3.1.8 \(\int \frac {2-3 x^2}{4-9 x^4} \, dx\)

Optimal. Leaf size=16 \[ \frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {6}} \]

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {26, 203} \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 3*x^2)/(4 - 9*x^4),x]

[Out]

ArcTan[Sqrt[3/2]*x]/Sqrt[6]

Rule 26

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(j_))^(p_.), x_Symbol] :> Dist[(-(b^2/d))^m, Int[
u/(a - b*x^n)^m, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[j, 2*n] && EqQ[p, -m] && EqQ[b^2*c + a^2*d,
0] && GtQ[a, 0] && LtQ[d, 0]

Rule 203

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

Rubi steps

\begin {align*} \int \frac {2-3 x^2}{4-9 x^4} \, dx &=\int \frac {1}{2+3 x^2} \, dx\\ &=\frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {6}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 3*x^2)/(4 - 9*x^4),x]

[Out]

ArcTan[Sqrt[3/2]*x]/Sqrt[6]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2-3 x^2}{4-9 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(2 - 3*x^2)/(4 - 9*x^4),x]

[Out]

IntegrateAlgebraic[(2 - 3*x^2)/(4 - 9*x^4), x]

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fricas [A]  time = 0.86, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{6} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2+2)/(-9*x^4+4),x, algorithm="fricas")

[Out]

1/6*sqrt(6)*arctan(1/2*sqrt(6)*x)

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giac [A]  time = 0.16, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{6} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2+2)/(-9*x^4+4),x, algorithm="giac")

[Out]

1/6*sqrt(6)*arctan(1/2*sqrt(6)*x)

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maple [A]  time = 0.00, size = 13, normalized size = 0.81 \begin {gather*} \frac {\sqrt {6}\, \arctan \left (\frac {\sqrt {6}\, x}{2}\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^2+2)/(-9*x^4+4),x)

[Out]

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

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maxima [A]  time = 2.31, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{6} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2+2)/(-9*x^4+4),x, algorithm="maxima")

[Out]

1/6*sqrt(6)*arctan(1/2*sqrt(6)*x)

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mupad [B]  time = 0.03, size = 12, normalized size = 0.75 \begin {gather*} \frac {\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x}{2}\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2 - 2)/(9*x^4 - 4),x)

[Out]

(6^(1/2)*atan((6^(1/2)*x)/2))/6

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sympy [A]  time = 0.11, size = 15, normalized size = 0.94 \begin {gather*} \frac {\sqrt {6} \operatorname {atan}{\left (\frac {\sqrt {6} x}{2} \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x**2+2)/(-9*x**4+4),x)

[Out]

sqrt(6)*atan(sqrt(6)*x/2)/6

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