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

Optimal result

Integrand size = 8, antiderivative size = 4 \[ \int \frac {\sin \left (\frac {1}{x}\right )}{x^2} \, dx=\cos \left (\frac {1}{x}\right ) \]

[Out]

cos(1/x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3460, 2718} \[ \int \frac {\sin \left (\frac {1}{x}\right )}{x^2} \, dx=\cos \left (\frac {1}{x}\right ) \]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {\sin \left (\frac {1}{x}\right )}{x^2} \, dx=\cos \left (\frac {1}{x}\right ) \]

[In]

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

[Out]

Cos[x^(-1)]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\cos \left (\frac {1}{x}\right )\)	\(5\)
default	\(\cos \left (\frac {1}{x}\right )\)	\(5\)
risch	\(\cos \left (\frac {1}{x}\right )\)	\(5\)
parallelrisch	\(1+\cos \left (\frac {1}{x}\right )\)	\(7\)
norman	\(\frac {2}{1+\tan ^{2}\left (\frac {1}{2 x}\right )}\)	\(15\)
meijerg	\(-\sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (\frac {1}{x}\right )}{\sqrt {\pi }}\right )\)	\(19\)

[In]

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

[Out]

cos(1/x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {\sin \left (\frac {1}{x}\right )}{x^2} \, dx=\cos \left (\frac {1}{x}\right ) \]

[In]

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

[Out]

cos(1/x)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {\sin \left (\frac {1}{x}\right )}{x^2} \, dx=\cos {\left (\frac {1}{x} \right )} \]

[In]

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

[Out]

cos(1/x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {\sin \left (\frac {1}{x}\right )}{x^2} \, dx=\cos \left (\frac {1}{x}\right ) \]

[In]

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

[Out]

cos(1/x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {\sin \left (\frac {1}{x}\right )}{x^2} \, dx=\cos \left (\frac {1}{x}\right ) \]

[In]

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

[Out]

cos(1/x)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {\sin \left (\frac {1}{x}\right )}{x^2} \, dx=\cos \left (\frac {1}{x}\right ) \]

[In]

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

[Out]

cos(1/x)