\(\int x^3 \sqrt {2 r x-x^2} \, dx\) [263]

Optimal result

Integrand size = 18, antiderivative size = 113 \[ \int x^3 \sqrt {2 r x-x^2} \, dx=-\frac {7}{8} r^3 (r-x) \sqrt {2 r x-x^2}-\frac {7}{12} r^2 \left (2 r x-x^2\right )^{3/2}-\frac {7}{20} r x \left (2 r x-x^2\right )^{3/2}-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2}+\frac {7}{4} r^5 \arctan \left (\frac {x}{\sqrt {2 r x-x^2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {684, 654, 626, 634, 209} \[ \int x^3 \sqrt {2 r x-x^2} \, dx=\frac {7}{4} r^5 \arctan \left (\frac {x}{\sqrt {2 r x-x^2}}\right )-\frac {7}{8} r^3 (r-x) \sqrt {2 r x-x^2}-\frac {7}{12} r^2 \left (2 r x-x^2\right )^{3/2}-\frac {7}{20} r x \left (2 r x-x^2\right )^{3/2}-\frac {1}{5} x^2 \left (2 r x-x^2\right )^{3/2} \]

[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 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]

Rule 684

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*} \text {integral}& = -\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 ) \text {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 \arctan \left (\frac {x}{\sqrt {2 r x-x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74 \[ \int x^3 \sqrt {2 r x-x^2} \, dx=\frac {1}{120} \sqrt {-x (-2 r+x)} \left (-105 r^4-35 r^3 x-14 r^2 x^2-6 r x^3+24 x^4+\frac {210 r^5 \log \left (-\sqrt {x}+\sqrt {-2 r+x}\right )}{\sqrt {x} \sqrt {-2 r+x}}\right ) \]

[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^5*Log[-Sqrt[x] + Sqrt[-2*
r + x]])/(Sqrt[x]*Sqrt[-2*r + x])))/120

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.58

method	result	size
pseudoelliptic	\(-\frac {7 \arctan \left (\frac {\sqrt {x \left (2 r -x \right )}}{x}\right ) r^{5}}{4}-\frac {7 \sqrt {x \left (2 r -x \right )}\, \left (r^{4}+\frac {1}{3} r^{3} x +\frac {2}{15} r^{2} x^{2}+\frac {2}{35} r \,x^{3}-\frac {8}{35} x^{4}\right )}{8}\)	\(65\)
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	\(-\frac {x^{2} \left (2 r x -x^{2}\right )^{\frac {3}{2}}}{5}+\frac {7 r \left (-\frac {x \left (2 r x -x^{2}\right )^{\frac {3}{2}}}{4}+\frac {5 r \left (-\frac {\left (2 r x -x^{2}\right )^{\frac {3}{2}}}{3}+r \left (-\frac {\left (2 r -2 x \right ) \sqrt {2 r x -x^{2}}}{4}+\frac {r^{2} \arctan \left (\frac {x -r}{\sqrt {2 r x -x^{2}}}\right )}{2}\right )\right )}{4}\right )}{5}\)	\(104\)

[In]

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

[Out]

-7/4*arctan(1/x*(x*(2*r-x))^(1/2))*r^5-7/8*(x*(2*r-x))^(1/2)*(r^4+1/3*r^3*x+2/15*r^2*x^2+2/35*r*x^3-8/35*x^4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.60 \[ \int x^3 \sqrt {2 r x-x^2} \, dx=-\frac {7}{4} \, r^{5} \arctan \left (\frac {\sqrt {2 \, r x - x^{2}}}{x}\right ) - \frac {1}{120} \, {\left (105 \, r^{4} + 35 \, r^{3} x + 14 \, r^{2} x^{2} + 6 \, r x^{3} - 24 \, x^{4}\right )} \sqrt {2 \, r x - x^{2}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90 \[ \int x^3 \sqrt {2 r x-x^2} \, dx=\frac {7 r^{5} \left (\begin {cases} - i \log {\left (2 r - 2 x + 2 i \sqrt {2 r x - x^{2}} \right )} & \text {for}\: r^{2} \neq 0 \\\frac {\left (- r + x\right ) \log {\left (- r + x \right )}}{\sqrt {- \left (- r + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8} + \sqrt {2 r x - x^{2}} \left (- \frac {7 r^{4}}{8} - \frac {7 r^{3} x}{24} - \frac {7 r^{2} x^{2}}{60} - \frac {r x^{3}}{20} + \frac {x^{4}}{5}\right ) \]

[In]

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

[Out]

7*r**5*Piecewise((-I*log(2*r - 2*x + 2*I*sqrt(2*r*x - x**2)), Ne(r**2, 0)), ((-r + x)*log(-r + x)/sqrt(-(-r +
x)**2), True))/8 + sqrt(2*r*x - x**2)*(-7*r**4/8 - 7*r**3*x/24 - 7*r**2*x**2/60 - r*x**3/20 + x**4/5)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int x^3 \sqrt {2 r x-x^2} \, dx=-\frac {7}{8} \, r^{5} \arcsin \left (\frac {r - x}{r}\right ) - \frac {7}{8} \, \sqrt {2 \, r x - x^{2}} r^{4} + \frac {7}{8} \, \sqrt {2 \, r x - x^{2}} r^{3} x - \frac {7}{12} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} r^{2} - \frac {7}{20} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} r x - \frac {1}{5} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} x^{2} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.56 \[ \int x^3 \sqrt {2 r x-x^2} \, dx=-\frac {7}{8} \, r^{5} \arcsin \left (\frac {r - x}{r}\right ) \mathrm {sgn}\left (r\right ) - \frac {1}{120} \, {\left (105 \, r^{4} + {\left (35 \, r^{3} + 2 \, {\left (7 \, r^{2} + 3 \, {\left (r - 4 \, x\right )} x\right )} x\right )} x\right )} \sqrt {2 \, r x - x^{2}} \]

[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)

Mupad [B] (verification not implemented)

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

[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