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

Optimal result

Integrand size = 10, antiderivative size = 6 \[ \int \frac {\text {sech}^2\left (\frac {1}{x}\right )}{x^2} \, dx=-\tanh \left (\frac {1}{x}\right ) \]

[Out]

-tanh(1/x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5544, 3852, 8} \[ \int \frac {\text {sech}^2\left (\frac {1}{x}\right )}{x^2} \, dx=-\tanh \left (\frac {1}{x}\right ) \]

[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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}^2\left (\frac {1}{x}\right )}{x^2} \, dx=-\tanh \left (\frac {1}{x}\right ) \]

[In]

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

[Out]

-Tanh[x^(-1)]

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(-\tanh \left (\frac {1}{x}\right )\)	\(7\)
default	\(-\tanh \left (\frac {1}{x}\right )\)	\(7\)
risch	\(\frac {2}{{\mathrm e}^{\frac {2}{x}}+1}\)	\(13\)
parallelrisch	\(-\frac {2 \tanh \left (\frac {1}{2 x}\right )}{1+\tanh \left (\frac {1}{2 x}\right )^{2}}\)	\(21\)

[In]

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

[Out]

-tanh(1/x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 28 vs. \(2 (6) = 12\).

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 4.67 \[ \int \frac {\text {sech}^2\left (\frac {1}{x}\right )}{x^2} \, dx=\frac {2}{\cosh \left (\frac {1}{x}\right )^{2} + 2 \, \cosh \left (\frac {1}{x}\right ) \sinh \left (\frac {1}{x}\right ) + \sinh \left (\frac {1}{x}\right )^{2} + 1} \]

[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)

Sympy [F]

\[ \int \frac {\text {sech}^2\left (\frac {1}{x}\right )}{x^2} \, dx=\int \frac {\operatorname {sech}^{2}{\left (\frac {1}{x} \right )}}{x^{2}}\, dx \]

[In]

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

[Out]

Integral(sech(1/x)**2/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 2.00 \[ \int \frac {\text {sech}^2\left (\frac {1}{x}\right )}{x^2} \, dx=\frac {2}{e^{\frac {2}{x}} + 1} \]

[In]

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

[Out]

2/(e^(2/x) + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 2.00 \[ \int \frac {\text {sech}^2\left (\frac {1}{x}\right )}{x^2} \, dx=\frac {2}{e^{\frac {2}{x}} + 1} \]

[In]

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

[Out]

2/(e^(2/x) + 1)

Mupad [B] (verification not implemented)

Time = 2.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 2.00 \[ \int \frac {\text {sech}^2\left (\frac {1}{x}\right )}{x^2} \, dx=\frac {2}{{\mathrm {e}}^{2/x}+1} \]

[In]

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

[Out]

2/(exp(2/x) + 1)