Stuck on an integral

I think I am supposed to use integration by substitution or integration by parts, but I'm not sure. Can anyone help with this?

$\int{x}^{3}\sqrt{{x}^{2}-1}dx$

We apply twice substitution

We need the following formulae:

$\begin{array}{|rcll|} \hline \cosh^2(z) - \sinh^2(z) &=& 1 \\ \sinh^2(z) &=& \cosh^2(z) -1 \\ \sinh(z) &=& \sqrt{\cosh^2(z) -1} \\\\ \cosh^2(z) - \sinh^2(z) &=& 1 \\ \cosh^2(z) &=& 1+\sinh^2(z) \\\\ \frac{d}{dz}\sinh(z) &=& \cosh(z) \\ \frac{d}{dz}\cosh(z) &=& \sinh(z) \\ \hline \end{array}$

1. Substitution

$\begin{array}{|rcll|} \hline x &=& \cosh(z)\\ dx &=& \sinh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ x^3\cdot\sqrt{x^2-1}\ dx } \\ &=& \int{ \cosh^3(z)\cdot \sqrt{\cosh^2(z)-1} \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh(z) \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ \hline \end{array}$

2. Substitution

$\begin{array}{|rcll|} \hline u &=& \sinh(z)\\ du &=& \cosh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ &=& \int{ \cosh^2(z)\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ [1+\sinh^2(z)]\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ (1+u^2)\cdot u^2 \ du } \\ &=& \int{ (u^2+u^4)\ du } \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3}\cdot \frac55 + \frac{u^5}{5}\cdot \frac33 + c \\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ \hline \end{array}$

3. Back substitution

$\begin{array}{|rcll|} \hline u &=& \sinh(z) \\ &=& \sqrt{\cosh^2(z) -1} \\ &=& \sqrt{x^2 -1} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{\int{ x^3\cdot\sqrt{x^2-1}\ dx } }\\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ &=& \frac{(\sqrt{x^2 -1})^3}{15}\cdot \left[ 5+3(\sqrt{x^2 -1})^2 \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot \left[ 5+3(x^2 -1) \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot ( 5+3x^2-3 )+ c \\ &\mathbf{=}& \mathbf{ \dfrac{(x^2 -1)^{\frac32}}{15}\cdot ( 3x^2+2 )+ c }\\ \hline \end{array}$

Sorry this got posted 3 times, something was lagging and I couldn't see it so I thought it wasn't posting. 

I've gotten this far on it but after that I'm stuck:

Set u = x² and du = 2x dx

$(1/2)\int u\sqrt{u-1} du$

x = sec θ     dx  =  sec θ*tan θ dθ

∫  sec^3( θ )  √[ sec^2θ  - 1 ] secθ*tanθ dθ  =

∫ sec^3( θ ) √ [ tan^2θ ]  secθ*tanθ dθ  =

∫ sec^3( θ ) √ [ tanθ ]  secθ*tanθ dθ  =

∫ sec^4 θ *  tan^2θ   dθ

∫ [sec^2 θ] * [sec^2 θ] * tan^2θ   dθ  =

∫ tan^2θ * [ tan^2θ + 1] [sec^2 θ]   dθ  =

∫ [  tan^4θ  +  tan^2θ ]  [sec^2 θ]   dθ  =

∫ [ tanθ ]^4 *  [sec^2 θ]   dθ    +  ∫   [ tanθ ]^2 *  [sec^2 θ]  dθ  =

(1/5) [ tanθ ]^5    +  (1/3)  [tanθ ]^3   + C

Now  x  = sec θ     ...so....   x^2  = sec^2θ  =  1 +  [ tanθ]^2  → 

x^2  - 1  =   [tanθ]^2     ...so......

[x^2  - 1]^(5/2)  = [ [tanθ]^2 ] ^(5/2)  = [ tanθ ]^5

And

[x^2  - 1]^(3/2)  = [ [tanθ]^2 ] ^(3/2)  = [ tanθ ]^3

So we have

∫  x^3 √[ x^2  - 1 ]  dx  =

(1/5)[x^2  - 1]^(5/2)  + (1/3)[x^2  - 1]^(3/2)   + C

Proof :

http://www.wolframalpha.com/input/?i=derivative++(1%2F5)%5Bx%5E2++-+1%5D%5E(5%2F2)++%2B+(1%2F3)%5Bx%5E2++-+1%5D%5E(3%2F2)+++%2B+C