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

Optimal. Leaf size=8 \[ \frac{1}{x}-\tanh ^{-1}(x) \]

[Out]

x^(-1) - ArcTanh[x]

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Rubi [A]  time = 0.0052763, antiderivative size = 8, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1593, 325, 207} \[ \frac{1}{x}-\tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(-x^2 + x^4)^(-1),x]

[Out]

x^(-1) - ArcTanh[x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [B]  time = 0.0025107, size = 22, normalized size = 2.75 \[ \frac{1}{x}+\frac{1}{2} \log (1-x)-\frac{1}{2} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[(-x^2 + x^4)^(-1),x]

[Out]

x^(-1) + Log[1 - x]/2 - Log[1 + x]/2

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Maple [A]  time = 0.006, size = 17, normalized size = 2.1 \begin{align*} -{\frac{\ln \left ( 1+x \right ) }{2}}+{\frac{\ln \left ( -1+x \right ) }{2}}+{x}^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4-x^2),x)

[Out]

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

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Maxima [A]  time = 0.928798, size = 22, normalized size = 2.75 \begin{align*} \frac{1}{x} - \frac{1}{2} \, \log \left (x + 1\right ) + \frac{1}{2} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x - 1/2*log(x + 1) + 1/2*log(x - 1)

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Fricas [B]  time = 1.77213, size = 57, normalized size = 7.12 \begin{align*} -\frac{x \log \left (x + 1\right ) - x \log \left (x - 1\right ) - 2}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.089384, size = 15, normalized size = 1.88 \begin{align*} \frac{\log{\left (x - 1 \right )}}{2} - \frac{\log{\left (x + 1 \right )}}{2} + \frac{1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 1)/2 - log(x + 1)/2 + 1/x

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Giac [B]  time = 1.05507, size = 24, normalized size = 3. \begin{align*} \frac{1}{x} - \frac{1}{2} \, \log \left ({\left | x + 1 \right |}\right ) + \frac{1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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