3.1.10 \(\int \frac {1}{3-2 x+x^2} \, dx\) [10]

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

[Out]

-1/2*arctan(1/2*(1-x)*2^(1/2))*2^(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 = {632, 210} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {1-x}{\sqrt {2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 2*x + x^2)^(-1),x]

[Out]

-(ArcTan[(1 - x)/Sqrt[2]]/Sqrt[2])

Rule 210

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

Rule 632

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}{3-2 x+x^2} \, dx &=-\left (2 \text {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,-2+2 x\right )\right )\\ &=-\frac {\tan ^{-1}\left (\frac {1-x}{\sqrt {2}}\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 16, normalized size = 0.84 \begin {gather*} \frac {\tan ^{-1}\left (\frac {-1+x}{\sqrt {2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - 2*x + x^2)^(-1),x]

[Out]

ArcTan[(-1 + x)/Sqrt[2]]/Sqrt[2]

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Maple [A]
time = 0.21, size = 17, normalized size = 0.89

method	result	size
risch	\(\frac {\sqrt {2}\, \arctan \left (\frac {\left (-1+x \right ) \sqrt {2}}{2}\right )}{2}\)	\(15\)
default	\(\frac {\sqrt {2}\, \arctan \left (\frac {\left (2 x -2\right ) \sqrt {2}}{4}\right )}{2}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2-2*x+3),x,method=_RETURNVERBOSE)

[Out]

1/2*2^(1/2)*arctan(1/4*(2*x-2)*2^(1/2))

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Maxima [A]
time = 2.79, size = 14, normalized size = 0.74 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-2*x+3),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.65, size = 14, normalized size = 0.74 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-2*x+3),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.03, size = 22, normalized size = 1.16 \begin {gather*} \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} - \frac {\sqrt {2}}{2} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**2-2*x+3),x)

[Out]

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

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Giac [A]
time = 0.54, size = 14, normalized size = 0.74 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-2*x+3),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.03, size = 14, normalized size = 0.74 \begin {gather*} \frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (x-1\right )}{2}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2 - 2*x + 3),x)

[Out]

(2^(1/2)*atan((2^(1/2)*(x - 1))/2))/2

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