1 + s^2)^m - 2 \int s^2 (1 + s^2)^{m - 1} ds .$$
We can rewrite $s^2$ in the integral on the right as $(1 + s^2) - 1$ and use linearity to give
\begin{align}
\int (1 + s^2)^m
&= s (1 + s^2)^m - 2 \int [(1 + s^2) - 1](1 + s^2)^{m - 1} ds \\
&= s (1 + s^2)^m - 2 \int (1 + s^2)^m ds + 2 \int (1 + s^2)^{m - 1} ds
.
\end{align}
Now, $\int (1 + s^2)^m ds$ appears on both sides of the equation, and be rearranging we can produce an identity of the form
$$\int (1 + s^2)^m ds = f_m(s) + B \int (1 + s^2)^{m - 1} ds$$ for some constant $B$ and some closed-form function $f_m$ that depends on $m$. In particular, applying it lets us write the original integral in terms of similar integrals with smaller exponents. After applying it thrice to our original integral, the only integral that appears in our expression for the antiderivative is
$\int \sqrt{1 + s^2} ds ,$
which is standard, if slightly tricky.
