3.31 \(\int \frac{\text{sech}^2(\frac{1}{x})}{x^2} \, dx\)

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

[Out]

-Tanh[x^(-1)]

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Rubi [A]  time = 0.0221789, antiderivative size = 6, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5436, 3767, 8} \[ -\tanh \left (\frac{1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x^(-1)]^2/x^2,x]

[Out]

-Tanh[x^(-1)]

Rule 5436

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sech[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\text{sech}^2\left (\frac{1}{x}\right )}{x^2} \, dx &=-\operatorname{Subst}\left (\int \text{sech}^2(x) \, dx,x,\frac{1}{x}\right )\\ &=-\left (i \operatorname{Subst}\left (\int 1 \, dx,x,-i \tanh \left (\frac{1}{x}\right )\right )\right )\\ &=-\tanh \left (\frac{1}{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.01752, size = 6, normalized size = 1. \[ -\tanh \left (\frac{1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x^(-1)]^2/x^2,x]

[Out]

-Tanh[x^(-1)]

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Maple [A]  time = 0.012, size = 7, normalized size = 1.2 \begin{align*} -\tanh \left ({x}^{-1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-tanh(1/x)

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Maxima [A]  time = 1.1392, size = 16, normalized size = 2.67 \begin{align*} \frac{2}{e^{\frac{2}{x}} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(e^(2/x) + 1)

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Fricas [B]  time = 1.99213, size = 80, normalized size = 13.33 \begin{align*} \frac{2}{\cosh \left (\frac{1}{x}\right )^{2} + 2 \, \cosh \left (\frac{1}{x}\right ) \sinh \left (\frac{1}{x}\right ) + \sinh \left (\frac{1}{x}\right )^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(cosh(1/x)^2 + 2*cosh(1/x)*sinh(1/x) + sinh(1/x)^2 + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{2}{\left (\frac{1}{x} \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(1/x)**2/x**2, x)

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Giac [A]  time = 1.18398, size = 16, normalized size = 2.67 \begin{align*} \frac{2}{e^{\frac{2}{x}} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(e^(2/x) + 1)