3.263 \(\int x^3 \sqrt {2 r x-x^2} \, dx\)

Optimal. Leaf size=113 \[ \frac {7}{4} r^5 \tan ^{-1}\left (\frac {x}{\sqrt {2 r x-x^2}}\right )-\frac {7}{8} r^3 (r-x) \sqrt {2 r x-x^2}-\frac {7}{12} r^2 \left (2 r x-x^2\right )^{3/2}-\frac {7}{20} r x \left (2 r x-x^2\right )^{3/2}-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2} \]

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Rubi [A]  time = 0.04, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {7}{8} r^3 (r-x) \sqrt {2 r x-x^2}-\frac {7}{12} r^2 \left (2 r x-x^2\right )^{3/2}+\frac {7}{4} r^5 \tan ^{-1}\left (\frac {x}{\sqrt {2 r x-x^2}}\right )-\frac {7}{20} r x \left (2 r x-x^2\right )^{3/2}-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[2*r*x - x^2],x]

[Out]

(-7*r^3*(r - x)*Sqrt[2*r*x - x^2])/8 - (7*r^2*(2*r*x - x^2)^(3/2))/12 - (7*r*x*(2*r*x - x^2)^(3/2))/20 - (x^2*
(2*r*x - x^2)^(3/2))/5 + (7*r^5*ArcTan[x/Sqrt[2*r*x - x^2]])/4

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]

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]

Rubi steps

\begin {align*} \int x^3 \sqrt {2 r x-x^2} \, dx &=-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2}+\frac {1}{5} (7 r) \int x^2 \sqrt {2 r x-x^2} \, dx\\ &=-\frac {7}{20} r x \left (2 r x-x^2\right )^{3/2}-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2}+\frac {1}{4} \left (7 r^2\right ) \int x \sqrt {2 r x-x^2} \, dx\\ &=-\frac {7}{12} r^2 \left (2 r x-x^2\right )^{3/2}-\frac {7}{20} r x \left (2 r x-x^2\right )^{3/2}-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2}+\frac {1}{4} \left (7 r^3\right ) \int \sqrt {2 r x-x^2} \, dx\\ &=-\frac {7}{8} r^3 (r-x) \sqrt {2 r x-x^2}-\frac {7}{12} r^2 \left (2 r x-x^2\right )^{3/2}-\frac {7}{20} r x \left (2 r x-x^2\right )^{3/2}-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2}+\frac {1}{8} \left (7 r^5\right ) \int \frac {1}{\sqrt {2 r x-x^2}} \, dx\\ &=-\frac {7}{8} r^3 (r-x) \sqrt {2 r x-x^2}-\frac {7}{12} r^2 \left (2 r x-x^2\right )^{3/2}-\frac {7}{20} r x \left (2 r x-x^2\right )^{3/2}-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2}+\frac {1}{4} \left (7 r^5\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {2 r x-x^2}}\right )\\ &=-\frac {7}{8} r^3 (r-x) \sqrt {2 r x-x^2}-\frac {7}{12} r^2 \left (2 r x-x^2\right )^{3/2}-\frac {7}{20} r x \left (2 r x-x^2\right )^{3/2}-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2}+\frac {7}{4} r^5 \tan ^{-1}\left (\frac {x}{\sqrt {2 r x-x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 88, normalized size = 0.78 \[ \frac {1}{120} \sqrt {-x (x-2 r)} \left (\frac {210 r^{9/2} \sin ^{-1}\left (\frac {\sqrt {x}}{\sqrt {2} \sqrt {r}}\right )}{\sqrt {x} \sqrt {2-\frac {x}{r}}}-105 r^4-35 r^3 x-14 r^2 x^2-6 r x^3+24 x^4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[2*r*x - x^2],x]

[Out]

(Sqrt[-(x*(-2*r + x))]*(-105*r^4 - 35*r^3*x - 14*r^2*x^2 - 6*r*x^3 + 24*x^4 + (210*r^(9/2)*ArcSin[Sqrt[x]/(Sqr
t[2]*Sqrt[r])])/(Sqrt[x]*Sqrt[2 - x/r])))/120

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IntegrateAlgebraic [C]  time = 0.23, size = 129, normalized size = 1.14 \[ -\frac {7}{8} i r^5 \tanh ^{-1}\left (\frac {x}{r}+\frac {i \sqrt {2 r x-x^2}}{r}\right )+\frac {7}{16} i r^5 \log \left (r^2-2 i x \sqrt {2 r x-x^2}+2 r x-2 x^2\right )+\frac {1}{120} \sqrt {2 r x-x^2} \left (-105 r^4-35 r^3 x-14 r^2 x^2-6 r x^3+24 x^4\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^3*Sqrt[2*r*x - x^2],x]

[Out]

(Sqrt[2*r*x - x^2]*(-105*r^4 - 35*r^3*x - 14*r^2*x^2 - 6*r*x^3 + 24*x^4))/120 - ((7*I)/8)*r^5*ArcTanh[x/r + (I
*Sqrt[2*r*x - x^2])/r] + ((7*I)/16)*r^5*Log[r^2 + 2*r*x - 2*x^2 - (2*I)*x*Sqrt[2*r*x - x^2]]

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fricas [A]  time = 0.60, size = 68, normalized size = 0.60 \[ -\frac {7}{4} \, r^{5} \arctan \left (\frac {\sqrt {2 \, r x - x^{2}}}{x}\right ) - \frac {1}{120} \, {\left (105 \, r^{4} + 35 \, r^{3} x + 14 \, r^{2} x^{2} + 6 \, r x^{3} - 24 \, x^{4}\right )} \sqrt {2 \, r x - x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(2*r*x-x^2)^(1/2),x, algorithm="fricas")

[Out]

-7/4*r^5*arctan(sqrt(2*r*x - x^2)/x) - 1/120*(105*r^4 + 35*r^3*x + 14*r^2*x^2 + 6*r*x^3 - 24*x^4)*sqrt(2*r*x -
x^2)

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giac [A]  time = 0.63, size = 63, normalized size = 0.56 \[ -\frac {7}{8} \, r^{5} \arcsin \left (\frac {r - x}{r}\right ) \mathrm {sgn}\relax (r) - \frac {1}{120} \, {\left (105 \, r^{4} + {\left (35 \, r^{3} + 2 \, {\left (7 \, r^{2} + 3 \, {\left (r - 4 \, x\right )} x\right )} x\right )} x\right )} \sqrt {2 \, r x - x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(2*r*x-x^2)^(1/2),x, algorithm="giac")

[Out]

-7/8*r^5*arcsin((r - x)/r)*sgn(r) - 1/120*(105*r^4 + (35*r^3 + 2*(7*r^2 + 3*(r - 4*x)*x)*x)*x)*sqrt(2*r*x - x^
2)

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maple [A]  time = 0.29, size = 77, normalized size = 0.68

method	result	size
risch	\(-\frac {\left (105 r^{4}+35 r^{3} x +14 r^{2} x^{2}+6 r \,x^{3}-24 x^{4}\right ) x \left (2 r -x \right )}{120 \sqrt {-x \left (-2 r +x \right )}}+\frac {7 r^{5} \arctan \left (\frac {x -r}{\sqrt {2 r x -x^{2}}}\right )}{8}\)	\(77\)
default	\(-\frac {x^{2} \left (2 r x -x^{2}\right )^{\frac {3}{2}}}{5}+\frac {7 r \left (-\frac {x \left (2 r x -x^{2}\right )^{\frac {3}{2}}}{4}+\frac {5 r \left (-\frac {\left (2 r x -x^{2}\right )^{\frac {3}{2}}}{3}+r \left (-\frac {\left (2 r -2 x \right ) \sqrt {2 r x -x^{2}}}{4}+\frac {r^{2} \arctan \left (\frac {x -r}{\sqrt {2 r x -x^{2}}}\right )}{2}\right )\right )}{4}\right )}{5}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(2*r*x-x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/120*(105*r^4+35*r^3*x+14*r^2*x^2+6*r*x^3-24*x^4)*x*(2*r-x)/(-x*(-2*r+x))^(1/2)+7/8*r^5*arctan((x-r)/(2*r*x-
x^2)^(1/2))

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maxima [A]  time = 1.10, size = 101, normalized size = 0.89 \[ -\frac {7}{8} \, r^{5} \arcsin \left (\frac {r - x}{r}\right ) - \frac {7}{8} \, \sqrt {2 \, r x - x^{2}} r^{4} + \frac {7}{8} \, \sqrt {2 \, r x - x^{2}} r^{3} x - \frac {7}{12} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} r^{2} - \frac {7}{20} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} r x - \frac {1}{5} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(2*r*x-x^2)^(1/2),x, algorithm="maxima")

[Out]

-7/8*r^5*arcsin((r - x)/r) - 7/8*sqrt(2*r*x - x^2)*r^4 + 7/8*sqrt(2*r*x - x^2)*r^3*x - 7/12*(2*r*x - x^2)^(3/2
)*r^2 - 7/20*(2*r*x - x^2)^(3/2)*r*x - 1/5*(2*r*x - x^2)^(3/2)*x^2

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mupad [B]  time = 0.14, size = 96, normalized size = 0.85 \[ -\frac {7\,r\,\left (\frac {x\,{\left (2\,r\,x-x^2\right )}^{3/2}}{4}+\frac {5\,r\,\left (\frac {\sqrt {2\,r\,x-x^2}\,\left (12\,r^2+4\,r\,x-8\,x^2\right )}{24}+\frac {r^3\,\ln \left (x-r-\sqrt {x\,\left (2\,r-x\right )}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}\right )}{4}\right )}{5}-\frac {x^2\,{\left (2\,r\,x-x^2\right )}^{3/2}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(2*r*x - x^2)^(1/2),x)

[Out]

- (7*r*((x*(2*r*x - x^2)^(3/2))/4 + (5*r*(((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))/4))/5 - (x^2*(2*r*x - x^2)^(3/2))/5

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \sqrt {- x \left (- 2 r + x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(2*r*x-x**2)**(1/2),x)

[Out]

Integral(x**3*sqrt(-x*(-2*r + x)), x)

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