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]

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

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Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, 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 2637

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

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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 2.47, size = 25, normalized size = 0.71 \[ \frac {m \cos \relax (x) \sin \left (m x\right ) - \cos \left (m x\right ) \sin \relax (x)}{m^{2} - 1} \]

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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giac [A]  time = 0.14, size = 29, normalized size = 0.83 \[ \frac {\sin \left (m x + x\right )}{2 \, {\left (m + 1\right )}} + \frac {\sin \left (m x - x\right )}{2 \, {\left (m - 1\right )}} \]

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)

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maple [A]  time = 0.03, size = 28, normalized size = 0.80 \[ \frac {\sin \left (\left (-1+m \right ) x \right )}{-2+2 m}+\frac {\sin \left (\left (1+m \right ) x \right )}{2+2 m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 28, normalized size = 0.80 \[ \frac {\sin \left ({\left (m + 1\right )} x\right )}{2 \, {\left (m + 1\right )}} - \frac {\sin \left (-{\left (m - 1\right )} x\right )}{2 \, {\left (m - 1\right )}} \]

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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mupad [B]  time = 0.12, size = 39, normalized size = 1.11 \[ \left \{\begin {array}{cl} \frac {x}{2}+\frac {\sin \left (2\,x\right )}{4} & \text {\ if\ \ }m=-1\vee m=1\\ \frac {\sin \left (x\,\left (m-1\right )\right )}{2\,m-2}+\frac {\sin \left (x\,\left (m+1\right )\right )}{2\,m+2} & \text {\ if\ \ }m\neq -1\wedge m\neq 1 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(m == -1 | m == 1, x/2 + sin(2*x)/4, m ~= -1 & m ~= 1, sin(x*(m - 1))/(2*m - 2) + sin(x*(m + 1))/(2*m
+ 2))

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sympy [A]  time = 0.94, size = 56, normalized size = 1.60 \[ \begin {cases} \frac {x \sin ^{2}{\relax (x )}}{2} + \frac {x \cos ^{2}{\relax (x )}}{2} + \frac {\sin {\relax (x )} \cos {\relax (x )}}{2} & \text {for}\: m = -1 \vee m = 1 \\\frac {m \sin {\left (m x \right )} \cos {\relax (x )}}{m^{2} - 1} - \frac {\sin {\relax (x )} \cos {\left (m x \right )}}{m^{2} - 1} & \text {otherwise} \end {cases} \]

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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