\(\int \frac {\cos (a+\frac {b}{x^2})}{x^3} \, dx\) [43]

Optimal result

Integrand size = 12, antiderivative size = 15 \[ \int \frac {\cos \left (a+\frac {b}{x^2}\right )}{x^3} \, dx=-\frac {\sin \left (a+\frac {b}{x^2}\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3461, 2717} \[ \int \frac {\cos \left (a+\frac {b}{x^2}\right )}{x^3} \, dx=-\frac {\sin \left (a+\frac {b}{x^2}\right )}{2 b} \]

[In]

Int[Cos[a + b/x^2]/x^3,x]

[Out]

-1/2*Sin[a + b/x^2]/b

Rule 2717

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {\cos \left (a+\frac {b}{x^2}\right )}{x^3} \, dx=-\frac {\sin \left (a+\frac {b}{x^2}\right )}{2 b} \]

[In]

Integrate[Cos[a + b/x^2]/x^3,x]

[Out]

-1/2*Sin[a + b/x^2]/b

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(-\frac {\sin \left (a +\frac {b}{x^{2}}\right )}{2 b}\)	\(14\)
default	\(-\frac {\sin \left (a +\frac {b}{x^{2}}\right )}{2 b}\)	\(14\)
risch	\(-\frac {\sin \left (\frac {a \,x^{2}+b}{x^{2}}\right )}{2 b}\)	\(18\)
parallelrisch	\(-\frac {\sin \left (\frac {a \,x^{2}+b}{x^{2}}\right )}{2 b}\)	\(18\)
norman	\(-\frac {\tan \left (\frac {a}{2}+\frac {b}{2 x^{2}}\right )}{b \left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b}{2 x^{2}}\right )\right )}\)	\(34\)
meijerg	\(-\frac {\cos \left (a \right ) \sin \left (\frac {b}{x^{2}}\right )}{2 b}+\frac {\sqrt {\pi }\, \sin \left (a \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (\frac {b}{x^{2}}\right )}{\sqrt {\pi }}\right )}{2 b}\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {\cos \left (a+\frac {b}{x^2}\right )}{x^3} \, dx=-\frac {\sin \left (\frac {a x^{2} + b}{x^{2}}\right )}{2 \, b} \]

[In]

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

[Out]

-1/2*sin((a*x^2 + b)/x^2)/b

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {\cos \left (a+\frac {b}{x^2}\right )}{x^3} \, dx=\begin {cases} - \frac {\sin {\left (a + \frac {b}{x^{2}} \right )}}{2 b} & \text {for}\: b \neq 0 \\- \frac {\cos {\left (a \right )}}{2 x^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(a+b/x**2)/x**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {\cos \left (a+\frac {b}{x^2}\right )}{x^3} \, dx=-\frac {\sin \left (a + \frac {b}{x^{2}}\right )}{2 \, b} \]

[In]

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

[Out]

-1/2*sin(a + b/x^2)/b

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {\cos \left (a+\frac {b}{x^2}\right )}{x^3} \, dx=-\frac {\sin \left (\frac {a x^{2} + b}{x^{2}}\right )}{2 \, b} \]

[In]

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

[Out]

-1/2*sin((a*x^2 + b)/x^2)/b

Mupad [B] (verification not implemented)

Time = 13.98 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {\cos \left (a+\frac {b}{x^2}\right )}{x^3} \, dx=-\frac {\sin \left (a+\frac {b}{x^2}\right )}{2\,b} \]

[In]

int(cos(a + b/x^2)/x^3,x)

[Out]

-sin(a + b/x^2)/(2*b)