3.3.88 \(\int \frac {(2 x-x^2)^{3/2}}{2-2 x} \, dx\)

Optimal. Leaf size=53 \[ -\frac {1}{6} \left (2 x-x^2\right )^{3/2}-\frac {1}{2} \sqrt {2 x-x^2}+\frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x-x^2}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {685, 688, 207} \begin {gather*} -\frac {1}{6} \left (2 x-x^2\right )^{3/2}-\frac {1}{2} \sqrt {2 x-x^2}+\frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x-x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x - x^2)^(3/2)/(2 - 2*x),x]

[Out]

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

Rule 207

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

Rule 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] &&
NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 688

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[4*c, Subst[Int[1/(b^2*e
- 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {\left (2 x-x^2\right )^{3/2}}{2-2 x} \, dx &=-\frac {1}{6} \left (2 x-x^2\right )^{3/2}+\int \frac {\sqrt {2 x-x^2}}{2-2 x} \, dx\\ &=-\frac {1}{2} \sqrt {2 x-x^2}-\frac {1}{6} \left (2 x-x^2\right )^{3/2}+\int \frac {1}{(2-2 x) \sqrt {2 x-x^2}} \, dx\\ &=-\frac {1}{2} \sqrt {2 x-x^2}-\frac {1}{6} \left (2 x-x^2\right )^{3/2}-4 \operatorname {Subst}\left (\int \frac {1}{-8+8 x^2} \, dx,x,\sqrt {2 x-x^2}\right )\\ &=-\frac {1}{2} \sqrt {2 x-x^2}-\frac {1}{6} \left (2 x-x^2\right )^{3/2}+\frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x-x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 48, normalized size = 0.91 \begin {gather*} \frac {1}{6} \sqrt {-((x-2) x)} \left (x^2-2 x+\frac {6 \tan ^{-1}\left (\sqrt {\frac {x-2}{x}}\right )}{\sqrt {x-2} \sqrt {x}}-3\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(2*x - x^2)^(3/2)/(2 - 2*x),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.25, size = 44, normalized size = 0.83 \begin {gather*} \frac {1}{6} \sqrt {2 x-x^2} \left (x^2-2 x-3\right )+\tanh ^{-1}\left (\frac {\sqrt {2 x-x^2}}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2*x - x^2)^(3/2)/(2 - 2*x),x]

[Out]

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

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fricas [A]  time = 0.41, size = 65, normalized size = 1.23 \begin {gather*} \frac {1}{6} \, {\left (x^{2} - 2 \, x - 3\right )} \sqrt {-x^{2} + 2 \, x} + \frac {1}{2} \, \log \left (\frac {x + \sqrt {-x^{2} + 2 \, x}}{x}\right ) - \frac {1}{2} \, \log \left (-\frac {x - \sqrt {-x^{2} + 2 \, x}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(x^2 - 2*x - 3)*sqrt(-x^2 + 2*x) + 1/2*log((x + sqrt(-x^2 + 2*x))/x) - 1/2*log(-(x - sqrt(-x^2 + 2*x))/x)

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giac [A]  time = 0.19, size = 47, normalized size = 0.89 \begin {gather*} \frac {1}{6} \, {\left ({\left (x - 2\right )} x - 3\right )} \sqrt {-x^{2} + 2 \, x} - \frac {1}{2} \, \log \left (-\frac {2 \, {\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}}{{\left | -2 \, x + 2 \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((x - 2)*x - 3)*sqrt(-x^2 + 2*x) - 1/2*log(-2*(sqrt(-x^2 + 2*x) - 1)/abs(-2*x + 2))

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maple [A]  time = 0.04, size = 42, normalized size = 0.79 \begin {gather*} \frac {\arctanh \left (\frac {1}{\sqrt {-\left (x -1\right )^{2}+1}}\right )}{2}-\frac {\left (-\left (x -1\right )^{2}+1\right )^{\frac {3}{2}}}{6}-\frac {\sqrt {-\left (x -1\right )^{2}+1}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2+2*x)^(3/2)/(-2*x+2),x)

[Out]

-1/6*(-(x-1)^2+1)^(3/2)-1/2*(-(x-1)^2+1)^(1/2)+1/2*arctanh(1/(-(x-1)^2+1)^(1/2))

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maxima [A]  time = 3.00, size = 58, normalized size = 1.09 \begin {gather*} -\frac {1}{6} \, {\left (-x^{2} + 2 \, x\right )}^{\frac {3}{2}} - \frac {1}{2} \, \sqrt {-x^{2} + 2 \, x} + \frac {1}{2} \, \log \left (\frac {2 \, \sqrt {-x^{2} + 2 \, x}}{{\left | x - 1 \right |}} + \frac {2}{{\left | x - 1 \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(-x^2 + 2*x)^(3/2) - 1/2*sqrt(-x^2 + 2*x) + 1/2*log(2*sqrt(-x^2 + 2*x)/abs(x - 1) + 2/abs(x - 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} -\int \frac {{\left (2\,x-x^2\right )}^{3/2}}{2\,x-2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - x^2)^(3/2)/(2*x - 2),x)

[Out]

-int((2*x - x^2)^(3/2)/(2*x - 2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {2 x \sqrt {- x^{2} + 2 x}}{x - 1}\, dx + \int \left (- \frac {x^{2} \sqrt {- x^{2} + 2 x}}{x - 1}\right )\, dx}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2+2*x)**(3/2)/(2-2*x),x)

[Out]

-(Integral(2*x*sqrt(-x**2 + 2*x)/(x - 1), x) + Integral(-x**2*sqrt(-x**2 + 2*x)/(x - 1), x))/2

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