3.2.9 \(\int \frac {\sin (a+\frac {b}{x})}{x^4} \, dx\) [109]

Optimal. Leaf size=45 \[ -\frac {2 \cos \left (a+\frac {b}{x}\right )}{b^3}+\frac {\cos \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 \sin \left (a+\frac {b}{x}\right )}{b^2 x} \]

[Out]

-2*cos(a+b/x)/b^3+cos(a+b/x)/b/x^2-2*sin(a+b/x)/b^2/x

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Rubi [A]
time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3460, 3377, 2718} \begin {gather*} -\frac {2 \cos \left (a+\frac {b}{x}\right )}{b^3}-\frac {2 \sin \left (a+\frac {b}{x}\right )}{b^2 x}+\frac {\cos \left (a+\frac {b}{x}\right )}{b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x]/x^4,x]

[Out]

(-2*Cos[a + b/x])/b^3 + Cos[a + b/x]/(b*x^2) - (2*Sin[a + b/x])/(b^2*x)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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^4} \, dx &=-\text {Subst}\left (\int x^2 \sin (a+b x) \, dx,x,\frac {1}{x}\right )\\ &=\frac {\cos \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 \text {Subst}\left (\int x \cos (a+b x) \, dx,x,\frac {1}{x}\right )}{b}\\ &=\frac {\cos \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 \sin \left (a+\frac {b}{x}\right )}{b^2 x}+\frac {2 \text {Subst}\left (\int \sin (a+b x) \, dx,x,\frac {1}{x}\right )}{b^2}\\ &=-\frac {2 \cos \left (a+\frac {b}{x}\right )}{b^3}+\frac {\cos \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 \sin \left (a+\frac {b}{x}\right )}{b^2 x}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 38, normalized size = 0.84 \begin {gather*} \frac {\left (b^2-2 x^2\right ) \cos \left (a+\frac {b}{x}\right )-2 b x \sin \left (a+\frac {b}{x}\right )}{b^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x]/x^4,x]

[Out]

((b^2 - 2*x^2)*Cos[a + b/x] - 2*b*x*Sin[a + b/x])/(b^3*x^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(94\) vs. \(2(45)=90\).
time = 0.04, size = 95, normalized size = 2.11

method	result	size
risch	\(\frac {\left (b^{2}-2 x^{2}\right ) \cos \left (\frac {a x +b}{x}\right )}{b^{3} x^{2}}-\frac {2 \sin \left (\frac {a x +b}{x}\right )}{b^{2} x}\)	\(46\)
norman	\(\frac {\frac {x}{b}+\frac {4 x^{3} \left (\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right )}{b^{3}}-\frac {4 x^{2} \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )}{b^{2}}-\frac {x \left (\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right )}{b}}{\left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right ) x^{3}}\)	\(87\)
derivativedivides	\(-\frac {-a^{2} \cos \left (a +\frac {b}{x}\right )-2 a \left (\sin \left (a +\frac {b}{x}\right )-\left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )\right )-\left (a +\frac {b}{x}\right )^{2} \cos \left (a +\frac {b}{x}\right )+2 \cos \left (a +\frac {b}{x}\right )+2 \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{b^{3}}\)	\(95\)
default	\(-\frac {-a^{2} \cos \left (a +\frac {b}{x}\right )-2 a \left (\sin \left (a +\frac {b}{x}\right )-\left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )\right )-\left (a +\frac {b}{x}\right )^{2} \cos \left (a +\frac {b}{x}\right )+2 \cos \left (a +\frac {b}{x}\right )+2 \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{b^{3}}\)	\(95\)
meijerg	\(-\frac {4 \sqrt {\pi }\, \cos \left (a \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-\frac {b^{2}}{2 x^{2}}+1\right ) \cos \left (\frac {b}{x}\right )}{2 \sqrt {\pi }}+\frac {b \sin \left (\frac {b}{x}\right )}{2 \sqrt {\pi }\, x}\right )}{b^{3}}-\frac {4 \sqrt {\pi }\, \sin \left (a \right ) \sqrt {b^{2}}\, \left (\frac {\left (b^{2}\right )^{\frac {3}{2}} \cos \left (\frac {b}{x}\right )}{2 \sqrt {\pi }\, x \,b^{2}}-\frac {\left (b^{2}\right )^{\frac {3}{2}} \left (-\frac {3 b^{2}}{2 x^{2}}+3\right ) \sin \left (\frac {b}{x}\right )}{6 \sqrt {\pi }\, b^{3}}\right )}{b^{4}}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/x)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/b^3*(-a^2*cos(a+b/x)-2*a*(sin(a+b/x)-(a+b/x)*cos(a+b/x))-(a+b/x)^2*cos(a+b/x)+2*cos(a+b/x)+2*(a+b/x)*sin(a+
b/x))

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.34, size = 51, normalized size = 1.13 \begin {gather*} -\frac {{\left (\Gamma \left (3, \frac {i \, b}{x}\right ) + \Gamma \left (3, -\frac {i \, b}{x}\right )\right )} \cos \left (a\right ) - {\left (i \, \Gamma \left (3, \frac {i \, b}{x}\right ) - i \, \Gamma \left (3, -\frac {i \, b}{x}\right )\right )} \sin \left (a\right )}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x^4,x, algorithm="maxima")

[Out]

-1/2*((gamma(3, I*b/x) + gamma(3, -I*b/x))*cos(a) - (I*gamma(3, I*b/x) - I*gamma(3, -I*b/x))*sin(a))/b^3

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Fricas [A]
time = 0.34, size = 44, normalized size = 0.98 \begin {gather*} -\frac {2 \, b x \sin \left (\frac {a x + b}{x}\right ) - {\left (b^{2} - 2 \, x^{2}\right )} \cos \left (\frac {a x + b}{x}\right )}{b^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x^4,x, algorithm="fricas")

[Out]

-(2*b*x*sin((a*x + b)/x) - (b^2 - 2*x^2)*cos((a*x + b)/x))/(b^3*x^2)

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Sympy [A]
time = 0.77, size = 46, normalized size = 1.02 \begin {gather*} \begin {cases} \frac {\cos {\left (a + \frac {b}{x} \right )}}{b x^{2}} - \frac {2 \sin {\left (a + \frac {b}{x} \right )}}{b^{2} x} - \frac {2 \cos {\left (a + \frac {b}{x} \right )}}{b^{3}} & \text {for}\: b \neq 0 \\- \frac {\sin {\left (a \right )}}{3 x^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x**4,x)

[Out]

Piecewise((cos(a + b/x)/(b*x**2) - 2*sin(a + b/x)/(b**2*x) - 2*cos(a + b/x)/b**3, Ne(b, 0)), (-sin(a)/(3*x**3)
, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (45) = 90\).
time = 3.64, size = 106, normalized size = 2.36 \begin {gather*} \frac {a^{2} \cos \left (\frac {a x + b}{x}\right ) - \frac {2 \, {\left (a x + b\right )} a \cos \left (\frac {a x + b}{x}\right )}{x} + 2 \, a \sin \left (\frac {a x + b}{x}\right ) + \frac {{\left (a x + b\right )}^{2} \cos \left (\frac {a x + b}{x}\right )}{x^{2}} - \frac {2 \, {\left (a x + b\right )} \sin \left (\frac {a x + b}{x}\right )}{x} - 2 \, \cos \left (\frac {a x + b}{x}\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x^4,x, algorithm="giac")

[Out]

(a^2*cos((a*x + b)/x) - 2*(a*x + b)*a*cos((a*x + b)/x)/x + 2*a*sin((a*x + b)/x) + (a*x + b)^2*cos((a*x + b)/x)
/x^2 - 2*(a*x + b)*sin((a*x + b)/x)/x - 2*cos((a*x + b)/x))/b^3

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Mupad [B]
time = 4.63, size = 46, normalized size = 1.02 \begin {gather*} \frac {b^2\,\cos \left (a+\frac {b}{x}\right )-2\,b\,x\,\sin \left (a+\frac {b}{x}\right )}{b^3\,x^2}-\frac {2\,\cos \left (a+\frac {b}{x}\right )}{b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/x)/x^4,x)

[Out]

(b^2*cos(a + b/x) - 2*b*x*sin(a + b/x))/(b^3*x^2) - (2*cos(a + b/x))/b^3

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