\(\int \cos (3 x) \cos (5 x) \, dx\) [283]

Optimal result

Integrand size = 9, antiderivative size = 17 \[ \int \cos (3 x) \cos (5 x) \, dx=\frac {1}{4} \sin (2 x)+\frac {1}{16} \sin (8 x) \]

[Out]

1/4*sin(2*x)+1/16*sin(8*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 = {4368} \[ \int \cos (3 x) \cos (5 x) \, dx=\frac {1}{4} \sin (2 x)+\frac {1}{16} \sin (8 x) \]

[In]

Int[Cos[3*x]*Cos[5*x],x]

[Out]

Sin[2*x]/4 + Sin[8*x]/16

Rule 4368

Int[cos[(a_.) + (b_.)*(x_)]*cos[(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}{4} \sin (2 x)+\frac {1}{16} \sin (8 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \cos (3 x) \cos (5 x) \, dx=\frac {1}{4} \sin (2 x)+\frac {1}{16} \sin (8 x) \]

[In]

Integrate[Cos[3*x]*Cos[5*x],x]

[Out]

Sin[2*x]/4 + Sin[8*x]/16

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\sin \left (2 x \right )}{4}+\frac {\sin \left (8 x \right )}{16}\)	\(14\)
risch	\(\frac {\sin \left (2 x \right )}{4}+\frac {\sin \left (8 x \right )}{16}\)	\(14\)
parallelrisch	\(\frac {\sin \left (2 x \right )}{4}+\frac {\sin \left (8 x \right )}{16}\)	\(14\)
norman	\(\frac {\frac {3 \tan \left (\frac {3 x}{2}\right ) \left (\tan ^{2}\left (\frac {5 x}{2}\right )\right )}{8}-\frac {5 \left (\tan ^{2}\left (\frac {3 x}{2}\right )\right ) \tan \left (\frac {5 x}{2}\right )}{8}-\frac {3 \tan \left (\frac {3 x}{2}\right )}{8}+\frac {5 \tan \left (\frac {5 x}{2}\right )}{8}}{\left (1+\tan ^{2}\left (\frac {3 x}{2}\right )\right ) \left (1+\tan ^{2}\left (\frac {5 x}{2}\right )\right )}\)	\(59\)

[In]

int(cos(3*x)*cos(5*x),x,method=_RETURNVERBOSE)

[Out]

1/4*sin(2*x)+1/16*sin(8*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \cos (3 x) \cos (5 x) \, dx={\left (8 \, \cos \left (x\right )^{7} - 12 \, \cos \left (x\right )^{5} + 5 \, \cos \left (x\right )^{3}\right )} \sin \left (x\right ) \]

[In]

integrate(cos(3*x)*cos(5*x),x, algorithm="fricas")

[Out]

(8*cos(x)^7 - 12*cos(x)^5 + 5*cos(x)^3)*sin(x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \cos (3 x) \cos (5 x) \, dx=- \frac {3 \sin {\left (3 x \right )} \cos {\left (5 x \right )}}{16} + \frac {5 \sin {\left (5 x \right )} \cos {\left (3 x \right )}}{16} \]

[In]

integrate(cos(3*x)*cos(5*x),x)

[Out]

-3*sin(3*x)*cos(5*x)/16 + 5*sin(5*x)*cos(3*x)/16

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \cos (3 x) \cos (5 x) \, dx=\frac {1}{16} \, \sin \left (8 \, x\right ) + \frac {1}{4} \, \sin \left (2 \, x\right ) \]

[In]

integrate(cos(3*x)*cos(5*x),x, algorithm="maxima")

[Out]

1/16*sin(8*x) + 1/4*sin(2*x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \cos (3 x) \cos (5 x) \, dx=\frac {1}{16} \, \sin \left (8 \, x\right ) + \frac {1}{4} \, \sin \left (2 \, x\right ) \]

[In]

integrate(cos(3*x)*cos(5*x),x, algorithm="giac")

[Out]

1/16*sin(8*x) + 1/4*sin(2*x)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \cos (3 x) \cos (5 x) \, dx=\frac {\sin \left (2\,x\right )}{4}+\frac {\sin \left (8\,x\right )}{16} \]

[In]

int(cos(3*x)*cos(5*x),x)

[Out]

sin(2*x)/4 + sin(8*x)/16