3.39 \(\int x \sqrt{2 x-x^2} \, dx\)

Optimal. Leaf size=50 \[ -\frac{1}{3} \left (2 x-x^2\right )^{3/2}-\frac{1}{2} (1-x) \sqrt{2 x-x^2}-\frac{1}{2} \sin ^{-1}(1-x) \]

[Out]

-((1 - x)*Sqrt[2*x - x^2])/2 - (2*x - x^2)^(3/2)/3 - ArcSin[1 - x]/2

________________________________________________________________________________________

Rubi [A]  time = 0.0109595, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {640, 612, 619, 216} \[ -\frac{1}{3} \left (2 x-x^2\right )^{3/2}-\frac{1}{2} (1-x) \sqrt{2 x-x^2}-\frac{1}{2} \sin ^{-1}(1-x) \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[2*x - x^2],x]

[Out]

-((1 - x)*Sqrt[2*x - x^2])/2 - (2*x - x^2)^(3/2)/3 - ArcSin[1 - x]/2

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int x \sqrt{2 x-x^2} \, dx &=-\frac{1}{3} \left (2 x-x^2\right )^{3/2}+\int \sqrt{2 x-x^2} \, dx\\ &=-\frac{1}{2} (1-x) \sqrt{2 x-x^2}-\frac{1}{3} \left (2 x-x^2\right )^{3/2}+\frac{1}{2} \int \frac{1}{\sqrt{2 x-x^2}} \, dx\\ &=-\frac{1}{2} (1-x) \sqrt{2 x-x^2}-\frac{1}{3} \left (2 x-x^2\right )^{3/2}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4}}} \, dx,x,2-2 x\right )\\ &=-\frac{1}{2} (1-x) \sqrt{2 x-x^2}-\frac{1}{3} \left (2 x-x^2\right )^{3/2}-\frac{1}{2} \sin ^{-1}(1-x)\\ \end{align*}

Mathematica [A]  time = 0.0484477, size = 39, normalized size = 0.78 \[ \frac{1}{6} \sqrt{-(x-2) x} \left (2 x^2-x-3\right )-\sin ^{-1}\left (\sqrt{1-\frac{x}{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[2*x - x^2],x]

[Out]

(Sqrt[-((-2 + x)*x)]*(-3 - x + 2*x^2))/6 - ArcSin[Sqrt[1 - x/2]]

________________________________________________________________________________________

Maple [A]  time = 0.05, size = 39, normalized size = 0.8 \begin{align*} -{\frac{1}{3} \left ( -{x}^{2}+2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{2-2\,x}{4}\sqrt{-{x}^{2}+2\,x}}+{\frac{\arcsin \left ( -1+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-x^2+2*x)^(1/2),x)

[Out]

-1/3*(-x^2+2*x)^(3/2)-1/4*(2-2*x)*(-x^2+2*x)^(1/2)+1/2*arcsin(-1+x)

________________________________________________________________________________________

Maxima [A]  time = 1.78255, size = 66, normalized size = 1.32 \begin{align*} -\frac{1}{3} \,{\left (-x^{2} + 2 \, x\right )}^{\frac{3}{2}} + \frac{1}{2} \, \sqrt{-x^{2} + 2 \, x} x - \frac{1}{2} \, \sqrt{-x^{2} + 2 \, x} - \frac{1}{2} \, \arcsin \left (-x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-x^2+2*x)^(1/2),x, algorithm="maxima")

[Out]

-1/3*(-x^2 + 2*x)^(3/2) + 1/2*sqrt(-x^2 + 2*x)*x - 1/2*sqrt(-x^2 + 2*x) - 1/2*arcsin(-x + 1)

________________________________________________________________________________________

Fricas [A]  time = 1.80531, size = 90, normalized size = 1.8 \begin{align*} \frac{1}{6} \,{\left (2 \, x^{2} - x - 3\right )} \sqrt{-x^{2} + 2 \, x} - \arctan \left (\frac{\sqrt{-x^{2} + 2 \, x}}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-x^2+2*x)^(1/2),x, algorithm="fricas")

[Out]

1/6*(2*x^2 - x - 3)*sqrt(-x^2 + 2*x) - arctan(sqrt(-x^2 + 2*x)/x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{- x \left (x - 2\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-x**2+2*x)**(1/2),x)

[Out]

Integral(x*sqrt(-x*(x - 2)), x)

________________________________________________________________________________________

Giac [A]  time = 1.21428, size = 39, normalized size = 0.78 \begin{align*} \frac{1}{6} \,{\left ({\left (2 \, x - 1\right )} x - 3\right )} \sqrt{-x^{2} + 2 \, x} + \frac{1}{2} \, \arcsin \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-x^2+2*x)^(1/2),x, algorithm="giac")

[Out]

1/6*((2*x - 1)*x - 3)*sqrt(-x^2 + 2*x) + 1/2*arcsin(x - 1)