3.2.22 \(\int \frac {\sin (a+\frac {b}{x^2})}{x^3} \, dx\) [122]

Optimal. Leaf size=15 \[ \frac {\cos \left (a+\frac {b}{x^2}\right )}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 15, 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 = {3460, 2718} \begin {gather*} \frac {\cos \left (a+\frac {b}{x^2}\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x^2]/x^3,x]

[Out]

Cos[a + b/x^2]/(2*b)

Rule 2718

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

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

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Mathematica [A]
time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} \frac {\cos \left (a+\frac {b}{x^2}\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x^2]/x^3,x]

[Out]

Cos[a + b/x^2]/(2*b)

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Maple [A]
time = 0.03, size = 14, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {\cos \left (a +\frac {b}{x^{2}}\right )}{2 b}\)	\(14\)
default	\(\frac {\cos \left (a +\frac {b}{x^{2}}\right )}{2 b}\)	\(14\)
risch	\(\frac {\cos \left (\frac {a \,x^{2}+b}{x^{2}}\right )}{2 b}\)	\(18\)
norman	\(\frac {1}{b \left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x^{2}}\right )\right )}\)	\(22\)
meijerg	\(-\frac {\sqrt {\pi }\, \cos \left (a \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (\frac {b}{x^{2}}\right )}{\sqrt {\pi }}\right )}{2 b}-\frac {\sin \left (a \right ) \sin \left (\frac {b}{x^{2}}\right )}{2 b}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/x^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 13, normalized size = 0.87 \begin {gather*} \frac {\cos \left (a + \frac {b}{x^{2}}\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x^2)/x^3,x, algorithm="maxima")

[Out]

1/2*cos(a + b/x^2)/b

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Fricas [A]
time = 0.37, size = 17, normalized size = 1.13 \begin {gather*} \frac {\cos \left (\frac {a x^{2} + b}{x^{2}}\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x^2)/x^3,x, algorithm="fricas")

[Out]

1/2*cos((a*x^2 + b)/x^2)/b

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Sympy [A]
time = 0.63, size = 20, normalized size = 1.33 \begin {gather*} \begin {cases} \frac {\cos {\left (a + \frac {b}{x^{2}} \right )}}{2 b} & \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**2)/x**3,x)

[Out]

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

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Giac [A]
time = 4.68, size = 17, normalized size = 1.13 \begin {gather*} \frac {\cos \left (\frac {a x^{2} + b}{x^{2}}\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/x^2)/x^3,x, algorithm="giac")

[Out]

1/2*cos((a*x^2 + b)/x^2)/b

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Mupad [B]
time = 4.67, size = 13, normalized size = 0.87 \begin {gather*} \frac {\cos \left (a+\frac {b}{x^2}\right )}{2\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/x^2)/x^3,x)

[Out]

cos(a + b/x^2)/(2*b)

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