\(\int \sin (3 x) \sin (6 x) \, dx\) [109]

Optimal result

Integrand size = 9, antiderivative size = 17 \[ \int \sin (3 x) \sin (6 x) \, dx=\frac {1}{6} \sin (3 x)-\frac {1}{18} \sin (9 x) \]

[Out]

1/6*sin(3*x)-1/18*sin(9*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4367} \[ \int \sin (3 x) \sin (6 x) \, dx=\frac {1}{6} \sin (3 x)-\frac {1}{18} \sin (9 x) \]

[In]

Int[Sin[3*x]*Sin[6*x],x]

[Out]

Sin[3*x]/6 - Sin[9*x]/18

Rule 4367

Int[sin[(a_.) + (b_.)*(x_)]*sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[a - c + (b - d)*x]/(2*(b - d)), x]
- Simp[Sin[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 {1}{6} \sin (3 x)-\frac {1}{18} \sin (9 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \sin (3 x) \sin (6 x) \, dx=\frac {1}{6} \sin (3 x)-\frac {1}{18} \sin (9 x) \]

[In]

Integrate[Sin[3*x]*Sin[6*x],x]

[Out]

Sin[3*x]/6 - Sin[9*x]/18

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.53

method	result	size
derivativedivides	\(\frac {2 \left (\sin ^{3}\left (3 x \right )\right )}{9}\)	\(9\)
default	\(\frac {2 \left (\sin ^{3}\left (3 x \right )\right )}{9}\)	\(9\)
risch	\(\frac {\sin \left (3 x \right )}{6}-\frac {\sin \left (9 x \right )}{18}\)	\(14\)
parallelrisch	\(\frac {\sin \left (3 x \right )}{6}-\frac {\sin \left (9 x \right )}{18}\)	\(14\)
norman	\(\frac {-\frac {2 \tan \left (3 x \right ) \left (\tan ^{2}\left (\frac {3 x}{2}\right )\right )}{9}+\frac {4 \left (\tan ^{2}\left (3 x \right )\right ) \tan \left (\frac {3 x}{2}\right )}{9}+\frac {2 \tan \left (3 x \right )}{9}-\frac {4 \tan \left (\frac {3 x}{2}\right )}{9}}{\left (1+\tan ^{2}\left (\frac {3 x}{2}\right )\right ) \left (1+\tan ^{2}\left (3 x \right )\right )}\)	\(59\)

[In]

int(sin(3*x)*sin(6*x),x,method=_RETURNVERBOSE)

[Out]

2/9*sin(3*x)^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \sin (3 x) \sin (6 x) \, dx=-\frac {2}{9} \, {\left (\cos \left (3 \, x\right )^{2} - 1\right )} \sin \left (3 \, x\right ) \]

[In]

integrate(sin(3*x)*sin(6*x),x, algorithm="fricas")

[Out]

-2/9*(cos(3*x)^2 - 1)*sin(3*x)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int \sin (3 x) \sin (6 x) \, dx=- \frac {2 \sin {\left (3 x \right )} \cos {\left (6 x \right )}}{9} + \frac {\sin {\left (6 x \right )} \cos {\left (3 x \right )}}{9} \]

[In]

integrate(sin(3*x)*sin(6*x),x)

[Out]

-2*sin(3*x)*cos(6*x)/9 + sin(6*x)*cos(3*x)/9

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \sin (3 x) \sin (6 x) \, dx=-\frac {1}{18} \, \sin \left (9 \, x\right ) + \frac {1}{6} \, \sin \left (3 \, x\right ) \]

[In]

integrate(sin(3*x)*sin(6*x),x, algorithm="maxima")

[Out]

-1/18*sin(9*x) + 1/6*sin(3*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \sin (3 x) \sin (6 x) \, dx=\frac {2}{9} \, \sin \left (3 \, x\right )^{3} \]

[In]

integrate(sin(3*x)*sin(6*x),x, algorithm="giac")

[Out]

2/9*sin(3*x)^3

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \sin (3 x) \sin (6 x) \, dx=\frac {\sin \left (3\,x\right )}{6}-\frac {\sin \left (9\,x\right )}{18} \]

[In]

int(sin(3*x)*sin(6*x),x)

[Out]

sin(3*x)/6 - sin(9*x)/18