Integrand size = 18, antiderivative size = 89 \[ \int x^2 \sqrt {2 r x-x^2} \, dx=-\frac {5}{8} r^2 (r-x) \sqrt {2 r x-x^2}-\frac {5}{12} r \left (2 r x-x^2\right )^{3/2}-\frac {1}{4} x \left (2 r x-x^2\right )^{3/2}+\frac {5}{4} r^4 \arctan \left (\frac {x}{\sqrt {2 r x-x^2}}\right ) \] Output:
-5/12*r*(2*r*x-x^2)^(3/2)-1/4*x*(2*r*x-x^2)^(3/2)+5/4*r^4*arctan(x/(2*r*x- x^2)^(1/2))-5/8*r^2*(r-x)*(2*r*x-x^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int x^2 \sqrt {2 r x-x^2} \, dx=\frac {1}{24} \sqrt {-x (-2 r+x)} \left (-15 r^3-5 r^2 x-2 r x^2+6 x^3+\frac {30 r^4 \log \left (-\sqrt {x}+\sqrt {-2 r+x}\right )}{\sqrt {x} \sqrt {-2 r+x}}\right ) \] Input:
Integrate[x^2*Sqrt[2*r*x - x^2],x]
Output:
(Sqrt[-(x*(-2*r + x))]*(-15*r^3 - 5*r^2*x - 2*r*x^2 + 6*x^3 + (30*r^4*Log[ -Sqrt[x] + Sqrt[-2*r + x]])/(Sqrt[x]*Sqrt[-2*r + x])))/24
Time = 0.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1134, 1160, 1087, 1091, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \sqrt {2 r x-x^2} \, dx\) |
\(\Big \downarrow \) 1134 |
\(\displaystyle \frac {5}{4} r \int x \sqrt {2 r x-x^2}dx-\frac {1}{4} x \left (2 r x-x^2\right )^{3/2}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {5}{4} r \left (r \int \sqrt {2 r x-x^2}dx-\frac {1}{3} \left (2 r x-x^2\right )^{3/2}\right )-\frac {1}{4} x \left (2 r x-x^2\right )^{3/2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {5}{4} r \left (r \left (\frac {1}{2} r^2 \int \frac {1}{\sqrt {2 r x-x^2}}dx-\frac {1}{2} (r-x) \sqrt {2 r x-x^2}\right )-\frac {1}{3} \left (2 r x-x^2\right )^{3/2}\right )-\frac {1}{4} x \left (2 r x-x^2\right )^{3/2}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {5}{4} r \left (r \left (r^2 \int \frac {1}{\frac {x^2}{2 r x-x^2}+1}d\frac {x}{\sqrt {2 r x-x^2}}-\frac {1}{2} (r-x) \sqrt {2 r x-x^2}\right )-\frac {1}{3} \left (2 r x-x^2\right )^{3/2}\right )-\frac {1}{4} x \left (2 r x-x^2\right )^{3/2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {5}{4} r \left (r \left (r^2 \arctan \left (\frac {x}{\sqrt {2 r x-x^2}}\right )-\frac {1}{2} (r-x) \sqrt {2 r x-x^2}\right )-\frac {1}{3} \left (2 r x-x^2\right )^{3/2}\right )-\frac {1}{4} x \left (2 r x-x^2\right )^{3/2}\) |
Input:
Int[x^2*Sqrt[2*r*x - x^2],x]
Output:
-1/4*(x*(2*r*x - x^2)^(3/2)) + (5*r*(-1/3*(2*r*x - x^2)^(3/2) + r*(-1/2*(( r - x)*Sqrt[2*r*x - x^2]) + r^2*ArcTan[x/Sqrt[2*r*x - x^2]])))/4
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] - Simp[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] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[(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] && EqQ[ c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 *p]
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] + Simp[(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[p, -1]
Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.64
method | result | size |
pseudoelliptic | \(-\frac {5 \arctan \left (\frac {\sqrt {x \left (2 r -x \right )}}{x}\right ) r^{4}}{4}-\frac {5 \sqrt {x \left (2 r -x \right )}\, \left (r^{3}+\frac {1}{3} r^{2} x +\frac {2}{15} r \,x^{2}-\frac {2}{5} x^{3}\right )}{8}\) | \(57\) |
risch | \(-\frac {\left (15 r^{3}+5 r^{2} x +2 r \,x^{2}-6 x^{3}\right ) x \left (2 r -x \right )}{24 \sqrt {-x \left (-2 r +x \right )}}+\frac {5 r^{4} \arctan \left (\frac {x -r}{\sqrt {2 r x -x^{2}}}\right )}{8}\) | \(69\) |
default | \(-\frac {x \left (2 r x -x^{2}\right )^{\frac {3}{2}}}{4}+\frac {5 r \left (-\frac {\left (2 r x -x^{2}\right )^{\frac {3}{2}}}{3}+r \left (-\frac {\left (2 r -2 x \right ) \sqrt {2 r x -x^{2}}}{4}+\frac {r^{2} \arctan \left (\frac {x -r}{\sqrt {2 r x -x^{2}}}\right )}{2}\right )\right )}{4}\) | \(83\) |
Input:
int(x^2*(2*r*x-x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-5/4*arctan(1/x*(x*(2*r-x))^(1/2))*r^4-5/8*(x*(2*r-x))^(1/2)*(r^3+1/3*r^2* x+2/15*r*x^2-2/5*x^3)
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.75 \[ \int x^2 \sqrt {2 r x-x^2} \, dx=-\frac {5}{4} \, r^{4} \arctan \left (-\frac {\sqrt {2 \, r x - x^{2}}}{2 \, r - x}\right ) - \frac {1}{24} \, {\left (15 \, r^{3} + 5 \, r^{2} x + 2 \, r x^{2} - 6 \, x^{3}\right )} \sqrt {2 \, r x - x^{2}} \] Input:
integrate(x^2*(2*r*x-x^2)^(1/2),x, algorithm="fricas")
Output:
-5/4*r^4*arctan(-sqrt(2*r*x - x^2)/(2*r - x)) - 1/24*(15*r^3 + 5*r^2*x + 2 *r*x^2 - 6*x^3)*sqrt(2*r*x - x^2)
Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03 \[ \int x^2 \sqrt {2 r x-x^2} \, dx=\frac {5 r^{4} \left (\begin {cases} - i \log {\left (2 r - 2 x + 2 i \sqrt {2 r x - x^{2}} \right )} & \text {for}\: r^{2} \neq 0 \\\frac {\left (- r + x\right ) \log {\left (- r + x \right )}}{\sqrt {- \left (- r + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8} + \sqrt {2 r x - x^{2}} \left (- \frac {5 r^{3}}{8} - \frac {5 r^{2} x}{24} - \frac {r x^{2}}{12} + \frac {x^{3}}{4}\right ) \] Input:
integrate(x**2*(2*r*x-x**2)**(1/2),x)
Output:
5*r**4*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)* (-5*r**3/8 - 5*r**2*x/24 - r*x**2/12 + x**3/4)
Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.91 \[ \int x^2 \sqrt {2 r x-x^2} \, dx=-\frac {5}{8} \, r^{4} \arcsin \left (\frac {r - x}{r}\right ) - \frac {5}{8} \, \sqrt {2 \, r x - x^{2}} r^{3} + \frac {5}{8} \, \sqrt {2 \, r x - x^{2}} r^{2} x - \frac {5}{12} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} r - \frac {1}{4} \, {\left (2 \, r x - x^{2}\right )}^{\frac {3}{2}} x \] Input:
integrate(x^2*(2*r*x-x^2)^(1/2),x, algorithm="maxima")
Output:
-5/8*r^4*arcsin((r - x)/r) - 5/8*sqrt(2*r*x - x^2)*r^3 + 5/8*sqrt(2*r*x - x^2)*r^2*x - 5/12*(2*r*x - x^2)^(3/2)*r - 1/4*(2*r*x - x^2)^(3/2)*x
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.61 \[ \int x^2 \sqrt {2 r x-x^2} \, dx=-\frac {5}{8} \, r^{4} \arcsin \left (\frac {r - x}{r}\right ) \mathrm {sgn}\left (r\right ) - \frac {1}{24} \, {\left (15 \, r^{3} + {\left (5 \, r^{2} + 2 \, {\left (r - 3 \, x\right )} x\right )} x\right )} \sqrt {2 \, r x - x^{2}} \] Input:
integrate(x^2*(2*r*x-x^2)^(1/2),x, algorithm="giac")
Output:
-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)
Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84 \[ \int x^2 \sqrt {2 r x-x^2} \, dx=-\frac {x\,{\left (2\,r\,x-x^2\right )}^{3/2}}{4}-\frac {5\,r\,\left (\frac {\sqrt {2\,r\,x-x^2}\,\left (12\,r^2+4\,r\,x-8\,x^2\right )}{24}+\frac {r^3\,\ln \left (x-r-\sqrt {x\,\left (2\,r-x\right )}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}\right )}{4} \] Input:
int(x^2*(2*r*x - x^2)^(1/2),x)
Output:
- (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
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03 \[ \int x^2 \sqrt {2 r x-x^2} \, dx=-\frac {5 \sqrt {x}\, \sqrt {2 r -x}\, r^{3}}{8}-\frac {5 \sqrt {x}\, \sqrt {2 r -x}\, r^{2} x}{24}-\frac {\sqrt {x}\, \sqrt {2 r -x}\, r \,x^{2}}{12}+\frac {\sqrt {x}\, \sqrt {2 r -x}\, x^{3}}{4}-\frac {5 \,\mathrm {log}\left (\frac {\sqrt {2 r -x}+\sqrt {x}\, i}{\sqrt {r}\, \sqrt {2}}\right ) i \,r^{4}}{4} \] Input:
int(x^2*(2*r*x-x^2)^(1/2),x)
Output:
( - 15*sqrt(x)*sqrt(2*r - x)*r**3 - 5*sqrt(x)*sqrt(2*r - x)*r**2*x - 2*sqr t(x)*sqrt(2*r - x)*r*x**2 + 6*sqrt(x)*sqrt(2*r - x)*x**3 - 30*log((sqrt(2* r - x) + sqrt(x)*i)/(sqrt(r)*sqrt(2)))*i*r**4)/24