\(\int \cos (x) \sin (2 x) \, dx\) [97]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 7, antiderivative size = 15 \[ \int \cos (x) \sin (2 x) \, dx=-\frac {\cos (x)}{2}-\frac {1}{6} \cos (3 x) \]

[Out]

-1/2*cos(x)-1/6*cos(3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4369} \[ \int \cos (x) \sin (2 x) \, dx=-\frac {\cos (x)}{2}-\frac {1}{6} \cos (3 x) \]

[In]

Int[Cos[x]*Sin[2*x],x]

[Out]

-1/2*Cos[x] - Cos[3*x]/6

Rule 4369

Int[cos[(c_.) + (d_.)*(x_)]*sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[-Cos[a - c + (b - d)*x]/(2*(b - d)), x]
 - Simp[Cos[a + c + (b + d)*x]/(2*(b + d)), x] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - d^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (x)}{2}-\frac {1}{6} \cos (3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \cos (x) \sin (2 x) \, dx=-\frac {\cos (x)}{2}-\frac {1}{6} \cos (3 x) \]

[In]

Integrate[Cos[x]*Sin[2*x],x]

[Out]

-1/2*Cos[x] - Cos[3*x]/6

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80

method result size
default \(-\frac {\cos \left (x \right )}{2}-\frac {\cos \left (3 x \right )}{6}\) \(12\)
risch \(-\frac {\cos \left (x \right )}{2}-\frac {\cos \left (3 x \right )}{6}\) \(12\)
parallelrisch \(\frac {2}{3}-\frac {\cos \left (x \right )}{2}-\frac {\cos \left (3 x \right )}{6}\) \(13\)
norman \(\frac {\frac {4 \tan \left (x \right )^{2}}{3}+\frac {4 \tan \left (\frac {x}{2}\right )^{2}}{3}-\frac {4 \tan \left (\frac {x}{2}\right ) \tan \left (x \right )}{3}}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right ) \left (1+\tan \left (x \right )^{2}\right )}\) \(43\)

[In]

int(cos(x)*sin(2*x),x,method=_RETURNVERBOSE)

[Out]

-1/2*cos(x)-1/6*cos(3*x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.40 \[ \int \cos (x) \sin (2 x) \, dx=-\frac {2}{3} \, \cos \left (x\right )^{3} \]

[In]

integrate(cos(x)*sin(2*x),x, algorithm="fricas")

[Out]

-2/3*cos(x)^3

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \cos (x) \sin (2 x) \, dx=- \frac {\sin {\left (x \right )} \sin {\left (2 x \right )}}{3} - \frac {2 \cos {\left (x \right )} \cos {\left (2 x \right )}}{3} \]

[In]

integrate(cos(x)*sin(2*x),x)

[Out]

-sin(x)*sin(2*x)/3 - 2*cos(x)*cos(2*x)/3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \cos (x) \sin (2 x) \, dx=-\frac {1}{6} \, \cos \left (3 \, x\right ) - \frac {1}{2} \, \cos \left (x\right ) \]

[In]

integrate(cos(x)*sin(2*x),x, algorithm="maxima")

[Out]

-1/6*cos(3*x) - 1/2*cos(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.40 \[ \int \cos (x) \sin (2 x) \, dx=-\frac {2}{3} \, \cos \left (x\right )^{3} \]

[In]

integrate(cos(x)*sin(2*x),x, algorithm="giac")

[Out]

-2/3*cos(x)^3

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.40 \[ \int \cos (x) \sin (2 x) \, dx=-\frac {2\,{\cos \left (x\right )}^3}{3} \]

[In]

int(sin(2*x)*cos(x),x)

[Out]

-(2*cos(x)^3)/3