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

Optimal. Leaf size=19 \[ -\frac{2 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{7}}\right )}{\sqrt{7}} \]

[Out]

(-2*ArcTan[(1 - 2*x)/Sqrt[7]])/Sqrt[7]

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Rubi [A]  time = 0.0111904, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {618, 204} \[ -\frac{2 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{7}}\right )}{\sqrt{7}} \]

Antiderivative was successfully verified.

[In]

Int[(2 - x + x^2)^(-1),x]

[Out]

(-2*ArcTan[(1 - 2*x)/Sqrt[7]])/Sqrt[7]

Rule 618

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{2-x+x^2} \, dx &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,-1+2 x\right )\right )\\ &=-\frac{2 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{7}}\right )}{\sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0062284, size = 19, normalized size = 1. \[ \frac{2 \tan ^{-1}\left (\frac{2 x-1}{\sqrt{7}}\right )}{\sqrt{7}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 - x + x^2)^(-1),x]

[Out]

(2*ArcTan[(-1 + 2*x)/Sqrt[7]])/Sqrt[7]

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Maple [A]  time = 0.003, size = 17, normalized size = 0.9 \begin{align*}{\frac{2\,\sqrt{7}}{7}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{7}}{7}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*7^(1/2)*arctan(1/7*(2*x-1)*7^(1/2))

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Maxima [A]  time = 1.60247, size = 22, normalized size = 1.16 \begin{align*} \frac{2}{7} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (2 \, x - 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*sqrt(7)*arctan(1/7*sqrt(7)*(2*x - 1))

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Fricas [A]  time = 1.23032, size = 58, normalized size = 3.05 \begin{align*} \frac{2}{7} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (2 \, x - 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*sqrt(7)*arctan(1/7*sqrt(7)*(2*x - 1))

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Sympy [A]  time = 0.096842, size = 26, normalized size = 1.37 \begin{align*} \frac{2 \sqrt{7} \operatorname{atan}{\left (\frac{2 \sqrt{7} x}{7} - \frac{\sqrt{7}}{7} \right )}}{7} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(7)*atan(2*sqrt(7)*x/7 - sqrt(7)/7)/7

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Giac [A]  time = 1.12003, size = 22, normalized size = 1.16 \begin{align*} \frac{2}{7} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (2 \, x - 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*sqrt(7)*arctan(1/7*sqrt(7)*(2*x - 1))