3.2.10 \(\int \frac {\sin (a+\frac {b}{x})}{x^5} \, dx\) [110]

Optimal. Leaf size=61 \[ \frac {\cos \left (a+\frac {b}{x}\right )}{b x^3}-\frac {6 \cos \left (a+\frac {b}{x}\right )}{b^3 x}+\frac {6 \sin \left (a+\frac {b}{x}\right )}{b^4}-\frac {3 \sin \left (a+\frac {b}{x}\right )}{b^2 x^2} \]

[Out]

cos(a+b/x)/b/x^3-6*cos(a+b/x)/b^3/x+6*sin(a+b/x)/b^4-3*sin(a+b/x)/b^2/x^2

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Rubi [A]
time = 0.05, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {6 \sin \left (a+\frac {b}{x}\right )}{b^4}-\frac {6 \cos \left (a+\frac {b}{x}\right )}{b^3 x}-\frac {3 \sin \left (a+\frac {b}{x}\right )}{b^2 x^2}+\frac {\cos \left (a+\frac {b}{x}\right )}{b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x]/x^5,x]

[Out]

Cos[a + b/x]/(b*x^3) - (6*Cos[a + b/x])/(b^3*x) + (6*Sin[a + b/x])/b^4 - (3*Sin[a + b/x])/(b^2*x^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^5} \, dx &=-\text {Subst}\left (\int x^3 \sin (a+b x) \, dx,x,\frac {1}{x}\right )\\ &=\frac {\cos \left (a+\frac {b}{x}\right )}{b x^3}-\frac {3 \text {Subst}\left (\int x^2 \cos (a+b x) \, dx,x,\frac {1}{x}\right )}{b}\\ &=\frac {\cos \left (a+\frac {b}{x}\right )}{b x^3}-\frac {3 \sin \left (a+\frac {b}{x}\right )}{b^2 x^2}+\frac {6 \text {Subst}\left (\int x \sin (a+b x) \, dx,x,\frac {1}{x}\right )}{b^2}\\ &=\frac {\cos \left (a+\frac {b}{x}\right )}{b x^3}-\frac {6 \cos \left (a+\frac {b}{x}\right )}{b^3 x}-\frac {3 \sin \left (a+\frac {b}{x}\right )}{b^2 x^2}+\frac {6 \text {Subst}\left (\int \cos (a+b x) \, dx,x,\frac {1}{x}\right )}{b^3}\\ &=\frac {\cos \left (a+\frac {b}{x}\right )}{b x^3}-\frac {6 \cos \left (a+\frac {b}{x}\right )}{b^3 x}+\frac {6 \sin \left (a+\frac {b}{x}\right )}{b^4}-\frac {3 \sin \left (a+\frac {b}{x}\right )}{b^2 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 61, normalized size = 1.00 \begin {gather*} \frac {\cos \left (a+\frac {b}{x}\right )}{b x^3}-\frac {6 \cos \left (a+\frac {b}{x}\right )}{b^3 x}+\frac {6 \sin \left (a+\frac {b}{x}\right )}{b^4}-\frac {3 \sin \left (a+\frac {b}{x}\right )}{b^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x]/x^5,x]

[Out]

Cos[a + b/x]/(b*x^3) - (6*Cos[a + b/x])/(b^3*x) + (6*Sin[a + b/x])/b^4 - (3*Sin[a + b/x])/(b^2*x^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(164\) vs. \(2(61)=122\).
time = 0.06, size = 165, normalized size = 2.70

method	result	size
risch	\(\frac {\left (b^{2}-6 x^{2}\right ) \cos \left (\frac {a x +b}{x}\right )}{b^{3} x^{3}}-\frac {3 \left (b^{2}-2 x^{2}\right ) \sin \left (\frac {a x +b}{x}\right )}{x^{2} b^{4}}\)	\(55\)
norman	\(\frac {\frac {x}{b}-\frac {6 x^{3}}{b^{3}}+\frac {12 x^{4} \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )}{b^{4}}+\frac {6 x^{3} \left (\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right )}{b^{3}}-\frac {6 x^{2} \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )}{b^{2}}-\frac {x \left (\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right )}{b}}{\left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right ) x^{4}}\)	\(114\)
meijerg	\(-\frac {8 \sqrt {\pi }\, \cos \left (a \right ) \left (\frac {b \left (-\frac {5 b^{2}}{2 x^{2}}+15\right ) \cos \left (\frac {b}{x}\right )}{20 \sqrt {\pi }\, x}-\frac {\left (-\frac {15 b^{2}}{2 x^{2}}+15\right ) \sin \left (\frac {b}{x}\right )}{20 \sqrt {\pi }}\right )}{b^{4}}-\frac {8 \sqrt {\pi }\, \sin \left (a \right ) \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (-\frac {3 b^{2}}{2 x^{2}}+3\right ) \cos \left (\frac {b}{x}\right )}{4 \sqrt {\pi }}-\frac {b \left (-\frac {b^{2}}{2 x^{2}}+3\right ) \sin \left (\frac {b}{x}\right )}{4 \sqrt {\pi }\, x}\right )}{b^{4}}\)	\(121\)
derivativedivides	\(-\frac {a^{3} \cos \left (a +\frac {b}{x}\right )+3 a^{2} \left (\sin \left (a +\frac {b}{x}\right )-\left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )\right )-3 a \left (-\left (a +\frac {b}{x}\right )^{2} \cos \left (a +\frac {b}{x}\right )+2 \cos \left (a +\frac {b}{x}\right )+2 \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )\right )-\left (a +\frac {b}{x}\right )^{3} \cos \left (a +\frac {b}{x}\right )+3 \left (a +\frac {b}{x}\right )^{2} \sin \left (a +\frac {b}{x}\right )-6 \sin \left (a +\frac {b}{x}\right )+6 \left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )}{b^{4}}\)	\(165\)
default	\(-\frac {a^{3} \cos \left (a +\frac {b}{x}\right )+3 a^{2} \left (\sin \left (a +\frac {b}{x}\right )-\left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )\right )-3 a \left (-\left (a +\frac {b}{x}\right )^{2} \cos \left (a +\frac {b}{x}\right )+2 \cos \left (a +\frac {b}{x}\right )+2 \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )\right )-\left (a +\frac {b}{x}\right )^{3} \cos \left (a +\frac {b}{x}\right )+3 \left (a +\frac {b}{x}\right )^{2} \sin \left (a +\frac {b}{x}\right )-6 \sin \left (a +\frac {b}{x}\right )+6 \left (a +\frac {b}{x}\right ) \cos \left (a +\frac {b}{x}\right )}{b^{4}}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/x)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/b^4*(a^3*cos(a+b/x)+3*a^2*(sin(a+b/x)-(a+b/x)*cos(a+b/x))-3*a*(-(a+b/x)^2*cos(a+b/x)+2*cos(a+b/x)+2*(a+b/x)
*sin(a+b/x))-(a+b/x)^3*cos(a+b/x)+3*(a+b/x)^2*sin(a+b/x)-6*sin(a+b/x)+6*(a+b/x)*cos(a+b/x))

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.35, size = 50, normalized size = 0.82 \begin {gather*} \frac {{\left (i \, \Gamma \left (4, \frac {i \, b}{x}\right ) - i \, \Gamma \left (4, -\frac {i \, b}{x}\right )\right )} \cos \left (a\right ) + {\left (\Gamma \left (4, \frac {i \, b}{x}\right ) + \Gamma \left (4, -\frac {i \, b}{x}\right )\right )} \sin \left (a\right )}{2 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x^5,x, algorithm="maxima")

[Out]

1/2*((I*gamma(4, I*b/x) - I*gamma(4, -I*b/x))*cos(a) + (gamma(4, I*b/x) + gamma(4, -I*b/x))*sin(a))/b^4

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Fricas [A]
time = 0.36, size = 52, normalized size = 0.85 \begin {gather*} \frac {{\left (b^{3} - 6 \, b x^{2}\right )} \cos \left (\frac {a x + b}{x}\right ) - 3 \, {\left (b^{2} x - 2 \, x^{3}\right )} \sin \left (\frac {a x + b}{x}\right )}{b^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x^5,x, algorithm="fricas")

[Out]

((b^3 - 6*b*x^2)*cos((a*x + b)/x) - 3*(b^2*x - 2*x^3)*sin((a*x + b)/x))/(b^4*x^3)

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Sympy [A]
time = 1.04, size = 61, normalized size = 1.00 \begin {gather*} \begin {cases} \frac {\cos {\left (a + \frac {b}{x} \right )}}{b x^{3}} - \frac {3 \sin {\left (a + \frac {b}{x} \right )}}{b^{2} x^{2}} - \frac {6 \cos {\left (a + \frac {b}{x} \right )}}{b^{3} x} + \frac {6 \sin {\left (a + \frac {b}{x} \right )}}{b^{4}} & \text {for}\: b \neq 0 \\- \frac {\sin {\left (a \right )}}{4 x^{4}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x**5,x)

[Out]

Piecewise((cos(a + b/x)/(b*x**3) - 3*sin(a + b/x)/(b**2*x**2) - 6*cos(a + b/x)/(b**3*x) + 6*sin(a + b/x)/b**4,
Ne(b, 0)), (-sin(a)/(4*x**4), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (61) = 122\).
time = 4.06, size = 191, normalized size = 3.13 \begin {gather*} -\frac {a^{3} \cos \left (\frac {a x + b}{x}\right ) - \frac {3 \, {\left (a x + b\right )} a^{2} \cos \left (\frac {a x + b}{x}\right )}{x} + 3 \, a^{2} \sin \left (\frac {a x + b}{x}\right ) - 6 \, a \cos \left (\frac {a x + b}{x}\right ) + \frac {3 \, {\left (a x + b\right )}^{2} a \cos \left (\frac {a x + b}{x}\right )}{x^{2}} - \frac {6 \, {\left (a x + b\right )} a \sin \left (\frac {a x + b}{x}\right )}{x} - \frac {{\left (a x + b\right )}^{3} \cos \left (\frac {a x + b}{x}\right )}{x^{3}} + \frac {6 \, {\left (a x + b\right )} \cos \left (\frac {a x + b}{x}\right )}{x} + \frac {3 \, {\left (a x + b\right )}^{2} \sin \left (\frac {a x + b}{x}\right )}{x^{2}} - 6 \, \sin \left (\frac {a x + b}{x}\right )}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x)/x^5,x, algorithm="giac")

[Out]

-(a^3*cos((a*x + b)/x) - 3*(a*x + b)*a^2*cos((a*x + b)/x)/x + 3*a^2*sin((a*x + b)/x) - 6*a*cos((a*x + b)/x) +
3*(a*x + b)^2*a*cos((a*x + b)/x)/x^2 - 6*(a*x + b)*a*sin((a*x + b)/x)/x - (a*x + b)^3*cos((a*x + b)/x)/x^3 + 6
*(a*x + b)*cos((a*x + b)/x)/x + 3*(a*x + b)^2*sin((a*x + b)/x)/x^2 - 6*sin((a*x + b)/x))/b^4

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Mupad [B]
time = 4.77, size = 64, normalized size = 1.05 \begin {gather*} \frac {6\,\sin \left (a+\frac {b}{x}\right )}{b^4}-\frac {6\,b\,x^2\,\cos \left (a+\frac {b}{x}\right )-b^3\,\cos \left (a+\frac {b}{x}\right )+3\,b^2\,x\,\sin \left (a+\frac {b}{x}\right )}{b^4\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/x)/x^5,x)

[Out]

(6*sin(a + b/x))/b^4 - (6*b*x^2*cos(a + b/x) - b^3*cos(a + b/x) + 3*b^2*x*sin(a + b/x))/(b^4*x^3)

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