3.314 \(\int \frac {1}{(2-2 x) (2 x-x^2)^{5/2}} \, dx\)

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

[Out]

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

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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 = {687, 688, 207} \[ -\frac {1}{2 \sqrt {2 x-x^2}}-\frac {1}{6 \left (2 x-x^2\right )^{3/2}}+\frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/((2 - 2*x)*(2*x - x^2)^(5/2)),x]

[Out]

-1/(6*(2*x - x^2)^(3/2)) - 1/(2*Sqrt[2*x - x^2]) + 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 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), 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] && LtQ[p, -1] &&  !GtQ[m, 1] && 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 {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx &=-\frac {1}{6 \left (2 x-x^2\right )^{3/2}}+\int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{3/2}} \, dx\\ &=-\frac {1}{6 \left (2 x-x^2\right )^{3/2}}-\frac {1}{2 \sqrt {2 x-x^2}}+\int \frac {1}{(2-2 x) \sqrt {2 x-x^2}} \, dx\\ &=-\frac {1}{6 \left (2 x-x^2\right )^{3/2}}-\frac {1}{2 \sqrt {2 x-x^2}}-4 \operatorname {Subst}\left (\int \frac {1}{-8+8 x^2} \, dx,x,\sqrt {2 x-x^2}\right )\\ &=-\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 {align*}

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Mathematica [A]  time = 0.04, size = 50, normalized size = 0.94 \[ \frac {6 (x-2)^{3/2} x^{3/2} \tan ^{-1}\left (\sqrt {\frac {x-2}{x}}\right )+3 x^2-6 x-1}{6 (-((x-2) x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((2 - 2*x)*(2*x - x^2)^(5/2)),x]

[Out]

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

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fricas [B]  time = 1.02, size = 112, normalized size = 2.11 \[ \frac {3 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (\frac {x + \sqrt {-x^{2} + 2 \, x}}{x}\right ) - 3 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (-\frac {x - \sqrt {-x^{2} + 2 \, x}}{x}\right ) + {\left (3 \, x^{2} - 6 \, x - 1\right )} \sqrt {-x^{2} + 2 \, x}}{6 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2-2*x)/(-x^2+2*x)^(5/2),x, algorithm="fricas")

[Out]

1/6*(3*(x^4 - 4*x^3 + 4*x^2)*log((x + sqrt(-x^2 + 2*x))/x) - 3*(x^4 - 4*x^3 + 4*x^2)*log(-(x - sqrt(-x^2 + 2*x
))/x) + (3*x^2 - 6*x - 1)*sqrt(-x^2 + 2*x))/(x^4 - 4*x^3 + 4*x^2)

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giac [A]  time = 0.25, size = 57, normalized size = 1.08 \[ \frac {{\left (3 \, {\left (x - 2\right )} x - 1\right )} \sqrt {-x^{2} + 2 \, x}}{6 \, {\left (x^{2} - 2 \, x\right )}^{2}} - \frac {1}{2} \, \log \left (-\frac {2 \, {\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}}{{\left | -2 \, x + 2 \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2-2*x)/(-x^2+2*x)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 42, normalized size = 0.79 \[ \frac {\arctanh \left (\frac {1}{\sqrt {-\left (x -1\right )^{2}+1}}\right )}{2}-\frac {1}{6 \left (-\left (x -1\right )^{2}+1\right )^{\frac {3}{2}}}-\frac {1}{2 \sqrt {-\left (x -1\right )^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-2*x+2)/(-x^2+2*x)^(5/2),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 = 1.32, size = 58, normalized size = 1.09 \[ -\frac {1}{2 \, \sqrt {-x^{2} + 2 \, x}} - \frac {1}{6 \, {\left (-x^{2} + 2 \, x\right )}^{\frac {3}{2}}} + \frac {1}{2} \, \log \left (\frac {2 \, \sqrt {-x^{2} + 2 \, x}}{{\left | x - 1 \right |}} + \frac {2}{{\left | x - 1 \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2-2*x)/(-x^2+2*x)^(5/2),x, algorithm="maxima")

[Out]

-1/2/sqrt(-x^2 + 2*x) - 1/6/(-x^2 + 2*x)^(3/2) + 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 \[ -\int \frac {1}{\left (2\,x-2\right )\,{\left (2\,x-x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/((2*x - 2)*(2*x - x^2)^(5/2)),x)

[Out]

-int(1/((2*x - 2)*(2*x - x^2)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {1}{x^{5} \sqrt {- x^{2} + 2 x} - 5 x^{4} \sqrt {- x^{2} + 2 x} + 8 x^{3} \sqrt {- x^{2} + 2 x} - 4 x^{2} \sqrt {- x^{2} + 2 x}}\, dx}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2-2*x)/(-x**2+2*x)**(5/2),x)

[Out]

-Integral(1/(x**5*sqrt(-x**2 + 2*x) - 5*x**4*sqrt(-x**2 + 2*x) + 8*x**3*sqrt(-x**2 + 2*x) - 4*x**2*sqrt(-x**2
+ 2*x)), x)/2

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