3.557 \(\int \frac{81+36 x^2+16 x^4}{729-64 x^6} \, dx\)

Optimal. Leaf size=10 \[ \frac{1}{6} \tanh ^{-1}\left (\frac{2 x}{3}\right ) \]

[Out]

ArcTanh[(2*x)/3]/6

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Rubi [A]  time = 0.009933, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1586, 206} \[ \frac{1}{6} \tanh ^{-1}\left (\frac{2 x}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(81 + 36*x^2 + 16*x^4)/(729 - 64*x^6),x]

[Out]

ArcTanh[(2*x)/3]/6

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{81+36 x^2+16 x^4}{729-64 x^6} \, dx &=\int \frac{1}{9-4 x^2} \, dx\\ &=\frac{1}{6} \tanh ^{-1}\left (\frac{2 x}{3}\right )\\ \end{align*}

Mathematica [B]  time = 0.0026819, size = 21, normalized size = 2.1 \[ \frac{1}{12} \log (2 x+3)-\frac{1}{12} \log (3-2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[(81 + 36*x^2 + 16*x^4)/(729 - 64*x^6),x]

[Out]

-Log[3 - 2*x]/12 + Log[3 + 2*x]/12

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Maple [B]  time = 0.006, size = 18, normalized size = 1.8 \begin{align*}{\frac{\ln \left ( 3+2\,x \right ) }{12}}-{\frac{\ln \left ( -3+2\,x \right ) }{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^4+36*x^2+81)/(-64*x^6+729),x)

[Out]

1/12*ln(3+2*x)-1/12*ln(-3+2*x)

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Maxima [B]  time = 0.895089, size = 23, normalized size = 2.3 \begin{align*} \frac{1}{12} \, \log \left (2 \, x + 3\right ) - \frac{1}{12} \, \log \left (2 \, x - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^4+36*x^2+81)/(-64*x^6+729),x, algorithm="maxima")

[Out]

1/12*log(2*x + 3) - 1/12*log(2*x - 3)

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Fricas [B]  time = 1.35319, size = 53, normalized size = 5.3 \begin{align*} \frac{1}{12} \, \log \left (2 \, x + 3\right ) - \frac{1}{12} \, \log \left (2 \, x - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^4+36*x^2+81)/(-64*x^6+729),x, algorithm="fricas")

[Out]

1/12*log(2*x + 3) - 1/12*log(2*x - 3)

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Sympy [B]  time = 0.0933, size = 15, normalized size = 1.5 \begin{align*} - \frac{\log{\left (x - \frac{3}{2} \right )}}{12} + \frac{\log{\left (x + \frac{3}{2} \right )}}{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x**4+36*x**2+81)/(-64*x**6+729),x)

[Out]

-log(x - 3/2)/12 + log(x + 3/2)/12

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Giac [B]  time = 1.04777, size = 20, normalized size = 2. \begin{align*} \frac{1}{12} \, \log \left ({\left | x + \frac{3}{2} \right |}\right ) - \frac{1}{12} \, \log \left ({\left | x - \frac{3}{2} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^4+36*x^2+81)/(-64*x^6+729),x, algorithm="giac")

[Out]

1/12*log(abs(x + 3/2)) - 1/12*log(abs(x - 3/2))