3.1.91 \(\int \frac {1}{(2+4 x+3 x^2)^2} \, dx\) [91]

Optimal. Leaf size=43 \[ \frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {2+3 x}{\sqrt {2}}\right )}{4 \sqrt {2}} \]

[Out]

1/4*(2+3*x)/(3*x^2+4*x+2)+3/8*arctan(1/2*(2+3*x)*2^(1/2))*2^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {628, 632, 210} \begin {gather*} \frac {3 \text {ArcTan}\left (\frac {3 x+2}{\sqrt {2}}\right )}{4 \sqrt {2}}+\frac {3 x+2}{4 \left (3 x^2+4 x+2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 4*x + 3*x^2)^(-2),x]

[Out]

(2 + 3*x)/(4*(2 + 4*x + 3*x^2)) + (3*ArcTan[(2 + 3*x)/Sqrt[2]])/(4*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 628

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

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}{\left (2+4 x+3 x^2\right )^2} \, dx &=\frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}+\frac {3}{4} \int \frac {1}{2+4 x+3 x^2} \, dx\\ &=\frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}-\frac {3}{2} \text {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,4+6 x\right )\\ &=\frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {2+3 x}{\sqrt {2}}\right )}{4 \sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 43, normalized size = 1.00 \begin {gather*} \frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {2+3 x}{\sqrt {2}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 4*x + 3*x^2)^(-2),x]

[Out]

(2 + 3*x)/(4*(2 + 4*x + 3*x^2)) + (3*ArcTan[(2 + 3*x)/Sqrt[2]])/(4*Sqrt[2])

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Maple [A]
time = 0.64, size = 37, normalized size = 0.86

method	result	size
risch	\(\frac {\frac {x}{4}+\frac {1}{6}}{x^{2}+\frac {4}{3} x +\frac {2}{3}}+\frac {3 \arctan \left (\frac {\left (2+3 x \right ) \sqrt {2}}{2}\right ) \sqrt {2}}{8}\)	\(34\)
default	\(\frac {6 x +4}{24 x^{2}+32 x +16}+\frac {3 \sqrt {2}\, \arctan \left (\frac {\left (6 x +4\right ) \sqrt {2}}{4}\right )}{8}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x^2+4*x+2)^2,x,method=_RETURNVERBOSE)

[Out]

1/8/(3*x^2+4*x+2)*(6*x+4)+3/8*2^(1/2)*arctan(1/4*(6*x+4)*2^(1/2))

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Maxima [A]
time = 0.51, size = 36, normalized size = 0.84 \begin {gather*} \frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) + \frac {3 \, x + 2}{4 \, {\left (3 \, x^{2} + 4 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2+4*x+2)^2,x, algorithm="maxima")

[Out]

3/8*sqrt(2)*arctan(1/2*sqrt(2)*(3*x + 2)) + 1/4*(3*x + 2)/(3*x^2 + 4*x + 2)

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Fricas [A]
time = 1.19, size = 45, normalized size = 1.05 \begin {gather*} \frac {3 \, \sqrt {2} {\left (3 \, x^{2} + 4 \, x + 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) + 6 \, x + 4}{8 \, {\left (3 \, x^{2} + 4 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2+4*x+2)^2,x, algorithm="fricas")

[Out]

1/8*(3*sqrt(2)*(3*x^2 + 4*x + 2)*arctan(1/2*sqrt(2)*(3*x + 2)) + 6*x + 4)/(3*x^2 + 4*x + 2)

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Sympy [A]
time = 0.05, size = 39, normalized size = 0.91 \begin {gather*} \frac {3 x + 2}{12 x^{2} + 16 x + 8} + \frac {3 \sqrt {2} \operatorname {atan}{\left (\frac {3 \sqrt {2} x}{2} + \sqrt {2} \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x**2+4*x+2)**2,x)

[Out]

(3*x + 2)/(12*x**2 + 16*x + 8) + 3*sqrt(2)*atan(3*sqrt(2)*x/2 + sqrt(2))/8

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Giac [A]
time = 0.86, size = 36, normalized size = 0.84 \begin {gather*} \frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) + \frac {3 \, x + 2}{4 \, {\left (3 \, x^{2} + 4 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2+4*x+2)^2,x, algorithm="giac")

[Out]

3/8*sqrt(2)*arctan(1/2*sqrt(2)*(3*x + 2)) + 1/4*(3*x + 2)/(3*x^2 + 4*x + 2)

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Mupad [B]
time = 0.04, size = 33, normalized size = 0.77 \begin {gather*} \frac {\frac {x}{4}+\frac {1}{6}}{x^2+\frac {4\,x}{3}+\frac {2}{3}}+\frac {3\,\sqrt {2}\,\mathrm {atan}\left (\frac {3\,\sqrt {2}\,x}{2}+\sqrt {2}\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*x + 3*x^2 + 2)^2,x)

[Out]

(x/4 + 1/6)/((4*x)/3 + x^2 + 2/3) + (3*2^(1/2)*atan((3*2^(1/2)*x)/2 + 2^(1/2)))/8

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