3.1.52 \(\int \frac {\sin (\frac {1}{x})}{x^2} \, dx\) [52]

Optimal. Leaf size=4 \[ \cos \left (\frac {1}{x}\right ) \]

[Out]

cos(1/x)

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Rubi [A]
time = 0.01, antiderivative size = 4, 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 = {3460, 2718} \begin {gather*} \cos \left (\frac {1}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Cos[x^(-1)]

Rule 2718

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

Rule 3460

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 4, normalized size = 1.00 \begin {gather*} \cos \left (\frac {1}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Cos[x^(-1)]

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Maple [A]
time = 0.01, size = 5, normalized size = 1.25

method	result	size
derivativedivides	\(\cos \left (\frac {1}{x}\right )\)	\(5\)
default	\(\cos \left (\frac {1}{x}\right )\)	\(5\)
risch	\(\cos \left (\frac {1}{x}\right )\)	\(5\)
norman	\(\frac {2}{1+\tan ^{2}\left (\frac {1}{2 x}\right )}\)	\(15\)
meijerg	\(-\sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (\frac {1}{x}\right )}{\sqrt {\pi }}\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(1/x)

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Maxima [A]
time = 1.74, size = 4, normalized size = 1.00 \begin {gather*} \cos \left (\frac {1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(1/x)

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Fricas [A]
time = 0.69, size = 4, normalized size = 1.00 \begin {gather*} \cos \left (\frac {1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(1/x)

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Sympy [A]
time = 0.17, size = 3, normalized size = 0.75 \begin {gather*} \cos {\left (\frac {1}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(1/x)

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Giac [A]
time = 0.46, size = 4, normalized size = 1.00 \begin {gather*} \cos \left (\frac {1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(1/x)

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Mupad [B]
time = 0.10, size = 4, normalized size = 1.00 \begin {gather*} \cos \left (\frac {1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(1/x)

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