3.1.71 \(\int \frac {x^2}{\sqrt {2 x-x^2}} \, dx\) [71]

Optimal. Leaf size=46 \[ -\frac {3}{2} \sqrt {2 x-x^2}-\frac {1}{2} x \sqrt {2 x-x^2}-\frac {3}{2} \sin ^{-1}(1-x) \]

[Out]

3/2*arcsin(-1+x)-3/2*(-x^2+2*x)^(1/2)-1/2*x*(-x^2+2*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {684, 654, 633, 222} \begin {gather*} -\frac {3}{2} \text {ArcSin}(1-x)-\frac {1}{2} \sqrt {2 x-x^2} x-\frac {3}{2} \sqrt {2 x-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[2*x - x^2],x]

[Out]

(-3*Sqrt[2*x - x^2])/2 - (x*Sqrt[2*x - x^2])/2 - (3*ArcSin[1 - x])/2

Rule 222

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

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

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*} \int \frac {x^2}{\sqrt {2 x-x^2}} \, dx &=-\frac {1}{2} x \sqrt {2 x-x^2}+\frac {3}{2} \int \frac {x}{\sqrt {2 x-x^2}} \, dx\\ &=-\frac {3}{2} \sqrt {2 x-x^2}-\frac {1}{2} x \sqrt {2 x-x^2}+\frac {3}{2} \int \frac {1}{\sqrt {2 x-x^2}} \, dx\\ &=-\frac {3}{2} \sqrt {2 x-x^2}-\frac {1}{2} x \sqrt {2 x-x^2}-\frac {3}{4} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{4}}} \, dx,x,2-2 x\right )\\ &=-\frac {3}{2} \sqrt {2 x-x^2}-\frac {1}{2} x \sqrt {2 x-x^2}-\frac {3}{2} \sin ^{-1}(1-x)\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 49, normalized size = 1.07 \begin {gather*} \frac {x \left (-6+x+x^2\right )+6 \sqrt {-2+x} \sqrt {x} \tanh ^{-1}\left (\frac {1}{\sqrt {\frac {-2+x}{x}}}\right )}{2 \sqrt {-((-2+x) x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2/Sqrt[2*x - x^2],x]

[Out]

(x*(-6 + x + x^2) + 6*Sqrt[-2 + x]*Sqrt[x]*ArcTanh[1/Sqrt[(-2 + x)/x]])/(2*Sqrt[-((-2 + x)*x)])

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Maple [A]
time = 0.40, size = 35, normalized size = 0.76

method	result	size
risch	\(\frac {\left (3+x \right ) x \left (x -2\right )}{2 \sqrt {-x \left (x -2\right )}}+\frac {3 \arcsin \left (x -1\right )}{2}\)	\(25\)
default	\(\frac {3 \arcsin \left (x -1\right )}{2}-\frac {3 \sqrt {-x^{2}+2 x}}{2}-\frac {x \sqrt {-x^{2}+2 x}}{2}\)	\(35\)
meijerg	\(-\frac {4 i \left (-\frac {i \sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (5 x +15\right ) \sqrt {1-\frac {x}{2}}}{40}+\frac {3 i \sqrt {\pi }\, \arcsin \left (\frac {\sqrt {2}\, \sqrt {x}}{2}\right )}{4}\right )}{\sqrt {\pi }}\)	\(47\)
trager	\(\left (-\frac {3}{2}-\frac {x}{2}\right ) \sqrt {-x^{2}+2 x}+\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+2 x}+x -1\right )}{2}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*arcsin(x-1)-3/2*(-x^2+2*x)^(1/2)-1/2*x*(-x^2+2*x)^(1/2)

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Maxima [A]
time = 0.51, size = 36, normalized size = 0.78 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 2 \, x} x - \frac {3}{2} \, \sqrt {-x^{2} + 2 \, x} - \frac {3}{2} \, \arcsin \left (-x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(-x^2 + 2*x)*x - 3/2*sqrt(-x^2 + 2*x) - 3/2*arcsin(-x + 1)

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Fricas [A]
time = 1.48, size = 35, normalized size = 0.76 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 2 \, x} {\left (x + 3\right )} - 3 \, \arctan \left (\frac {\sqrt {-x^{2} + 2 \, x}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\sqrt {- x \left (x - 2\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/sqrt(-x*(x - 2)), x)

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Giac [A]
time = 0.84, size = 23, normalized size = 0.50 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 2 \, x} {\left (x + 3\right )} + \frac {3}{2} \, \arcsin \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(-x^2 + 2*x)*(x + 3) + 3/2*arcsin(x - 1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{\sqrt {2\,x-x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(2*x - x^2)^(1/2),x)

[Out]

int(x^2/(2*x - x^2)^(1/2), x)

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