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]

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

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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])

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]

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*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \[ \frac {2 \tan ^{-1}\left (\frac {2 x-1}{\sqrt {7}}\right )}{\sqrt {7}} \]

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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fricas [A]  time = 0.60, size = 16, normalized size = 0.84 \[ \frac {2}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x - 1\right )}\right ) \]

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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giac [A]  time = 0.38, size = 16, normalized size = 0.84 \[ \frac {2}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x - 1\right )}\right ) \]

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))

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maple [A]  time = 0.00, size = 17, normalized size = 0.89 \[ \frac {2 \sqrt {7}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {7}}{7}\right )}{7} \]

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 = 0.95, size = 16, normalized size = 0.84 \[ \frac {2}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x - 1\right )}\right ) \]

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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mupad [B]  time = 0.03, size = 16, normalized size = 0.84 \[ \frac {2\,\sqrt {7}\,\mathrm {atan}\left (\frac {\sqrt {7}\,\left (2\,x-1\right )}{7}\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2 - x + 2),x)

[Out]

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

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sympy [A]  time = 0.11, size = 26, normalized size = 1.37 \[ \frac {2 \sqrt {7} \operatorname {atan}{\left (\frac {2 \sqrt {7} x}{7} - \frac {\sqrt {7}}{7} \right )}}{7} \]

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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