3.1.78 \(\int \frac {1}{2+4 x+3 x^2} \, dx\) [78]

Optimal. Leaf size=18 \[ \frac {\tan ^{-1}\left (\frac {2+3 x}{\sqrt {2}}\right )}{\sqrt {2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {632, 210} \begin {gather*} \frac {\text {ArcTan}\left (\frac {3 x+2}{\sqrt {2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 4*x + 3*x^2)^(-1),x]

[Out]

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

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Mathematica [A]
time = 0.00, size = 18, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {2+3 x}{\sqrt {2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 4*x + 3*x^2)^(-1),x]

[Out]

ArcTan[(2 + 3*x)/Sqrt[2]]/Sqrt[2]

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Maple [A]
time = 0.82, size = 17, normalized size = 0.94

method	result	size
default	\(\frac {\sqrt {2}\, \arctan \left (\frac {\left (6 x +4\right ) \sqrt {2}}{4}\right )}{2}\)	\(17\)
risch	\(\frac {\arctan \left (\frac {\left (2+3 x \right ) \sqrt {2}}{2}\right ) \sqrt {2}}{2}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 16, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(3*x + 2))

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Fricas [A]
time = 1.34, size = 16, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(3*x + 2))

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Sympy [A]
time = 0.05, size = 22, normalized size = 1.22 \begin {gather*} \frac {\sqrt {2} \operatorname {atan}{\left (\frac {3 \sqrt {2} x}{2} + \sqrt {2} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*atan(3*sqrt(2)*x/2 + sqrt(2))/2

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Giac [A]
time = 1.59, size = 16, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(3*x + 2))

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Mupad [B]
time = 0.14, size = 16, normalized size = 0.89 \begin {gather*} \frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (3\,x+2\right )}{2}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*x + 3*x^2 + 2),x)

[Out]

(2^(1/2)*atan((2^(1/2)*(3*x + 2))/2))/2

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