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

Optimal. Leaf size=27 \[ -2 \sqrt{-x^2+x+2}-2 \sin ^{-1}\left (\frac{1}{3} (1-2 x)\right ) \]

[Out]

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

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Rubi [A]  time = 0.0099554, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {640, 619, 216} \[ -2 \sqrt{-x^2+x+2}-2 \sin ^{-1}\left (\frac{1}{3} (1-2 x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x)/Sqrt[2 + x - x^2],x]

[Out]

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

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]

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

Mathematica [A]  time = 0.0063088, size = 27, normalized size = 1. \[ -2 \sqrt{-x^2+x+2}-2 \sin ^{-1}\left (\frac{1}{3} (1-2 x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 22, normalized size = 0.8 \begin{align*} 2\,\arcsin \left ( -1/3+2/3\,x \right ) -2\,\sqrt{-{x}^{2}+x+2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.42728, size = 28, normalized size = 1.04 \begin{align*} -2 \, \sqrt{-x^{2} + x + 2} - 2 \, \arcsin \left (-\frac{2}{3} \, x + \frac{1}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.91765, size = 111, normalized size = 4.11 \begin{align*} -2 \, \sqrt{-x^{2} + x + 2} - 2 \, \arctan \left (\frac{\sqrt{-x^{2} + x + 2}{\left (2 \, x - 1\right )}}{2 \,{\left (x^{2} - x - 2\right )}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{2 x + 1}{\sqrt{- \left (x - 2\right ) \left (x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((2*x + 1)/sqrt(-(x - 2)*(x + 1)), x)

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Giac [A]  time = 1.07244, size = 28, normalized size = 1.04 \begin{align*} -2 \, \sqrt{-x^{2} + x + 2} + 2 \, \arcsin \left (\frac{2}{3} \, x - \frac{1}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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