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

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

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Rubi [A]  time = 0.02, 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 = {670, 640, 619, 216} \begin {gather*} -\frac {1}{2} \sqrt {2 x-x^2} x-\frac {3}{2} \sqrt {2 x-x^2}-\frac {3}{2} \sin ^{-1}(1-x) \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 216

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 619

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 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 \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} \operatorname {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 = 0.04, size = 47, normalized size = 1.02 \begin {gather*} \frac {1}{2} \left (-\sqrt {2-x} x^{3/2}-3 \sqrt {-((x-2) x)}-6 \sin ^{-1}\left (\sqrt {1-\frac {x}{2}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[2*x - x^2],x]

[Out]

(-(Sqrt[2 - x]*x^(3/2)) - 3*Sqrt[-((-2 + x)*x)] - 6*ArcSin[Sqrt[1 - x/2]])/2

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IntegrateAlgebraic [A]  time = 0.13, size = 43, normalized size = 0.93 \begin {gather*} \frac {1}{2} (-x-3) \sqrt {2 x-x^2}-3 \tan ^{-1}\left (\frac {\sqrt {2 x-x^2}}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/Sqrt[2*x - x^2],x]

[Out]

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

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fricas [A]  time = 0.39, 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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giac [A]  time = 0.20, 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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maple [A]  time = 0.04, size = 35, normalized size = 0.76 \begin {gather*} -\frac {\sqrt {-x^{2}+2 x}\, x}{2}+\frac {3 \arcsin \left (x -1\right )}{2}-\frac {3 \sqrt {-x^{2}+2 x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 2.92, 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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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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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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