3.3.9 \(\int \frac {-3+x}{(4+2 x+x^2)^2} \, dx\) [209]

Optimal. Leaf size=39 \[ \frac {-7-4 x}{6 \left (4+2 x+x^2\right )}-\frac {2 \tan ^{-1}\left (\frac {1+x}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

[Out]

1/6*(-7-4*x)/(x^2+2*x+4)-2/9*arctan(1/3*(1+x)*3^(1/2))*3^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 = {652, 632, 210} \begin {gather*} -\frac {4 x+7}{6 \left (x^2+2 x+4\right )}-\frac {2 \tan ^{-1}\left (\frac {x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + x)/(4 + 2*x + x^2)^2,x]

[Out]

-1/6*(7 + 4*x)/(4 + 2*x + x^2) - (2*ArcTan[(1 + x)/Sqrt[3]])/(3*Sqrt[3])

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]

Rule 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

\begin {align*} \int \frac {-3+x}{\left (4+2 x+x^2\right )^2} \, dx &=-\frac {7+4 x}{6 \left (4+2 x+x^2\right )}-\frac {2}{3} \int \frac {1}{4+2 x+x^2} \, dx\\ &=-\frac {7+4 x}{6 \left (4+2 x+x^2\right )}+\frac {4}{3} \text {Subst}\left (\int \frac {1}{-12-x^2} \, dx,x,2+2 x\right )\\ &=-\frac {7+4 x}{6 \left (4+2 x+x^2\right )}-\frac {2 \tan ^{-1}\left (\frac {1+x}{\sqrt {3}}\right )}{3 \sqrt {3}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 39, normalized size = 1.00 \begin {gather*} \frac {-7-4 x}{6 \left (4+2 x+x^2\right )}-\frac {2 \tan ^{-1}\left (\frac {1+x}{\sqrt {3}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + x)/(4 + 2*x + x^2)^2,x]

[Out]

(-7 - 4*x)/(6*(4 + 2*x + x^2)) - (2*ArcTan[(1 + x)/Sqrt[3]])/(3*Sqrt[3])

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Mathics [A]
time = 1.99, size = 39, normalized size = 1.00 \begin {gather*} \frac {-21-12 x-4 \sqrt {3} \text {ArcTan}\left [\frac {\sqrt {3} \left (1+x\right )}{3}\right ] \left (4+2 x+x^2\right )}{72+36 x+18 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(-3 + x)/(4 + 2*x + x^2)^2,x]')

[Out]

(-21 - 12 x - 4 Sqrt[3] ArcTan[Sqrt[3] (1 + x) / 3] (4 + 2 x + x ^ 2)) / (18 (4 + 2 x + x ^ 2))

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Maple [A]
time = 0.14, size = 35, normalized size = 0.90

method	result	size
risch	\(\frac {-\frac {2 x}{3}-\frac {7}{6}}{x^{2}+2 x +4}-\frac {2 \arctan \left (\frac {\left (1+x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}\)	\(32\)
default	\(\frac {-8 x -14}{12 x^{2}+24 x +48}-\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2+2 x \right ) \sqrt {3}}{6}\right )}{9}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3+x)/(x^2+2*x+4)^2,x,method=_RETURNVERBOSE)

[Out]

1/12*(-8*x-14)/(x^2+2*x+4)-2/9*3^(1/2)*arctan(1/6*(2+2*x)*3^(1/2))

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Maxima [A]
time = 0.34, size = 32, normalized size = 0.82 \begin {gather*} -\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x + 1\right )}\right ) - \frac {4 \, x + 7}{6 \, {\left (x^{2} + 2 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3+x)/(x^2+2*x+4)^2,x, algorithm="maxima")

[Out]

-2/9*sqrt(3)*arctan(1/3*sqrt(3)*(x + 1)) - 1/6*(4*x + 7)/(x^2 + 2*x + 4)

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Fricas [A]
time = 0.35, size = 39, normalized size = 1.00 \begin {gather*} -\frac {4 \, \sqrt {3} {\left (x^{2} + 2 \, x + 4\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x + 1\right )}\right ) + 12 \, x + 21}{18 \, {\left (x^{2} + 2 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3+x)/(x^2+2*x+4)^2,x, algorithm="fricas")

[Out]

-1/18*(4*sqrt(3)*(x^2 + 2*x + 4)*arctan(1/3*sqrt(3)*(x + 1)) + 12*x + 21)/(x^2 + 2*x + 4)

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Sympy [A]
time = 0.06, size = 41, normalized size = 1.05 \begin {gather*} \frac {- 4 x - 7}{6 x^{2} + 12 x + 24} - \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3+x)/(x**2+2*x+4)**2,x)

[Out]

(-4*x - 7)/(6*x**2 + 12*x + 24) - 2*sqrt(3)*atan(sqrt(3)*x/3 + sqrt(3)/3)/9

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Giac [A]
time = 0.00, size = 42, normalized size = 1.08 \begin {gather*} \frac {-4 x-7}{6 \left (x^{2}+2 x+4\right )}-\frac {4 \arctan \left (\frac {x+1}{\sqrt {3}}\right )}{2\cdot 3 \sqrt {3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3+x)/(x^2+2*x+4)^2,x)

[Out]

-2/9*sqrt(3)*arctan(1/3*sqrt(3)*(x + 1)) - 1/6*(4*x + 7)/(x^2 + 2*x + 4)

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Mupad [B]
time = 0.04, size = 36, normalized size = 0.92 \begin {gather*} -\frac {\frac {2\,x}{3}+\frac {7}{6}}{x^2+2\,x+4}-\frac {2\,\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,x}{3}+\frac {\sqrt {3}}{3}\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 3)/(2*x + x^2 + 4)^2,x)

[Out]

- ((2*x)/3 + 7/6)/(2*x + x^2 + 4) - (2*3^(1/2)*atan((3^(1/2)*x)/3 + 3^(1/2)/3))/9

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