Question: ArcLength

Problem: find the length of the curve y=(x/2)^{3/2}. Using VectorCalculus Maple11 says

VectorCalculus[`-`](VectorCalculus[`*`](1/24, VectorCalculus[`*`](VectorCalculus[`*`](sqrt(3), VectorCalculus[`+`](VectorCalculus[`+`](VectorCalculus[`+`](VectorCalculus[`-`](VectorCalculus[`*`](VectorCalculus[`*`](VectorCalculus[`*`](VectorCalculus[`*`](2, 1/27), Pi^VectorCalculus[`*`](5, 1/2)), sqrt(3)), hypergeom([1/6, 1/2, VectorCalculus[`*`](5, 1/6)], [VectorCalculus[`*`](4, 1/3), VectorCalculus[`*`](5, 1/3)], VectorCalculus[`-`](1/729)))), VectorCalculus[`*`](VectorCalculus[`*`](VectorCalculus[`*`](16, 1/27), Pi^VectorCalculus[`*`](5, 1/2)), sqrt(3))), VectorCalculus[`-`](VectorCalculus[`*`](VectorCalculus[`*`](VectorCalculus[`*`](VectorCalculus[`*`](8, 1/3), Pi^VectorCalculus[`*`](5, 1/2)), sqrt(3)), hypergeom([VectorCalculus[`-`](1/6), 1/6, 1/2], [VectorCalculus[`*`](2, 1/3), VectorCalculus[`*`](4, 1/3)], VectorCalculus[`-`](1/729))))), VectorCalculus[`-`](VectorCalculus[`*`](VectorCalculus[`*`](VectorCalculus[`*`](16, Pi^VectorCalculus[`*`](5, 1/2)), sqrt(3)), hypergeom([VectorCalculus[`-`](1/2), VectorCalculus[`-`](1/6), 1/6], [1/3, VectorCalculus[`*`](2, 1/3)], VectorCalculus[`-`](1/729)))))), 1/Pi^VectorCalculus[`*`](5, 1/2))))

Sorry. How could I copy math2D?

Anyway, the correct answer is 2/27 ( 10 sqrt(10) -1 )

Sandor