3.1.37 \(\int \frac {\cos (a+\frac {b}{x})}{x^2} \, dx\) [37]

Optimal. Leaf size=13 \[ -\frac {\sin \left (a+\frac {b}{x}\right )}{b} \]

[Out]

-sin(a+b/x)/b

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Rubi [A]
time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3461, 2717} \begin {gather*} -\frac {\sin \left (a+\frac {b}{x}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b/x]/x^2,x]

[Out]

-(Sin[a + b/x]/b)

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3461

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

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Mathematica [A]
time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} -\frac {\sin \left (a+\frac {b}{x}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b/x]/x^2,x]

[Out]

-(Sin[a + b/x]/b)

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Maple [A]
time = 0.04, size = 14, normalized size = 1.08

method	result	size
derivativedivides	\(-\frac {\sin \left (a +\frac {b}{x}\right )}{b}\)	\(14\)
default	\(-\frac {\sin \left (a +\frac {b}{x}\right )}{b}\)	\(14\)
risch	\(-\frac {\sin \left (\frac {a x +b}{x}\right )}{b}\)	\(16\)
norman	\(-\frac {2 \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )}{b \left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right )}\)	\(34\)
meijerg	\(-\frac {\cos \left (a \right ) \sin \left (\frac {b}{x}\right )}{b}+\frac {\sqrt {\pi }\, \sin \left (a \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (\frac {b}{x}\right )}{\sqrt {\pi }}\right )}{b}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a+b/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-sin(a+b/x)/b

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Maxima [A]
time = 0.29, size = 13, normalized size = 1.00 \begin {gather*} -\frac {\sin \left (a + \frac {b}{x}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b/x)/x^2,x, algorithm="maxima")

[Out]

-sin(a + b/x)/b

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Fricas [A]
time = 0.39, size = 15, normalized size = 1.15 \begin {gather*} -\frac {\sin \left (\frac {a x + b}{x}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b/x)/x^2,x, algorithm="fricas")

[Out]

-sin((a*x + b)/x)/b

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Sympy [A]
time = 0.34, size = 15, normalized size = 1.15 \begin {gather*} \begin {cases} - \frac {\sin {\left (a + \frac {b}{x} \right )}}{b} & \text {for}\: b \neq 0 \\- \frac {\cos {\left (a \right )}}{x} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b/x)/x**2,x)

[Out]

Piecewise((-sin(a + b/x)/b, Ne(b, 0)), (-cos(a)/x, True))

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Giac [A]
time = 0.41, size = 15, normalized size = 1.15 \begin {gather*} -\frac {\sin \left (\frac {a x + b}{x}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b/x)/x^2,x, algorithm="giac")

[Out]

-sin((a*x + b)/x)/b

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Mupad [B]
time = 0.26, size = 13, normalized size = 1.00 \begin {gather*} -\frac {\sin \left (a+\frac {b}{x}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b/x)/x^2,x)

[Out]

-sin(a + b/x)/b

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