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

Optimal. Leaf size=21 \[ \frac{x}{2 \left (1-x^2\right )}+\frac{1}{2} \tanh ^{-1}(x) \]

[Out]

x/(2*(1 - x^2)) + ArcTanh[x]/2

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Rubi [A]  time = 0.0028625, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {199, 207} \[ \frac{x}{2 \left (1-x^2\right )}+\frac{1}{2} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(-1 + x^2)^(-2),x]

[Out]

x/(2*(1 - x^2)) + ArcTanh[x]/2

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [A]  time = 0.0073377, size = 27, normalized size = 1.29 \[ \frac{1}{4} \left (-\frac{2 x}{x^2-1}-\log (1-x)+\log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^2)^(-2),x]

[Out]

((-2*x)/(-1 + x^2) - Log[1 - x] + Log[1 + x])/4

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Maple [A]  time = 0.008, size = 28, normalized size = 1.3 \begin{align*} -{\frac{1}{4\,x-4}}-{\frac{\ln \left ( x-1 \right ) }{4}}-{\frac{1}{4+4\,x}}+{\frac{\ln \left ( 1+x \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(x-1)-1/4*ln(x-1)-1/4/(1+x)+1/4*ln(1+x)

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Maxima [A]  time = 1.06343, size = 31, normalized size = 1.48 \begin{align*} -\frac{x}{2 \,{\left (x^{2} - 1\right )}} + \frac{1}{4} \, \log \left (x + 1\right ) - \frac{1}{4} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(x^2 - 1) + 1/4*log(x + 1) - 1/4*log(x - 1)

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Fricas [B]  time = 0.952458, size = 90, normalized size = 4.29 \begin{align*} \frac{{\left (x^{2} - 1\right )} \log \left (x + 1\right ) -{\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 2 \, x}{4 \,{\left (x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((x^2 - 1)*log(x + 1) - (x^2 - 1)*log(x - 1) - 2*x)/(x^2 - 1)

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Sympy [A]  time = 0.099548, size = 20, normalized size = 0.95 \begin{align*} - \frac{x}{2 x^{2} - 2} - \frac{\log{\left (x - 1 \right )}}{4} + \frac{\log{\left (x + 1 \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(2*x**2 - 2) - log(x - 1)/4 + log(x + 1)/4

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Giac [A]  time = 1.18725, size = 34, normalized size = 1.62 \begin{align*} -\frac{x}{2 \,{\left (x^{2} - 1\right )}} + \frac{1}{4} \, \log \left ({\left | x + 1 \right |}\right ) - \frac{1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(x^2 - 1) + 1/4*log(abs(x + 1)) - 1/4*log(abs(x - 1))