3.826 \(\int \frac {\csc (\frac {1}{x})}{x^2} \, dx\)

Optimal. Leaf size=5 \[ \tanh ^{-1}\left (\cos \left (\frac {1}{x}\right )\right ) \]

[Out]

arctanh(cos(1/x))

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Rubi [A]  time = 0.01, antiderivative size = 5, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4205, 3770} \[ \tanh ^{-1}\left (\cos \left (\frac {1}{x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Cos[x^(-1)]]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4205

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Csc[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 {\csc \left (\frac {1}{x}\right )}{x^2} \, dx &=-\operatorname {Subst}\left (\int \csc (x) \, dx,x,\frac {1}{x}\right )\\ &=\tanh ^{-1}\left (\cos \left (\frac {1}{x}\right )\right )\\ \end {align*}

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Mathematica [B]  time = 0.01, size = 21, normalized size = 4.20 \[ \log \left (\cos \left (\frac {1}{2 x}\right )\right )-\log \left (\sin \left (\frac {1}{2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Cos[1/(2*x)]] - Log[Sin[1/(2*x)]]

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fricas [B]  time = 0.65, size = 23, normalized size = 4.60 \[ \frac {1}{2} \, \log \left (\frac {1}{2} \, \cos \left (\frac {1}{x}\right ) + \frac {1}{2}\right ) - \frac {1}{2} \, \log \left (-\frac {1}{2} \, \cos \left (\frac {1}{x}\right ) + \frac {1}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.14, size = 43, normalized size = 8.60 \[ -\frac {1}{2} \, \log \left (\frac {4 \, \tan \left (\frac {1}{2 \, x}\right )^{2}}{\tan \left (\frac {1}{2 \, x}\right )^{2} + 1}\right ) + \frac {1}{2} \, \log \left (\frac {4}{\tan \left (\frac {1}{2 \, x}\right )^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 11, normalized size = 2.20 \[ \ln \left (\csc \left (\frac {1}{x}\right )+\cot \left (\frac {1}{x}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(csc(1/x)+cot(1/x))

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maxima [A]  time = 0.33, size = 10, normalized size = 2.00 \[ \log \left (\cot \left (\frac {1}{x}\right ) + \csc \left (\frac {1}{x}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(cot(1/x) + csc(1/x))

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mupad [B]  time = 3.68, size = 31, normalized size = 6.20 \[ \ln \left (-{\mathrm {e}}^{1{}\mathrm {i}/x}\,2{}\mathrm {i}-2{}\mathrm {i}\right )-\ln \left (-{\mathrm {e}}^{1{}\mathrm {i}/x}\,2{}\mathrm {i}+2{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(- exp(1i/x)*2i - 2i) - log(2i - exp(1i/x)*2i)

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sympy [A]  time = 1.26, size = 10, normalized size = 2.00 \[ \log {\left (\cot {\left (\frac {1}{x} \right )} + \csc {\left (\frac {1}{x} \right )} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(cot(1/x) + csc(1/x))

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