3.1.39 \(\int x \sqrt {2 x-x^2} \, dx\) [39]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 = {654, 626, 633, 222} \begin {gather*} -\frac {1}{2} \text {ArcSin}(1-x)-\frac {1}{3} \left (2 x-x^2\right )^{3/2}-\frac {1}{2} (1-x) \sqrt {2 x-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[2*x - x^2],x]

[Out]

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

Rule 222

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]

Rule 626

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 633

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 654

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]

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} \text {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*}

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Mathematica [A]
time = 0.07, size = 50, normalized size = 1.00 \begin {gather*} \frac {1}{6} \sqrt {-((-2+x) x)} \left (-3-x+2 x^2-\frac {6 \tanh ^{-1}\left (\frac {1}{\sqrt {\frac {-2+x}{x}}}\right )}{\sqrt {-2+x} \sqrt {x}}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x*Sqrt[2*x - x^2],x]

[Out]

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

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Maple [A]
time = 0.42, size = 39, normalized size = 0.78

method	result	size
risch	\(-\frac {\left (2 x^{2}-x -3\right ) x \left (x -2\right )}{6 \sqrt {-x \left (x -2\right )}}+\frac {\arcsin \left (x -1\right )}{2}\)	\(32\)
default	\(-\frac {\left (-x^{2}+2 x \right )^{\frac {3}{2}}}{3}-\frac {\left (2-2 x \right ) \sqrt {-x^{2}+2 x}}{4}+\frac {\arcsin \left (x -1\right )}{2}\)	\(39\)
meijerg	\(\frac {4 i \left (\frac {i \sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-10 x^{2}+5 x +15\right ) \sqrt {1-\frac {x}{2}}}{120}-\frac {i \sqrt {\pi }\, \arcsin \left (\frac {\sqrt {2}\, \sqrt {x}}{2}\right )}{4}\right )}{\sqrt {\pi }}\)	\(52\)
trager	\(\left (\frac {1}{3} x^{2}-\frac {1}{6} x -\frac {1}{2}\right ) \sqrt {-x^{2}+2 x}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+2 x}+x -1\right )}{2}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.54, size = 49, normalized size = 0.98 \begin {gather*} -\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 {gather*}

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)

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Fricas [A]
time = 1.65, size = 42, normalized size = 0.84 \begin {gather*} \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 {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {- x \left (x - 2\right )}\, dx \end {gather*}

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)

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Giac [A]
time = 1.31, size = 29, normalized size = 0.58 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.08, size = 42, normalized size = 0.84 \begin {gather*} -\frac {\sqrt {2\,x-x^2}\,\left (-8\,x^2+4\,x+12\right )}{24}-\frac {\ln \left (x-1-\sqrt {-x\,\left (x-2\right )}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(2*x - x^2)^(1/2),x)

[Out]

- (log(x - (-x*(x - 2))^(1/2)*1i - 1)*1i)/2 - ((2*x - x^2)^(1/2)*(4*x - 8*x^2 + 12))/24

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