3.240 \(\int \cos (a+b x) \cos (c+d x) \, dx\)

Optimal. Leaf size=43 \[ \frac{\sin (a+x (b-d)-c)}{2 (b-d)}+\frac{\sin (a+x (b+d)+c)}{2 (b+d)} \]

[Out]

Sin[a - c + (b - d)*x]/(2*(b - d)) + Sin[a + c + (b + d)*x]/(2*(b + d))

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Rubi [A]  time = 0.0334101, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4570, 2637} \[ \frac{\sin (a+x (b-d)-c)}{2 (b-d)}+\frac{\sin (a+x (b+d)+c)}{2 (b+d)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Cos[c + d*x],x]

[Out]

Sin[a - c + (b - d)*x]/(2*(b - d)) + Sin[a + c + (b + d)*x]/(2*(b + d))

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 (a+b x) \cos (c+d x) \, dx &=\int \left (\frac{1}{2} \cos (a-c+(b-d) x)+\frac{1}{2} \cos (a+c+(b+d) x)\right ) \, dx\\ &=\frac{1}{2} \int \cos (a-c+(b-d) x) \, dx+\frac{1}{2} \int \cos (a+c+(b+d) x) \, dx\\ &=\frac{\sin (a-c+(b-d) x)}{2 (b-d)}+\frac{\sin (a+c+(b+d) x)}{2 (b+d)}\\ \end{align*}

Mathematica [A]  time = 0.180851, size = 43, normalized size = 1. \[ \frac{\sin (a+x (b-d)-c)}{2 (b-d)}+\frac{\sin (a+x (b+d)+c)}{2 (b+d)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Cos[c + d*x],x]

[Out]

Sin[a - c + (b - d)*x]/(2*(b - d)) + Sin[a + c + (b + d)*x]/(2*(b + d))

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Maple [A]  time = 0.019, size = 40, normalized size = 0.9 \begin{align*}{\frac{\sin \left ( a-c+ \left ( b-d \right ) x \right ) }{2\,b-2\,d}}+{\frac{\sin \left ( a+c+ \left ( b+d \right ) x \right ) }{2\,b+2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*cos(d*x+c),x)

[Out]

1/2*sin(a-c+(b-d)*x)/(b-d)+1/2*sin(a+c+(b+d)*x)/(b+d)

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Maxima [A]  time = 1.07282, size = 54, normalized size = 1.26 \begin{align*} \frac{\sin \left (b x + d x + a + c\right )}{2 \,{\left (b + d\right )}} - \frac{\sin \left (-b x + d x - a + c\right )}{2 \,{\left (b - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*cos(d*x+c),x, algorithm="maxima")

[Out]

1/2*sin(b*x + d*x + a + c)/(b + d) - 1/2*sin(-b*x + d*x - a + c)/(b - d)

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Fricas [A]  time = 0.473431, size = 99, normalized size = 2.3 \begin{align*} \frac{b \cos \left (d x + c\right ) \sin \left (b x + a\right ) - d \cos \left (b x + a\right ) \sin \left (d x + c\right )}{b^{2} - d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*cos(d*x+c),x, algorithm="fricas")

[Out]

(b*cos(d*x + c)*sin(b*x + a) - d*cos(b*x + a)*sin(d*x + c))/(b^2 - d^2)

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Sympy [A]  time = 2.41227, size = 153, normalized size = 3.56 \begin{align*} \begin{cases} x \cos{\left (a \right )} \cos{\left (c \right )} & \text{for}\: b = 0 \wedge d = 0 \\- \frac{x \sin{\left (a - d x \right )} \sin{\left (c + d x \right )}}{2} + \frac{x \cos{\left (a - d x \right )} \cos{\left (c + d x \right )}}{2} - \frac{\sin{\left (a - d x \right )} \cos{\left (c + d x \right )}}{2 d} & \text{for}\: b = - d \\\frac{x \sin{\left (a + d x \right )} \sin{\left (c + d x \right )}}{2} + \frac{x \cos{\left (a + d x \right )} \cos{\left (c + d x \right )}}{2} + \frac{\sin{\left (c + d x \right )} \cos{\left (a + d x \right )}}{2 d} & \text{for}\: b = d \\\frac{b \sin{\left (a + b x \right )} \cos{\left (c + d x \right )}}{b^{2} - d^{2}} - \frac{d \sin{\left (c + d x \right )} \cos{\left (a + b x \right )}}{b^{2} - d^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*cos(d*x+c),x)

[Out]

Piecewise((x*cos(a)*cos(c), Eq(b, 0) & Eq(d, 0)), (-x*sin(a - d*x)*sin(c + d*x)/2 + x*cos(a - d*x)*cos(c + d*x
)/2 - sin(a - d*x)*cos(c + d*x)/(2*d), Eq(b, -d)), (x*sin(a + d*x)*sin(c + d*x)/2 + x*cos(a + d*x)*cos(c + d*x
)/2 + sin(c + d*x)*cos(a + d*x)/(2*d), Eq(b, d)), (b*sin(a + b*x)*cos(c + d*x)/(b**2 - d**2) - d*sin(c + d*x)*
cos(a + b*x)/(b**2 - d**2), True))

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Giac [A]  time = 1.09072, size = 54, normalized size = 1.26 \begin{align*} \frac{\sin \left (b x + d x + a + c\right )}{2 \,{\left (b + d\right )}} + \frac{\sin \left (b x - d x + a - c\right )}{2 \,{\left (b - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*cos(d*x+c),x, algorithm="giac")

[Out]

1/2*sin(b*x + d*x + a + c)/(b + d) + 1/2*sin(b*x - d*x + a - c)/(b - d)