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

Optimal. Leaf size=89 \[ -\frac{5}{8} r^2 (r-x) \sqrt{2 r x-x^2}+\frac{5}{4} r^4 \tan ^{-1}\left (\frac{x}{\sqrt{2 r x-x^2}}\right )-\frac{5}{12} r \left (2 r x-x^2\right )^{3/2}-\frac{1}{4} x \left (2 r x-x^2\right )^{3/2} \]

[Out]

(-5*r^2*(r - x)*Sqrt[2*r*x - x^2])/8 - (5*r*(2*r*x - x^2)^(3/2))/12 - (x*(2*r*x - x^2)^(3/2))/4 + (5*r^4*ArcTa
n[x/Sqrt[2*r*x - x^2]])/4

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Rubi [A]  time = 0.0296294, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {670, 640, 612, 620, 203} \[ -\frac{5}{8} r^2 (r-x) \sqrt{2 r x-x^2}+\frac{5}{4} r^4 \tan ^{-1}\left (\frac{x}{\sqrt{2 r x-x^2}}\right )-\frac{5}{12} r \left (2 r x-x^2\right )^{3/2}-\frac{1}{4} x \left (2 r x-x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[2*r*x - x^2],x]

[Out]

(-5*r^2*(r - x)*Sqrt[2*r*x - x^2])/8 - (5*r*(2*r*x - x^2)^(3/2))/12 - (x*(2*r*x - x^2)^(3/2))/4 + (5*r^4*ArcTa
n[x/Sqrt[2*r*x - x^2]])/4

Rule 670

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

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 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 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])

Rubi steps

\begin{align*} \int x^2 \sqrt{2 r x-x^2} \, dx &=-\frac{1}{4} x \left (2 r x-x^2\right )^{3/2}+\frac{1}{4} (5 r) \int x \sqrt{2 r x-x^2} \, dx\\ &=-\frac{5}{12} r \left (2 r x-x^2\right )^{3/2}-\frac{1}{4} x \left (2 r x-x^2\right )^{3/2}+\frac{1}{4} \left (5 r^2\right ) \int \sqrt{2 r x-x^2} \, dx\\ &=-\frac{5}{8} r^2 (r-x) \sqrt{2 r x-x^2}-\frac{5}{12} r \left (2 r x-x^2\right )^{3/2}-\frac{1}{4} x \left (2 r x-x^2\right )^{3/2}+\frac{1}{8} \left (5 r^4\right ) \int \frac{1}{\sqrt{2 r x-x^2}} \, dx\\ &=-\frac{5}{8} r^2 (r-x) \sqrt{2 r x-x^2}-\frac{5}{12} r \left (2 r x-x^2\right )^{3/2}-\frac{1}{4} x \left (2 r x-x^2\right )^{3/2}+\frac{1}{4} \left (5 r^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{x}{\sqrt{2 r x-x^2}}\right )\\ &=-\frac{5}{8} r^2 (r-x) \sqrt{2 r x-x^2}-\frac{5}{12} r \left (2 r x-x^2\right )^{3/2}-\frac{1}{4} x \left (2 r x-x^2\right )^{3/2}+\frac{5}{4} r^4 \tan ^{-1}\left (\frac{x}{\sqrt{2 r x-x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.102408, size = 80, normalized size = 0.9 \[ \frac{1}{24} \sqrt{-x (x-2 r)} \left (-5 r^2 x+\frac{30 r^{7/2} \sin ^{-1}\left (\frac{\sqrt{x}}{\sqrt{2} \sqrt{r}}\right )}{\sqrt{x} \sqrt{2-\frac{x}{r}}}-15 r^3-2 r x^2+6 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[2*r*x - x^2],x]

[Out]

(Sqrt[-(x*(-2*r + x))]*(-15*r^3 - 5*r^2*x - 2*r*x^2 + 6*x^3 + (30*r^(7/2)*ArcSin[Sqrt[x]/(Sqrt[2]*Sqrt[r])])/(
Sqrt[x]*Sqrt[2 - x/r])))/24

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Maple [A]  time = 0.003, size = 91, normalized size = 1. \begin{align*} -{\frac{x}{4} \left ( 2\,rx-{x}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{5\,r}{12} \left ( 2\,rx-{x}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{r}^{2}x}{8}\sqrt{2\,rx-{x}^{2}}}-{\frac{5\,{r}^{3}}{8}\sqrt{2\,rx-{x}^{2}}}+{\frac{5\,{r}^{4}}{8}\arctan \left ({(x-r){\frac{1}{\sqrt{2\,rx-{x}^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x*(2*r*x-x^2)^(3/2)-5/12*r*(2*r*x-x^2)^(3/2)+5/8*r^2*(2*r*x-x^2)^(1/2)*x-5/8*(2*r*x-x^2)^(1/2)*r^3+5/8*r^
4*arctan((x-r)/(2*r*x-x^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.1459, size = 135, normalized size = 1.52 \begin{align*} -\frac{5}{4} \, r^{4} \arctan \left (\frac{\sqrt{2 \, r x - x^{2}}}{x}\right ) - \frac{1}{24} \,{\left (15 \, r^{3} + 5 \, r^{2} x + 2 \, r x^{2} - 6 \, x^{3}\right )} \sqrt{2 \, r x - x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/4*r^4*arctan(sqrt(2*r*x - x^2)/x) - 1/24*(15*r^3 + 5*r^2*x + 2*r*x^2 - 6*x^3)*sqrt(2*r*x - x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(-x*(-2*r + x)), x)

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Giac [A]  time = 1.08082, size = 73, normalized size = 0.82 \begin{align*} -\frac{5}{8} \, r^{4} \arcsin \left (\frac{r - x}{r}\right ) \mathrm{sgn}\left (r\right ) - \frac{1}{24} \,{\left (15 \, r^{3} +{\left (5 \, r^{2} + 2 \,{\left (r - 3 \, x\right )} x\right )} x\right )} \sqrt{2 \, r x - x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*r^4*arcsin((r - x)/r)*sgn(r) - 1/24*(15*r^3 + (5*r^2 + 2*(r - 3*x)*x)*x)*sqrt(2*r*x - x^2)