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

Optimal. Leaf size=39 \[ -\frac{1}{4} \sqrt{x^2-x} (1-2 x)-\frac{1}{4} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-x}}\right ) \]

[Out]

-((1 - 2*x)*Sqrt[-x + x^2])/4 - ArcTanh[x/Sqrt[-x + x^2]]/4

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Rubi [A]  time = 0.006844, antiderivative size = 39, 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} \[ -\frac{1}{4} \sqrt{x^2-x} (1-2 x)-\frac{1}{4} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-x + x^2],x]

[Out]

-((1 - 2*x)*Sqrt[-x + x^2])/4 - ArcTanh[x/Sqrt[-x + x^2]]/4

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

Mathematica [A]  time = 0.0291701, size = 46, normalized size = 1.18 \[ \frac{2 x^3-3 x^2+x+\sqrt{-(x-1) x} \sin ^{-1}\left (\sqrt{1-x}\right )}{4 \sqrt{(x-1) x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[-x + x^2],x]

[Out]

(x - 3*x^2 + 2*x^3 + Sqrt[-((-1 + x)*x)]*ArcSin[Sqrt[1 - x]])/(4*Sqrt[(-1 + x)*x])

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Maple [A]  time = 0.049, size = 33, normalized size = 0.9 \begin{align*}{\frac{2\,x-1}{4}\sqrt{{x}^{2}-x}}-{\frac{1}{8}\ln \left ( x-{\frac{1}{2}}+\sqrt{{x}^{2}-x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.22866, size = 58, normalized size = 1.49 \begin{align*} \frac{1}{2} \, \sqrt{x^{2} - x} x - \frac{1}{4} \, \sqrt{x^{2} - x} - \frac{1}{8} \, \log \left (2 \, x + 2 \, \sqrt{x^{2} - x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.17426, size = 90, normalized size = 2.31 \begin{align*} \frac{1}{4} \, \sqrt{x^{2} - x}{\left (2 \, x - 1\right )} + \frac{1}{8} \, \log \left (-2 \, x + 2 \, \sqrt{x^{2} - x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**2 - x), x)

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Giac [A]  time = 1.32737, size = 50, normalized size = 1.28 \begin{align*} \frac{1}{4} \, \sqrt{x^{2} - x}{\left (2 \, x - 1\right )} + \frac{1}{8} \, \log \left ({\left | -2 \, x + 2 \, \sqrt{x^{2} - x} + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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