3.2.8 \(\int \frac {\sin (a+\frac {b}{x})}{x^3} \, dx\) [108]

Optimal. Leaf size=29 \[ \frac {\cos \left (a+\frac {b}{x}\right )}{b x}-\frac {\sin \left (a+\frac {b}{x}\right )}{b^2} \]

[Out]

cos(a+b/x)/b/x-sin(a+b/x)/b^2

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Rubi [A]
time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3460, 3377, 2717} \begin {gather*} \frac {\cos \left (a+\frac {b}{x}\right )}{b x}-\frac {\sin \left (a+\frac {b}{x}\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x]/x^3,x]

[Out]

Cos[a + b/x]/(b*x) - Sin[a + b/x]/b^2

Rule 2717

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

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 (a+\frac {b}{x}\right )}{x^3} \, dx &=-\text {Subst}\left (\int x \sin (a+b x) \, dx,x,\frac {1}{x}\right )\\ &=\frac {\cos \left (a+\frac {b}{x}\right )}{b x}-\frac {\text {Subst}\left (\int \cos (a+b x) \, dx,x,\frac {1}{x}\right )}{b}\\ &=\frac {\cos \left (a+\frac {b}{x}\right )}{b x}-\frac {\sin \left (a+\frac {b}{x}\right )}{b^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 29, normalized size = 1.00 \begin {gather*} \frac {\cos \left (a+\frac {b}{x}\right )}{b x}-\frac {\sin \left (a+\frac {b}{x}\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x]/x^3,x]

[Out]

Cos[a + b/x]/(b*x) - Sin[a + b/x]/b^2

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Maple [A]
time = 0.04, size = 42, normalized size = 1.45

method	result	size
risch	\(\frac {\cos \left (\frac {a x +b}{x}\right )}{b x}-\frac {\sin \left (\frac {a x +b}{x}\right )}{b^{2}}\)	\(34\)
derivativedivides	\(-\frac {\sin \left (a +\frac {b}{x}\right )-\left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )+a \cos \left (a +\frac {b}{x}\right )}{b^{2}}\)	\(42\)
default	\(-\frac {\sin \left (a +\frac {b}{x}\right )-\left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )+a \cos \left (a +\frac {b}{x}\right )}{b^{2}}\)	\(42\)
norman	\(\frac {\frac {x}{b}-\frac {2 x^{2} \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )}{b^{2}}-\frac {x \left (\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right )}{b}}{\left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right ) x^{2}}\)	\(66\)
meijerg	\(-\frac {2 \sqrt {\pi }\, \cos \left (a \right ) \left (-\frac {b \cos \left (\frac {b}{x}\right )}{2 \sqrt {\pi }\, x}+\frac {\sin \left (\frac {b}{x}\right )}{2 \sqrt {\pi }}\right )}{b^{2}}-\frac {2 \sqrt {\pi }\, \sin \left (a \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (\frac {b}{x}\right )}{2 \sqrt {\pi }}+\frac {b \sin \left (\frac {b}{x}\right )}{2 \sqrt {\pi }\, x}\right )}{b^{2}}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/x)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.33, size = 50, normalized size = 1.72 \begin {gather*} -\frac {{\left (i \, \Gamma \left (2, \frac {i \, b}{x}\right ) - i \, \Gamma \left (2, -\frac {i \, b}{x}\right )\right )} \cos \left (a\right ) + {\left (\Gamma \left (2, \frac {i \, b}{x}\right ) + \Gamma \left (2, -\frac {i \, b}{x}\right )\right )} \sin \left (a\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x^3,x, algorithm="maxima")

[Out]

-1/2*((I*gamma(2, I*b/x) - I*gamma(2, -I*b/x))*cos(a) + (gamma(2, I*b/x) + gamma(2, -I*b/x))*sin(a))/b^2

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Fricas [A]
time = 0.35, size = 33, normalized size = 1.14 \begin {gather*} \frac {b \cos \left (\frac {a x + b}{x}\right ) - x \sin \left (\frac {a x + b}{x}\right )}{b^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x^3,x, algorithm="fricas")

[Out]

(b*cos((a*x + b)/x) - x*sin((a*x + b)/x))/(b^2*x)

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Sympy [A]
time = 0.51, size = 29, normalized size = 1.00 \begin {gather*} \begin {cases} \frac {\cos {\left (a + \frac {b}{x} \right )}}{b x} - \frac {\sin {\left (a + \frac {b}{x} \right )}}{b^{2}} & \text {for}\: b \neq 0 \\- \frac {\sin {\left (a \right )}}{2 x^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x**3,x)

[Out]

Piecewise((cos(a + b/x)/(b*x) - sin(a + b/x)/b**2, Ne(b, 0)), (-sin(a)/(2*x**2), True))

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Giac [A]
time = 2.67, size = 48, normalized size = 1.66 \begin {gather*} -\frac {a \cos \left (\frac {a x + b}{x}\right ) - \frac {{\left (a x + b\right )} \cos \left (\frac {a x + b}{x}\right )}{x} + \sin \left (\frac {a x + b}{x}\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x^3,x, algorithm="giac")

[Out]

-(a*cos((a*x + b)/x) - (a*x + b)*cos((a*x + b)/x)/x + sin((a*x + b)/x))/b^2

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Mupad [B]
time = 4.54, size = 29, normalized size = 1.00 \begin {gather*} \frac {\cos \left (a+\frac {b}{x}\right )}{b\,x}-\frac {\sin \left (a+\frac {b}{x}\right )}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/x)/x^3,x)

[Out]

cos(a + b/x)/(b*x) - sin(a + b/x)/b^2

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