3.353 $$\int \sqrt{5-4 x-x^2} \, dx$$

Optimal. Leaf size=36 $\frac{1}{2} (x+2) \sqrt{-x^2-4 x+5}-\frac{9}{2} \sin ^{-1}\left (\frac{1}{3} (-x-2)\right )$

[Out]

((2 + x)*Sqrt[5 - 4*x - x^2])/2 - (9*ArcSin[(-2 - x)/3])/2

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Rubi [A]  time = 0.0102789, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.214, Rules used = {612, 619, 216} $\frac{1}{2} (x+2) \sqrt{-x^2-4 x+5}-\frac{9}{2} \sin ^{-1}\left (\frac{1}{3} (-x-2)\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[5 - 4*x - x^2],x]

[Out]

((2 + x)*Sqrt[5 - 4*x - x^2])/2 - (9*ArcSin[(-2 - x)/3])/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 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 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]

Rubi steps

\begin{align*} \int \sqrt{5-4 x-x^2} \, dx &=\frac{1}{2} (2+x) \sqrt{5-4 x-x^2}+\frac{9}{2} \int \frac{1}{\sqrt{5-4 x-x^2}} \, dx\\ &=\frac{1}{2} (2+x) \sqrt{5-4 x-x^2}-\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{36}}} \, dx,x,-4-2 x\right )\\ &=\frac{1}{2} (2+x) \sqrt{5-4 x-x^2}-\frac{9}{2} \sin ^{-1}\left (\frac{1}{3} (-2-x)\right )\\ \end{align*}

Mathematica [A]  time = 0.015808, size = 33, normalized size = 0.92 $\frac{1}{2} \left (\sqrt{-x^2-4 x+5} (x+2)+9 \sin ^{-1}\left (\frac{x+2}{3}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[5 - 4*x - x^2],x]

[Out]

((2 + x)*Sqrt[5 - 4*x - x^2] + 9*ArcSin[(2 + x)/3])/2

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Maple [A]  time = 0.003, size = 29, normalized size = 0.8 \begin{align*} -{\frac{-2\,x-4}{4}\sqrt{-{x}^{2}-4\,x+5}}+{\frac{9}{2}\arcsin \left ({\frac{2}{3}}+{\frac{x}{3}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2-4*x+5)^(1/2),x)

[Out]

-1/4*(-2*x-4)*(-x^2-4*x+5)^(1/2)+9/2*arcsin(2/3+1/3*x)

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Maxima [A]  time = 1.43099, size = 49, normalized size = 1.36 \begin{align*} \frac{1}{2} \, \sqrt{-x^{2} - 4 \, x + 5} x + \sqrt{-x^{2} - 4 \, x + 5} - \frac{9}{2} \, \arcsin \left (-\frac{1}{3} \, x - \frac{2}{3}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-4*x+5)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(-x^2 - 4*x + 5)*x + sqrt(-x^2 - 4*x + 5) - 9/2*arcsin(-1/3*x - 2/3)

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Fricas [A]  time = 2.05362, size = 126, normalized size = 3.5 \begin{align*} \frac{1}{2} \, \sqrt{-x^{2} - 4 \, x + 5}{\left (x + 2\right )} - \frac{9}{2} \, \arctan \left (\frac{\sqrt{-x^{2} - 4 \, x + 5}{\left (x + 2\right )}}{x^{2} + 4 \, x - 5}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-4*x+5)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(-x^2 - 4*x + 5)*(x + 2) - 9/2*arctan(sqrt(-x^2 - 4*x + 5)*(x + 2)/(x^2 + 4*x - 5))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{- x^{2} - 4 x + 5}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2-4*x+5)**(1/2),x)

[Out]

Integral(sqrt(-x**2 - 4*x + 5), x)

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Giac [A]  time = 1.06128, size = 35, normalized size = 0.97 \begin{align*} \frac{1}{2} \, \sqrt{-x^{2} - 4 \, x + 5}{\left (x + 2\right )} + \frac{9}{2} \, \arcsin \left (\frac{1}{3} \, x + \frac{2}{3}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-4*x+5)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(-x^2 - 4*x + 5)*(x + 2) + 9/2*arcsin(1/3*x + 2/3)