3.21.96 \(\int x \sqrt {3-2 x-x^2} \, dx\)

Optimal. Leaf size=52 \[ -\frac {1}{3} \left (-x^2-2 x+3\right )^{3/2}-\frac {1}{2} (x+1) \sqrt {-x^2-2 x+3}+2 \sin ^{-1}\left (\frac {1}{2} (-x-1)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {640, 612, 619, 216} \begin {gather*} -\frac {1}{3} \left (-x^2-2 x+3\right )^{3/2}-\frac {1}{2} (x+1) \sqrt {-x^2-2 x+3}+2 \sin ^{-1}\left (\frac {1}{2} (-x-1)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[3 - 2*x - x^2],x]

[Out]

-((1 + x)*Sqrt[3 - 2*x - x^2])/2 - (3 - 2*x - x^2)^(3/2)/3 + 2*ArcSin[(-1 - 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 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 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 640

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

Rubi steps

\begin {align*} \int x \sqrt {3-2 x-x^2} \, dx &=-\frac {1}{3} \left (3-2 x-x^2\right )^{3/2}-\int \sqrt {3-2 x-x^2} \, dx\\ &=-\frac {1}{2} (1+x) \sqrt {3-2 x-x^2}-\frac {1}{3} \left (3-2 x-x^2\right )^{3/2}-2 \int \frac {1}{\sqrt {3-2 x-x^2}} \, dx\\ &=-\frac {1}{2} (1+x) \sqrt {3-2 x-x^2}-\frac {1}{3} \left (3-2 x-x^2\right )^{3/2}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{16}}} \, dx,x,-2-2 x\right )\\ &=-\frac {1}{2} (1+x) \sqrt {3-2 x-x^2}-\frac {1}{3} \left (3-2 x-x^2\right )^{3/2}+2 \sin ^{-1}\left (\frac {1}{2} (-1-x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 37, normalized size = 0.71 \begin {gather*} \frac {1}{6} \sqrt {-x^2-2 x+3} \left (2 x^2+x-9\right )-2 \sin ^{-1}\left (\frac {x+1}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[3 - 2*x - x^2],x]

[Out]

(Sqrt[3 - 2*x - x^2]*(-9 + x + 2*x^2))/6 - 2*ArcSin[(1 + x)/2]

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IntegrateAlgebraic [A]  time = 0.10, size = 50, normalized size = 0.96 \begin {gather*} \frac {1}{6} \sqrt {-x^2-2 x+3} \left (2 x^2+x-9\right )+4 \tan ^{-1}\left (\frac {\sqrt {-x^2-2 x+3}}{x+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*Sqrt[3 - 2*x - x^2],x]

[Out]

(Sqrt[3 - 2*x - x^2]*(-9 + x + 2*x^2))/6 + 4*ArcTan[Sqrt[3 - 2*x - x^2]/(3 + x)]

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fricas [A]  time = 0.41, size = 52, normalized size = 1.00 \begin {gather*} \frac {1}{6} \, {\left (2 \, x^{2} + x - 9\right )} \sqrt {-x^{2} - 2 \, x + 3} + 2 \, \arctan \left (\frac {\sqrt {-x^{2} - 2 \, x + 3} {\left (x + 1\right )}}{x^{2} + 2 \, x - 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-x^2-2*x+3)^(1/2),x, algorithm="fricas")

[Out]

1/6*(2*x^2 + x - 9)*sqrt(-x^2 - 2*x + 3) + 2*arctan(sqrt(-x^2 - 2*x + 3)*(x + 1)/(x^2 + 2*x - 3))

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giac [A]  time = 0.16, size = 32, normalized size = 0.62 \begin {gather*} \frac {1}{6} \, {\left ({\left (2 \, x + 1\right )} x - 9\right )} \sqrt {-x^{2} - 2 \, x + 3} - 2 \, \arcsin \left (\frac {1}{2} \, x + \frac {1}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-x^2-2*x+3)^(1/2),x, algorithm="giac")

[Out]

1/6*((2*x + 1)*x - 9)*sqrt(-x^2 - 2*x + 3) - 2*arcsin(1/2*x + 1/2)

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maple [A]  time = 0.04, size = 43, normalized size = 0.83 \begin {gather*} -2 \arcsin \left (\frac {x}{2}+\frac {1}{2}\right )-\frac {\left (-x^{2}-2 x +3\right )^{\frac {3}{2}}}{3}+\frac {\left (-2 x -2\right ) \sqrt {-x^{2}-2 x +3}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-x^2-2*x+3)^(1/2),x)

[Out]

-1/3*(-x^2-2*x+3)^(3/2)+1/4*(-2-2*x)*(-x^2-2*x+3)^(1/2)-2*arcsin(1/2*x+1/2)

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maxima [A]  time = 1.98, size = 52, normalized size = 1.00 \begin {gather*} -\frac {1}{3} \, {\left (-x^{2} - 2 \, x + 3\right )}^{\frac {3}{2}} - \frac {1}{2} \, \sqrt {-x^{2} - 2 \, x + 3} x - \frac {1}{2} \, \sqrt {-x^{2} - 2 \, x + 3} + 2 \, \arcsin \left (-\frac {1}{2} \, x - \frac {1}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-x^2-2*x+3)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 47, normalized size = 0.90 \begin {gather*} \frac {\sqrt {-x^2-2\,x+3}\,\left (8\,x^2+4\,x-36\right )}{24}+\ln \left (x+1-\sqrt {-x^2-2\,x+3}\,1{}\mathrm {i}\right )\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(3 - x^2 - 2*x)^(1/2),x)

[Out]

log(x - (3 - x^2 - 2*x)^(1/2)*1i + 1)*2i + ((3 - x^2 - 2*x)^(1/2)*(4*x + 8*x^2 - 36))/24

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {- \left (x - 1\right ) \left (x + 3\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-x**2-2*x+3)**(1/2),x)

[Out]

Integral(x*sqrt(-(x - 1)*(x + 3)), x)

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