Arc length of vector function - the integral seems impossible

[overpen57mm]

Homework Statement

Find the arc length from 0-3pi for v(x)=(e^x cos(2x), e^x sin(2x), e^x)

Relevant Equations

Arc length formula for vector equations

The vector equation is $v(x)=(e^x cos(2x), e^x sin(2x), e^x)$

I know the arc-length formula is $S=\int_a^b \|v(x)\| \,dx$

I found the derivative from a previous question dealing with this same function, but the when I plug it into the arc-length function I get an integral that I've tried and tried but just can't get anywhere with. The complexity of the problem also makes me think that I might be approaching it from the wrong direction. Here is the integral as I understand it: $$ \int{\sqrt{ (e^{2x}) ( (-2\sin(2x) + \cos(2x))^2 + (2\cos(2x)+\sin(2x))^2 + 1 ) }} \, dx $$

I would appreciate any tips on the integral or the problem as a whole, if there's another way to solve it that I haven't seen.

 

[fresh_42]

You have an integral over $\|(e^x \cdot F(x))'\|=\|e^x(F'(x)+F(x))\|=2e^x \sqrt{(\cos 2x -\sin 2x)^2+(\sin 2x+\cos 2x)^2+1^2}$ where $F(x)=(\cos 2x,\sin 2x,1).$

 

[overpen57mm]

You have an integral over $\|(e^x \cdot F(x))'\|=\|e^x(F'(x)+F(x))\|=2e^x \sqrt{(\cos 2x -\sin 2x)^2+(\sin 2x+\cos 2x)^2+1^2}$ where $F(x)=(\cos 2x,\sin 2x,1).$

Thanks, I guess I was blinded by the derivative I got from the previous question.