3.12 \(\int \frac{\cos (x) \sin (x)}{x^3} \, dx\)

Optimal. Leaf size=29 \[ -\text{Si}(2 x)-\frac{\sin (2 x)}{4 x^2}-\frac{\cos (2 x)}{2 x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0589954, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4406, 12, 3297, 3299} \[ -\text{Si}(2 x)-\frac{\sin (2 x)}{4 x^2}-\frac{\cos (2 x)}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4406

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]

Rule 12

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

Rule 3297

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 3299

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]

Rubi steps

\begin{align*} \int \frac{\cos (x) \sin (x)}{x^3} \, dx &=\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]  time = 0.0075177, size = 29, normalized size = 1. \[ -\text{Si}(2 x)-\frac{\sin (2 x)}{4 x^2}-\frac{\cos (2 x)}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.029, size = 26, normalized size = 0.9 \begin{align*} -{\frac{\cos \left ( 2\,x \right ) }{2\,x}}-{\it Si} \left ( 2\,x \right ) -{\frac{\sin \left ( 2\,x \right ) }{4\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [C]  time = 1.26484, size = 20, normalized size = 0.69 \begin{align*} i \, \Gamma \left (-2, 2 i \, x\right ) - i \, \Gamma \left (-2, -2 i \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.469186, size = 96, normalized size = 3.31 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 1.59581, size = 24, normalized size = 0.83 \begin{align*} - \operatorname{Si}{\left (2 x \right )} - \frac{\cos{\left (2 x \right )}}{2 x} - \frac{\sin{\left (2 x \right )}}{4 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.13422, size = 35, normalized size = 1.21 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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