3.928 \(\int \frac {\sec ^2(\frac {1}{x})}{x^2} \, dx\)

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

[Out]

-tan(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 = {4204, 3767, 8} \[ -\tan \left (\frac {1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Tan[x^(-1)]

Rule 8

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

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 4204

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 6, normalized size = 1.00 \[ -\tan \left (\frac {1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Tan[x^(-1)]

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fricas [A]  time = 1.04, size = 12, normalized size = 2.00 \[ -\frac {\sin \left (\frac {1}{x}\right )}{\cos \left (\frac {1}{x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sin(1/x)/cos(1/x)

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giac [B]  time = 0.13, size = 20, normalized size = 3.33 \[ \frac {2 \, \tan \left (\frac {1}{2 \, x}\right )}{\tan \left (\frac {1}{2 \, x}\right )^{2} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*tan(1/2/x)/(tan(1/2/x)^2 - 1)

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maple [A]  time = 0.04, size = 7, normalized size = 1.17 \[ -\tan \left (\frac {1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-tan(1/x)

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maxima [B]  time = 0.38, size = 36, normalized size = 6.00 \[ -\frac {2 \, \sin \left (\frac {2}{x}\right )}{\cos \left (\frac {2}{x}\right )^{2} + \sin \left (\frac {2}{x}\right )^{2} + 2 \, \cos \left (\frac {2}{x}\right ) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sin(2/x)/(cos(2/x)^2 + sin(2/x)^2 + 2*cos(2/x) + 1)

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mupad [B]  time = 3.03, size = 14, normalized size = 2.33 \[ -\frac {2{}\mathrm {i}}{{\mathrm {e}}^{\frac {2{}\mathrm {i}}{x}}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2i/(exp(2i/x) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (\frac {1}{x} \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(1/x)**2/x**2, x)

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