3.14 \(\int \sqrt{-8 x+x^2} \, dx\)

Optimal. Leaf size=37 \[ -\frac{1}{2} \sqrt{x^2-8 x} (4-x)-16 \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-8 x}}\right ) \]

[Out]

-((4 - x)*Sqrt[-8*x + x^2])/2 - 16*ArcTanh[x/Sqrt[-8*x + x^2]]

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Rubi [A]  time = 0.0066753, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {612, 620, 206} \[ -\frac{1}{2} \sqrt{x^2-8 x} (4-x)-16 \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-8 x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-8*x + x^2],x]

[Out]

-((4 - x)*Sqrt[-8*x + x^2])/2 - 16*ArcTanh[x/Sqrt[-8*x + x^2]]

Rule 612

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] - Dist[(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]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

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

Rubi steps

\begin{align*} \int \sqrt{-8 x+x^2} \, dx &=-\frac{1}{2} (4-x) \sqrt{-8 x+x^2}-8 \int \frac{1}{\sqrt{-8 x+x^2}} \, dx\\ &=-\frac{1}{2} (4-x) \sqrt{-8 x+x^2}-16 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x}{\sqrt{-8 x+x^2}}\right )\\ &=-\frac{1}{2} (4-x) \sqrt{-8 x+x^2}-16 \tanh ^{-1}\left (\frac{x}{\sqrt{-8 x+x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0376307, size = 48, normalized size = 1.3 \[ \frac{x \left (x^2-12 x+32\right )+32 \sqrt{-(x-8) x} \sin ^{-1}\left (\sqrt{1-\frac{x}{8}}\right )}{2 \sqrt{(x-8) x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[-8*x + x^2],x]

[Out]

(x*(32 - 12*x + x^2) + 32*Sqrt[-((-8 + x)*x)]*ArcSin[Sqrt[1 - x/8]])/(2*Sqrt[(-8 + x)*x])

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Maple [A]  time = 0.051, size = 33, normalized size = 0.9 \begin{align*}{\frac{2\,x-8}{4}\sqrt{{x}^{2}-8\,x}}-8\,\ln \left ( x-4+\sqrt{{x}^{2}-8\,x} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-8*x)^(1/2),x)

[Out]

1/4*(2*x-8)*(x^2-8*x)^(1/2)-8*ln(x-4+(x^2-8*x)^(1/2))

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Maxima [A]  time = 2.32576, size = 58, normalized size = 1.57 \begin{align*} \frac{1}{2} \, \sqrt{x^{2} - 8 \, x} x - 2 \, \sqrt{x^{2} - 8 \, x} - 8 \, \log \left (2 \, x + 2 \, \sqrt{x^{2} - 8 \, x} - 8\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-8*x)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(x^2 - 8*x)*x - 2*sqrt(x^2 - 8*x) - 8*log(2*x + 2*sqrt(x^2 - 8*x) - 8)

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Fricas [A]  time = 2.3514, size = 85, normalized size = 2.3 \begin{align*} \frac{1}{2} \, \sqrt{x^{2} - 8 \, x}{\left (x - 4\right )} + 8 \, \log \left (-x + \sqrt{x^{2} - 8 \, x} + 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-8*x)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(x^2 - 8*x)*(x - 4) + 8*log(-x + sqrt(x^2 - 8*x) + 4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{2} - 8 x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-8*x)**(1/2),x)

[Out]

Integral(sqrt(x**2 - 8*x), x)

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Giac [A]  time = 1.28506, size = 45, normalized size = 1.22 \begin{align*} \frac{1}{2} \, \sqrt{x^{2} - 8 \, x}{\left (x - 4\right )} + 8 \, \log \left ({\left | -x + \sqrt{x^{2} - 8 \, x} + 4 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-8*x)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(x^2 - 8*x)*(x - 4) + 8*log(abs(-x + sqrt(x^2 - 8*x) + 4))