3.44 \(\int \frac {x-x^2}{\sqrt {1-x^2}} \, dx\)

Optimal. Leaf size=27 \[ -\frac {1}{2} \sqrt {1-x^2} (2-x)-\frac {1}{2} \sin ^{-1}(x) \]

[Out]

-1/2*arcsin(x)-1/2*(2-x)*(-x^2+1)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1593, 780, 216} \[ -\frac {1}{2} \sqrt {1-x^2} (2-x)-\frac {1}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x - x^2)/Sqrt[1 - x^2],x]

[Out]

-((2 - x)*Sqrt[1 - x^2])/2 - ArcSin[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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {x-x^2}{\sqrt {1-x^2}} \, dx &=\int \frac {(1-x) x}{\sqrt {1-x^2}} \, dx\\ &=-\frac {1}{2} (2-x) \sqrt {1-x^2}-\frac {1}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx\\ &=-\frac {1}{2} (2-x) \sqrt {1-x^2}-\frac {1}{2} \sin ^{-1}(x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 0.89 \[ \frac {1}{2} \left ((x-2) \sqrt {1-x^2}-\sin ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x - x^2)/Sqrt[1 - x^2],x]

[Out]

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

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fricas [A]  time = 0.62, size = 31, normalized size = 1.15 \[ \frac {1}{2} \, \sqrt {-x^{2} + 1} {\left (x - 2\right )} + \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-x^2 + 1)*(x - 2) + arctan((sqrt(-x^2 + 1) - 1)/x)

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giac [A]  time = 0.42, size = 19, normalized size = 0.70 \[ \frac {1}{2} \, \sqrt {-x^{2} + 1} {\left (x - 2\right )} - \frac {1}{2} \, \arcsin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-x^2 + 1)*(x - 2) - 1/2*arcsin(x)

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maple [A]  time = 0.00, size = 29, normalized size = 1.07 \[ \frac {\sqrt {-x^{2}+1}\, x}{2}-\frac {\arcsin \relax (x )}{2}-\sqrt {-x^{2}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-x^2+1)^(1/2)*x-1/2*arcsin(x)-(-x^2+1)^(1/2)

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maxima [A]  time = 2.89, size = 28, normalized size = 1.04 \[ \frac {1}{2} \, \sqrt {-x^{2} + 1} x - \sqrt {-x^{2} + 1} - \frac {1}{2} \, \arcsin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-x^2 + 1)*x - sqrt(-x^2 + 1) - 1/2*arcsin(x)

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mupad [B]  time = 0.03, size = 20, normalized size = 0.74 \[ \left (\frac {x}{2}-1\right )\,\sqrt {1-x^2}-\frac {\mathrm {asin}\relax (x)}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - x^2)/(1 - x^2)^(1/2),x)

[Out]

(x/2 - 1)*(1 - x^2)^(1/2) - asin(x)/2

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sympy [A]  time = 0.29, size = 24, normalized size = 0.89 \[ \frac {x \sqrt {1 - x^{2}}}{2} - \sqrt {1 - x^{2}} - \frac {\operatorname {asin}{\relax (x )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*sqrt(1 - x**2)/2 - sqrt(1 - x**2) - asin(x)/2

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