3.11 \(\int \frac{1}{4-5 x^2+x^4} \, dx\)

Optimal. Leaf size=17 \[ \frac{1}{3} \tanh ^{-1}(x)-\frac{1}{6} \tanh ^{-1}\left (\frac{x}{2}\right ) \]

[Out]

-ArcTanh[x/2]/6 + ArcTanh[x]/3

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Rubi [A]  time = 0.0079561, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1093, 207} \[ \frac{1}{3} \tanh ^{-1}(x)-\frac{1}{6} \tanh ^{-1}\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(4 - 5*x^2 + x^4)^(-1),x]

[Out]

-ArcTanh[x/2]/6 + ArcTanh[x]/3

Rule 1093

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 207

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

Rubi steps

\begin{align*} \int \frac{1}{4-5 x^2+x^4} \, dx &=\frac{1}{3} \int \frac{1}{-4+x^2} \, dx-\frac{1}{3} \int \frac{1}{-1+x^2} \, dx\\ &=-\frac{1}{6} \tanh ^{-1}\left (\frac{x}{2}\right )+\frac{1}{3} \tanh ^{-1}(x)\\ \end{align*}

Mathematica [B]  time = 0.0072305, size = 37, normalized size = 2.18 \[ -\frac{1}{6} \log (1-x)+\frac{1}{12} \log (2-x)+\frac{1}{6} \log (x+1)-\frac{1}{12} \log (x+2) \]

Antiderivative was successfully verified.

[In]

Integrate[(4 - 5*x^2 + x^4)^(-1),x]

[Out]

-Log[1 - x]/6 + Log[2 - x]/12 + Log[1 + x]/6 - Log[2 + x]/12

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Maple [B]  time = 0.049, size = 26, normalized size = 1.5 \begin{align*} -{\frac{\ln \left ( 2+x \right ) }{12}}+{\frac{\ln \left ( 1+x \right ) }{6}}-{\frac{\ln \left ( -1+x \right ) }{6}}+{\frac{\ln \left ( -2+x \right ) }{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4-5*x^2+4),x)

[Out]

-1/12*ln(2+x)+1/6*ln(1+x)-1/6*ln(-1+x)+1/12*ln(-2+x)

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Maxima [B]  time = 0.97829, size = 34, normalized size = 2. \begin{align*} -\frac{1}{12} \, \log \left (x + 2\right ) + \frac{1}{6} \, \log \left (x + 1\right ) - \frac{1}{6} \, \log \left (x - 1\right ) + \frac{1}{12} \, \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4-5*x^2+4),x, algorithm="maxima")

[Out]

-1/12*log(x + 2) + 1/6*log(x + 1) - 1/6*log(x - 1) + 1/12*log(x - 2)

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Fricas [B]  time = 1.25218, size = 95, normalized size = 5.59 \begin{align*} -\frac{1}{12} \, \log \left (x + 2\right ) + \frac{1}{6} \, \log \left (x + 1\right ) - \frac{1}{6} \, \log \left (x - 1\right ) + \frac{1}{12} \, \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4-5*x^2+4),x, algorithm="fricas")

[Out]

-1/12*log(x + 2) + 1/6*log(x + 1) - 1/6*log(x - 1) + 1/12*log(x - 2)

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Sympy [B]  time = 0.169447, size = 26, normalized size = 1.53 \begin{align*} \frac{\log{\left (x - 2 \right )}}{12} - \frac{\log{\left (x - 1 \right )}}{6} + \frac{\log{\left (x + 1 \right )}}{6} - \frac{\log{\left (x + 2 \right )}}{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**4-5*x**2+4),x)

[Out]

log(x - 2)/12 - log(x - 1)/6 + log(x + 1)/6 - log(x + 2)/12

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Giac [B]  time = 1.14287, size = 39, normalized size = 2.29 \begin{align*} -\frac{1}{12} \, \log \left ({\left | x + 2 \right |}\right ) + \frac{1}{6} \, \log \left ({\left | x + 1 \right |}\right ) - \frac{1}{6} \, \log \left ({\left | x - 1 \right |}\right ) + \frac{1}{12} \, \log \left ({\left | x - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4-5*x^2+4),x, algorithm="giac")

[Out]

-1/12*log(abs(x + 2)) + 1/6*log(abs(x + 1)) - 1/6*log(abs(x - 1)) + 1/12*log(abs(x - 2))