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

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

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Rubi [A]  time = 0.02, 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 = {640, 612, 620, 203} \[ r^3 \tan ^{-1}\left (\frac {x}{\sqrt {2 r x-x^2}}\right )-\frac {1}{2} r (r-x) \sqrt {2 r x-x^2}-\frac {1}{3} \left (2 r x-x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[2*r*x - x^2],x]

[Out]

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

Rule 203

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

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 620

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 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]

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 \operatorname {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 = 0.12, size = 72, normalized size = 1.12 \[ \frac {1}{6} \sqrt {-x (x-2 r)} \left (\frac {6 r^{5/2} \sin ^{-1}\left (\frac {\sqrt {x}}{\sqrt {2} \sqrt {r}}\right )}{\sqrt {x} \sqrt {2-\frac {x}{r}}}-3 r^2-r x+2 x^2\right ) \]

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^(5/2)*ArcSin[Sqrt[x]/(Sqrt[2]*Sqrt[r])])/(Sqrt[x]*Sqrt[2 -
x/r])))/6

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IntegrateAlgebraic [C]  time = 0.18, size = 113, normalized size = 1.77 \[ -\frac {1}{2} i r^3 \tanh ^{-1}\left (\frac {x}{r}+\frac {i \sqrt {2 r x-x^2}}{r}\right )+\frac {1}{6} \sqrt {2 r x-x^2} \left (-3 r^2-r x+2 x^2\right )+\frac {1}{4} i r^3 \log \left (r^2-2 i x \sqrt {2 r x-x^2}+2 r x-2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*Sqrt[2*r*x - x^2],x]

[Out]

(Sqrt[2*r*x - x^2]*(-3*r^2 - r*x + 2*x^2))/6 - (I/2)*r^3*ArcTanh[x/r + (I*Sqrt[2*r*x - x^2])/r] + (I/4)*r^3*Lo
g[r^2 + 2*r*x - 2*x^2 - (2*I)*x*Sqrt[2*r*x - x^2]]

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fricas [A]  time = 0.66, size = 51, normalized size = 0.80 \[ -r^{3} \arctan \left (\frac {\sqrt {2 \, r x - x^{2}}}{x}\right ) - \frac {1}{6} \, {\left (3 \, r^{2} + r x - 2 \, x^{2}\right )} \sqrt {2 \, r x - x^{2}} \]

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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giac [A]  time = 0.65, size = 45, normalized size = 0.70 \[ -\frac {1}{2} \, r^{3} \arcsin \left (\frac {r - x}{r}\right ) \mathrm {sgn}\relax (r) - \frac {1}{6} \, {\left (3 \, r^{2} + {\left (r - 2 \, x\right )} x\right )} \sqrt {2 \, r x - x^{2}} \]

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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maple [A]  time = 0.31, size = 60, normalized size = 0.94

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]

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

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maxima [A]  time = 1.17, size = 63, normalized size = 0.98 \[ -\frac {1}{2} \, r^{3} \arcsin \left (\frac {r - x}{r}\right ) - \frac {1}{2} \, \sqrt {2 \, r x - x^{2}} r^{2} + \frac {1}{2} \, \sqrt {2 \, r x - x^{2}} r x - \frac {1}{3} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} \]

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

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

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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