3.2428 \(\int x \sqrt {8+2 x-x^2} \, dx\)

Optimal. Leaf size=56 \[ -\frac {1}{3} \left (-x^2+2 x+8\right )^{3/2}-\frac {1}{2} (1-x) \sqrt {-x^2+2 x+8}-\frac {9}{2} \sin ^{-1}\left (\frac {1-x}{3}\right ) \]

[Out]

-1/3*(-x^2+2*x+8)^(3/2)+9/2*arcsin(-1/3+1/3*x)-1/2*(1-x)*(-x^2+2*x+8)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 56, 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} \[ -\frac {1}{3} \left (-x^2+2 x+8\right )^{3/2}-\frac {1}{2} (1-x) \sqrt {-x^2+2 x+8}-\frac {9}{2} \sin ^{-1}\left (\frac {1-x}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[8 + 2*x - x^2],x]

[Out]

-((1 - x)*Sqrt[8 + 2*x - x^2])/2 - (8 + 2*x - x^2)^(3/2)/3 - (9*ArcSin[(1 - x)/3])/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 {8+2 x-x^2} \, dx &=-\frac {1}{3} \left (8+2 x-x^2\right )^{3/2}+\int \sqrt {8+2 x-x^2} \, dx\\ &=-\frac {1}{2} (1-x) \sqrt {8+2 x-x^2}-\frac {1}{3} \left (8+2 x-x^2\right )^{3/2}+\frac {9}{2} \int \frac {1}{\sqrt {8+2 x-x^2}} \, dx\\ &=-\frac {1}{2} (1-x) \sqrt {8+2 x-x^2}-\frac {1}{3} \left (8+2 x-x^2\right )^{3/2}-\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{36}}} \, dx,x,2-2 x\right )\\ &=-\frac {1}{2} (1-x) \sqrt {8+2 x-x^2}-\frac {1}{3} \left (8+2 x-x^2\right )^{3/2}-\frac {9}{2} \sin ^{-1}\left (\frac {1-x}{3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.75 \[ \frac {1}{6} \left (\sqrt {-x^2+2 x+8} \left (2 x^2-x-19\right )-27 \sin ^{-1}\left (\frac {1}{3}-\frac {x}{3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[8 + 2*x - x^2],x]

[Out]

(Sqrt[8 + 2*x - x^2]*(-19 - x + 2*x^2) - 27*ArcSin[1/3 - x/3])/6

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fricas [A]  time = 0.76, size = 54, normalized size = 0.96 \[ \frac {1}{6} \, {\left (2 \, x^{2} - x - 19\right )} \sqrt {-x^{2} + 2 \, x + 8} - \frac {9}{2} \, \arctan \left (\frac {\sqrt {-x^{2} + 2 \, x + 8} {\left (x - 1\right )}}{x^{2} - 2 \, x - 8}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*x^2 - x - 19)*sqrt(-x^2 + 2*x + 8) - 9/2*arctan(sqrt(-x^2 + 2*x + 8)*(x - 1)/(x^2 - 2*x - 8))

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giac [A]  time = 0.16, size = 32, normalized size = 0.57 \[ \frac {1}{6} \, {\left ({\left (2 \, x - 1\right )} x - 19\right )} \sqrt {-x^{2} + 2 \, x + 8} + \frac {9}{2} \, \arcsin \left (\frac {1}{3} \, x - \frac {1}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((2*x - 1)*x - 19)*sqrt(-x^2 + 2*x + 8) + 9/2*arcsin(1/3*x - 1/3)

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maple [A]  time = 0.06, size = 43, normalized size = 0.77 \[ \frac {9 \arcsin \left (\frac {x}{3}-\frac {1}{3}\right )}{2}-\frac {\left (-x^{2}+2 x +8\right )^{\frac {3}{2}}}{3}-\frac {\left (-2 x +2\right ) \sqrt {-x^{2}+2 x +8}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-x^2+2*x+8)^(3/2)-1/4*(-2*x+2)*(-x^2+2*x+8)^(1/2)+9/2*arcsin(-1/3+1/3*x)

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maxima [A]  time = 1.98, size = 52, normalized size = 0.93 \[ -\frac {1}{3} \, {\left (-x^{2} + 2 \, x + 8\right )}^{\frac {3}{2}} + \frac {1}{2} \, \sqrt {-x^{2} + 2 \, x + 8} x - \frac {1}{2} \, \sqrt {-x^{2} + 2 \, x + 8} - \frac {9}{2} \, \arcsin \left (-\frac {1}{3} \, x + \frac {1}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-x^2 + 2*x + 8)^(3/2) + 1/2*sqrt(-x^2 + 2*x + 8)*x - 1/2*sqrt(-x^2 + 2*x + 8) - 9/2*arcsin(-1/3*x + 1/3)

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mupad [B]  time = 0.08, size = 47, normalized size = 0.84 \[ -\frac {\sqrt {-x^2+2\,x+8}\,\left (-8\,x^2+4\,x+76\right )}{24}-\frac {\ln \left (x-1-\sqrt {-x^2+2\,x+8}\,1{}\mathrm {i}\right )\,9{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(2*x - x^2 + 8)^(1/2),x)

[Out]

- (log(x - (2*x - x^2 + 8)^(1/2)*1i - 1)*9i)/2 - ((2*x - x^2 + 8)^(1/2)*(4*x - 8*x^2 + 76))/24

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {- \left (x - 4\right ) \left (x + 2\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(-(x - 4)*(x + 2)), x)

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