3.1.31 \(\int \frac {\text {sech}^2(\frac {1}{x})}{x^2} \, dx\) [31]

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

[Out]

-tanh(1/x)

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Rubi [A]
time = 0.02, antiderivative size = 6, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5544, 3852, 8} \begin {gather*} -\tanh \left (\frac {1}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Tanh[x^(-1)]

Rule 8

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

Rule 3852

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 5544

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]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 6, normalized size = 1.00 \begin {gather*} -\tanh \left (\frac {1}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Tanh[x^(-1)]

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Maple [A]
time = 1.06, size = 13, normalized size = 2.17

method	result	size
risch	\(\frac {2}{{\mathrm e}^{\frac {2}{x}}+1}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(1/x)^2/x^2,x,method=_RETURNVERBOSE)

[Out]

2/(exp(2/x)+1)

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Maxima [A]
time = 0.27, size = 12, normalized size = 2.00 \begin {gather*} \frac {2}{e^{\frac {2}{x}} + 1} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 28 vs. \(2 (6) = 12\).
time = 0.34, size = 28, normalized size = 4.67 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {sech}^{2}{\left (\frac {1}{x} \right )}}{x^{2}}\, dx \end {gather*}

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 = 0.39, size = 12, normalized size = 2.00 \begin {gather*} \frac {2}{e^{\frac {2}{x}} + 1} \end {gather*}

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)

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Mupad [B]
time = 1.26, size = 12, normalized size = 2.00 \begin {gather*} \frac {2}{{\mathrm {e}}^{2/x}+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*cosh(1/x)^2),x)

[Out]

2/(exp(2/x) + 1)

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