3.1.39 \(\int \frac {\cos (a+\frac {b}{x})}{x^4} \, dx\) [39]

Optimal. Leaf size=46 \[ -\frac {2 \cos \left (a+\frac {b}{x}\right )}{b^2 x}+\frac {2 \sin \left (a+\frac {b}{x}\right )}{b^3}-\frac {\sin \left (a+\frac {b}{x}\right )}{b x^2} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 46, 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 = {3461, 3377, 2717} \begin {gather*} \frac {2 \sin \left (a+\frac {b}{x}\right )}{b^3}-\frac {2 \cos \left (a+\frac {b}{x}\right )}{b^2 x}-\frac {\sin \left (a+\frac {b}{x}\right )}{b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b/x]/x^4,x]

[Out]

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

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Mathematica [A]
time = 0.00, size = 46, normalized size = 1.00 \begin {gather*} -\frac {2 \cos \left (a+\frac {b}{x}\right )}{b^2 x}+\frac {2 \sin \left (a+\frac {b}{x}\right )}{b^3}-\frac {\sin \left (a+\frac {b}{x}\right )}{b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b/x]/x^4,x]

[Out]

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

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Maple [A]
time = 0.09, size = 92, normalized size = 2.00

method	result	size
risch	\(-\frac {2 \cos \left (\frac {a x +b}{x}\right )}{b^{2} x}-\frac {\left (b^{2}-2 x^{2}\right ) \sin \left (\frac {a x +b}{x}\right )}{b^{3} x^{2}}\)	\(47\)
norman	\(\frac {-\frac {2 x^{2}}{b^{2}}+\frac {4 x^{3} \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )}{b^{3}}+\frac {2 x^{2} \left (\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right )}{b^{2}}-\frac {2 x \tan \left (\frac {a}{2}+\frac {b}{2 x}\right )}{b}}{\left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x}\right )\right ) x^{3}}\)	\(88\)
derivativedivides	\(-\frac {a^{2} \sin \left (a +\frac {b}{x}\right )-2 a \left (\cos \left (a +\frac {b}{x}\right )+\left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )\right )+\left (a +\frac {b}{x}\right )^{2} \sin \left (a +\frac {b}{x}\right )-2 \sin \left (a +\frac {b}{x}\right )+2 \cos \left (a +\frac {b}{x}\right ) \left (a +\frac {b}{x}\right )}{b^{3}}\)	\(92\)
default	\(-\frac {a^{2} \sin \left (a +\frac {b}{x}\right )-2 a \left (\cos \left (a +\frac {b}{x}\right )+\left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )\right )+\left (a +\frac {b}{x}\right )^{2} \sin \left (a +\frac {b}{x}\right )-2 \sin \left (a +\frac {b}{x}\right )+2 \cos \left (a +\frac {b}{x}\right ) \left (a +\frac {b}{x}\right )}{b^{3}}\)	\(92\)
meijerg	\(-\frac {4 \sqrt {\pi }\, \cos \left (a \right ) \sqrt {b^{2}}\, \left (\frac {\left (b^{2}\right )^{\frac {3}{2}} \cos \left (\frac {b}{x}\right )}{2 \sqrt {\pi }\, x \,b^{2}}-\frac {\left (b^{2}\right )^{\frac {3}{2}} \left (-\frac {3 b^{2}}{2 x^{2}}+3\right ) \sin \left (\frac {b}{x}\right )}{6 \sqrt {\pi }\, b^{3}}\right )}{b^{4}}+\frac {4 \sqrt {\pi }\, \sin \left (a \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-\frac {b^{2}}{2 x^{2}}+1\right ) \cos \left (\frac {b}{x}\right )}{2 \sqrt {\pi }}+\frac {b \sin \left (\frac {b}{x}\right )}{2 \sqrt {\pi }\, x}\right )}{b^{3}}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.31, size = 50, normalized size = 1.09 \begin {gather*} \frac {{\left (i \, \Gamma \left (3, \frac {i \, b}{x}\right ) - i \, \Gamma \left (3, -\frac {i \, b}{x}\right )\right )} \cos \left (a\right ) + {\left (\Gamma \left (3, \frac {i \, b}{x}\right ) + \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(cos(a+b/x)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 43, normalized size = 0.93 \begin {gather*} -\frac {2 \, b x \cos \left (\frac {a x + b}{x}\right ) + {\left (b^{2} - 2 \, x^{2}\right )} \sin \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(cos(a+b/x)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.66, size = 46, normalized size = 1.00 \begin {gather*} \begin {cases} - \frac {\sin {\left (a + \frac {b}{x} \right )}}{b x^{2}} - \frac {2 \cos {\left (a + \frac {b}{x} \right )}}{b^{2} x} + \frac {2 \sin {\left (a + \frac {b}{x} \right )}}{b^{3}} & \text {for}\: b \neq 0 \\- \frac {\cos {\left (a \right )}}{3 x^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (46) = 92\).
time = 0.42, size = 107, normalized size = 2.33 \begin {gather*} -\frac {a^{2} \sin \left (\frac {a x + b}{x}\right ) - 2 \, a \cos \left (\frac {a x + b}{x}\right ) - \frac {2 \, {\left (a x + b\right )} a \sin \left (\frac {a x + b}{x}\right )}{x} + \frac {2 \, {\left (a x + b\right )} \cos \left (\frac {a x + b}{x}\right )}{x} + \frac {{\left (a x + b\right )}^{2} \sin \left (\frac {a x + b}{x}\right )}{x^{2}} - 2 \, \sin \left (\frac {a x + b}{x}\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.37, size = 47, normalized size = 1.02 \begin {gather*} \frac {2\,\sin \left (a+\frac {b}{x}\right )}{b^3}-\frac {b^2\,\sin \left (a+\frac {b}{x}\right )+2\,b\,x\,\cos \left (a+\frac {b}{x}\right )}{b^3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b/x)/x^4,x)

[Out]

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

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