3.3.61 \(\int x \sqrt {2 r x-x^2} \, dx\) [261]

Optimal. Leaf size=64 \[ -\frac {1}{2} r (r-x) \sqrt {2 r x-x^2}-\frac {1}{3} \left (2 r x-x^2\right )^{3/2}+r^3 \tan ^{-1}\left (\frac {x}{\sqrt {2 r x-x^2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {654, 626, 634, 209} \begin {gather*} r^3 \text {ArcTan}\left (\frac {x}{\sqrt {2 r x-x^2}}\right )-\frac {1}{2} r (r-x) \sqrt {2 r x-x^2}-\frac {1}{3} \left (2 r x-x^2\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[2*r*x - x^2],x]

[Out]

-1/2*(r*(r - x)*Sqrt[2*r*x - x^2]) - (2*r*x - x^2)^(3/2)/3 + r^3*ArcTan[x/Sqrt[2*r*x - x^2]]

Rule 209

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

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 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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 r x-x^2} \, dx &=-\frac {1}{3} \left (2 r x-x^2\right )^{3/2}+r \int \sqrt {2 r x-x^2} \, dx\\ &=-\frac {1}{2} r (r-x) \sqrt {2 r x-x^2}-\frac {1}{3} \left (2 r x-x^2\right )^{3/2}+\frac {1}{2} r^3 \int \frac {1}{\sqrt {2 r x-x^2}} \, dx\\ &=-\frac {1}{2} r (r-x) \sqrt {2 r x-x^2}-\frac {1}{3} \left (2 r x-x^2\right )^{3/2}+r^3 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {2 r x-x^2}}\right )\\ &=-\frac {1}{2} r (r-x) \sqrt {2 r x-x^2}-\frac {1}{3} \left (2 r x-x^2\right )^{3/2}+r^3 \tan ^{-1}\left (\frac {x}{\sqrt {2 r x-x^2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 66, normalized size = 1.03 \begin {gather*} \frac {1}{6} \sqrt {-x (-2 r+x)} \left (-3 r^2-r x+2 x^2-\frac {6 r^3 \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt {-2 r+x}}\right )}{\sqrt {x} \sqrt {-2 r+x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[2*r*x - x^2],x]

[Out]

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

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Maple [A]
time = 0.07, size = 64, normalized size = 1.00

method	result	size
risch	\(-\frac {\left (3 r^{2}+r x -2 x^{2}\right ) x \left (2 r -x \right )}{6 \sqrt {-x \left (-2 r +x \right )}}+\frac {r^{3} \arctan \left (\frac {x -r}{\sqrt {2 r x -x^{2}}}\right )}{2}\)	\(60\)
default	\(-\frac {\left (2 r x -x^{2}\right )^{\frac {3}{2}}}{3}+r \left (-\frac {\left (2 r -2 x \right ) \sqrt {2 r x -x^{2}}}{4}+\frac {r^{2} \arctan \left (\frac {x -r}{\sqrt {2 r x -x^{2}}}\right )}{2}\right )\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 63, normalized size = 0.98 \begin {gather*} -\frac {1}{2} \, r^{3} \arcsin \left (\frac {r - x}{r}\right ) - \frac {1}{2} \, \sqrt {2 \, r x - x^{2}} r^{2} + \frac {1}{2} \, \sqrt {2 \, r x - x^{2}} r x - \frac {1}{3} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.49, size = 51, normalized size = 0.80 \begin {gather*} -r^{3} \arctan \left (\frac {\sqrt {2 \, r x - x^{2}}}{x}\right ) - \frac {1}{6} \, {\left (3 \, r^{2} + r x - 2 \, x^{2}\right )} \sqrt {2 \, r x - x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(-x*(-2*r + x)), x)

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Giac [A]
time = 0.58, size = 45, normalized size = 0.70 \begin {gather*} -\frac {1}{2} \, r^{3} \arcsin \left (\frac {r - x}{r}\right ) \mathrm {sgn}\left (r\right ) - \frac {1}{6} \, {\left (3 \, r^{2} + {\left (r - 2 \, x\right )} x\right )} \sqrt {2 \, r x - x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.10, size = 56, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {2\,r\,x-x^2}\,\left (12\,r^2+4\,r\,x-8\,x^2\right )}{24}-\frac {r^3\,\ln \left (x-r-\sqrt {x\,\left (2\,r-x\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*r*x - x^2)^(1/2),x)

[Out]

- ((2*r*x - x^2)^(1/2)*(4*r*x + 12*r^2 - 8*x^2))/24 - (r^3*log(x - r - (x*(2*r - x))^(1/2)*1i)*1i)/2

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