\(\int \frac {\cos (x) \sin (x)}{x^3} \, dx\) [12]

Optimal result

Integrand size = 8, antiderivative size = 29 \[ \int \frac {\cos (x) \sin (x)}{x^3} \, dx=-\frac {\cos (2 x)}{2 x}-\frac {\sin (2 x)}{4 x^2}-\text {Si}(2 x) \]

[Out]

-1/2*cos(2*x)/x-Si(2*x)-1/4*sin(2*x)/x^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4491, 12, 3378, 3380} \[ \int \frac {\cos (x) \sin (x)}{x^3} \, dx=-\text {Si}(2 x)-\frac {\sin (2 x)}{4 x^2}-\frac {\cos (2 x)}{2 x} \]

[In]

Int[(Cos[x]*Sin[x])/x^3,x]

[Out]

-1/2*Cos[2*x]/x - Sin[2*x]/(4*x^2) - SinIntegral[2*x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3378

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m
+ 1))), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sin (2 x)}{2 x^3} \, dx \\ & = \frac {1}{2} \int \frac {\sin (2 x)}{x^3} \, dx \\ & = -\frac {\sin (2 x)}{4 x^2}+\frac {1}{2} \int \frac {\cos (2 x)}{x^2} \, dx \\ & = -\frac {\cos (2 x)}{2 x}-\frac {\sin (2 x)}{4 x^2}-\int \frac {\sin (2 x)}{x} \, dx \\ & = -\frac {\cos (2 x)}{2 x}-\frac {\sin (2 x)}{4 x^2}-\text {Si}(2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (x) \sin (x)}{x^3} \, dx=-\frac {\cos (2 x)}{2 x}-\frac {\sin (2 x)}{4 x^2}-\text {Si}(2 x) \]

[In]

Integrate[(Cos[x]*Sin[x])/x^3,x]

[Out]

-1/2*Cos[2*x]/x - Sin[2*x]/(4*x^2) - SinIntegral[2*x]

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {\cos \left (2 x \right )}{2 x}-\operatorname {Si}\left (2 x \right )-\frac {\sin \left (2 x \right )}{4 x^{2}}\)	\(26\)
risch	\(\frac {\pi \,\operatorname {csgn}\left (x \right )}{2}-\operatorname {Si}\left (2 x \right )-\frac {\cos \left (2 x \right )}{2 x}-\frac {\sin \left (2 x \right )}{4 x^{2}}\)	\(31\)
meijerg	\(\frac {\sqrt {\pi }\, \left (-\frac {2 \cos \left (2 x \right )}{x \sqrt {\pi }}-\frac {\sin \left (2 x \right )}{x^{2} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (2 x \right )}{\sqrt {\pi }}\right )}{4}\)	\(40\)

[In]

int(cos(x)*sin(x)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/2*cos(2*x)/x-Si(2*x)-1/4*sin(2*x)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (x) \sin (x)}{x^3} \, dx=-\frac {2 \, x \cos \left (x\right )^{2} + 2 \, x^{2} \operatorname {Si}\left (2 \, x\right ) + \cos \left (x\right ) \sin \left (x\right ) - x}{2 \, x^{2}} \]

[In]

integrate(cos(x)*sin(x)/x^3,x, algorithm="fricas")

[Out]

-1/2*(2*x*cos(x)^2 + 2*x^2*sin_integral(2*x) + cos(x)*sin(x) - x)/x^2

Sympy [A] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {\cos (x) \sin (x)}{x^3} \, dx=- \operatorname {Si}{\left (2 x \right )} - \frac {\cos {\left (2 x \right )}}{2 x} - \frac {\sin {\left (2 x \right )}}{4 x^{2}} \]

[In]

integrate(cos(x)*sin(x)/x**3,x)

[Out]

-Si(2*x) - cos(2*x)/(2*x) - sin(2*x)/(4*x**2)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.52 \[ \int \frac {\cos (x) \sin (x)}{x^3} \, dx=i \, \Gamma \left (-2, 2 i \, x\right ) - i \, \Gamma \left (-2, -2 i \, x\right ) \]

[In]

integrate(cos(x)*sin(x)/x^3,x, algorithm="maxima")

[Out]

I*gamma(-2, 2*I*x) - I*gamma(-2, -2*I*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (x) \sin (x)}{x^3} \, dx=-\frac {4 \, x^{2} \operatorname {Si}\left (2 \, x\right ) + 2 \, x \cos \left (2 \, x\right ) + \sin \left (2 \, x\right )}{4 \, x^{2}} \]

[In]

integrate(cos(x)*sin(x)/x^3,x, algorithm="giac")

[Out]

-1/4*(4*x^2*sin_integral(2*x) + 2*x*cos(2*x) + sin(2*x))/x^2

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (x) \sin (x)}{x^3} \, dx=\int \frac {\cos \left (x\right )\,\sin \left (x\right )}{x^3} \,d x \]

[In]

int((cos(x)*sin(x))/x^3,x)

[Out]

int((cos(x)*sin(x))/x^3, x)