3.104 \(\int \cos (x) \cos (m x) \, dx\)

Optimal. Leaf size=35 \[ \frac{\sin ((1-m) x)}{2 (1-m)}+\frac{\sin ((m+1) x)}{2 (m+1)} \]

[Out]

Sin[(1 - m)*x]/(2*(1 - m)) + Sin[(1 + m)*x]/(2*(1 + m))

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Rubi [A]  time = 0.0288736, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4570, 2637} \[ \frac{\sin ((1-m) x)}{2 (1-m)}+\frac{\sin ((m+1) x)}{2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]*Cos[m*x],x]

[Out]

Sin[(1 - m)*x]/(2*(1 - m)) + Sin[(1 + m)*x]/(2*(1 + m))

Rule 4570

Int[Cos[v_]^(p_.)*Cos[w_]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[Cos[v]^p*Cos[w]^q, x], x] /; ((PolynomialQ[
v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w], x])) && IGtQ[p, 0] && IGtQ[q
, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \cos (x) \cos (m x) \, dx &=\int \left (\frac{1}{2} \cos ((1-m) x)+\frac{1}{2} \cos ((1+m) x)\right ) \, dx\\ &=\frac{1}{2} \int \cos ((1-m) x) \, dx+\frac{1}{2} \int \cos ((1+m) x) \, dx\\ &=\frac{\sin ((1-m) x)}{2 (1-m)}+\frac{\sin ((1+m) x)}{2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0394965, size = 25, normalized size = 0.71 \[ \frac{m \cos (x) \sin (m x)-\sin (x) \cos (m x)}{m^2-1} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*Cos[m*x],x]

[Out]

(-(Cos[m*x]*Sin[x]) + m*Cos[x]*Sin[m*x])/(-1 + m^2)

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Maple [A]  time = 0.012, size = 28, normalized size = 0.8 \begin{align*}{\frac{\sin \left ( \left ( m-1 \right ) x \right ) }{2\,m-2}}+{\frac{\sin \left ( \left ( 1+m \right ) x \right ) }{2+2\,m}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*cos(m*x),x)

[Out]

1/2/(m-1)*sin((m-1)*x)+1/2*sin((1+m)*x)/(1+m)

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Maxima [A]  time = 1.01591, size = 38, normalized size = 1.09 \begin{align*} \frac{\sin \left ({\left (m + 1\right )} x\right )}{2 \,{\left (m + 1\right )}} - \frac{\sin \left (-{\left (m - 1\right )} x\right )}{2 \,{\left (m - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*cos(m*x),x, algorithm="maxima")

[Out]

1/2*sin((m + 1)*x)/(m + 1) - 1/2*sin(-(m - 1)*x)/(m - 1)

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Fricas [A]  time = 2.42915, size = 66, normalized size = 1.89 \begin{align*} \frac{m \cos \left (x\right ) \sin \left (m x\right ) - \cos \left (m x\right ) \sin \left (x\right )}{m^{2} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*cos(m*x),x, algorithm="fricas")

[Out]

(m*cos(x)*sin(m*x) - cos(m*x)*sin(x))/(m^2 - 1)

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Sympy [A]  time = 4.91142, size = 56, normalized size = 1.6 \begin{align*} \begin{cases} \frac{x \sin ^{2}{\left (x \right )}}{2} + \frac{x \cos ^{2}{\left (x \right )}}{2} + \frac{\sin{\left (x \right )} \cos{\left (x \right )}}{2} & \text{for}\: m = -1 \vee m = 1 \\\frac{m \sin{\left (m x \right )} \cos{\left (x \right )}}{m^{2} - 1} - \frac{\sin{\left (x \right )} \cos{\left (m x \right )}}{m^{2} - 1} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*cos(m*x),x)

[Out]

Piecewise((x*sin(x)**2/2 + x*cos(x)**2/2 + sin(x)*cos(x)/2, Eq(m, -1) | Eq(m, 1)), (m*sin(m*x)*cos(x)/(m**2 -
1) - sin(x)*cos(m*x)/(m**2 - 1), True))

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Giac [A]  time = 1.11229, size = 39, normalized size = 1.11 \begin{align*} \frac{\sin \left (m x + x\right )}{2 \,{\left (m + 1\right )}} + \frac{\sin \left (m x - x\right )}{2 \,{\left (m - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*cos(m*x),x, algorithm="giac")

[Out]

1/2*sin(m*x + x)/(m + 1) + 1/2*sin(m*x - x)/(m - 1)