The sum of energy in an isolated device

[Ludovic]

I have the indefinite integral for the work of the black arm:

$\int \sqrt{(1/6)^2+0.182355^2}\sqrt{(1.5\cos(2x))^2+(1.5+1.5\sin(2x))^2}(\sin(\pi/4-x+0.7404789456)-\cos(\pi/4-x+0.7404789456)) dx = \frac{0.5\sqrt{\sin(2x)+1}(-0.047065x+0.523532\sin(2x)+0.0235325\cos(2x))}{\sin(x)+\cos(x)}$

So, it's not possible a problem of accuracy.