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

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

[Out]

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

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Rubi [A]  time = 0.0357136, 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 = {4574, 2638} \[ -\frac{\cos (a+x (b-d)-c)}{2 (b-d)}-\frac{\cos (a+x (b+d)+c)}{2 (b+d)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Sin[a + b*x],x]

[Out]

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

Rule 4574

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

Rule 2638

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

Rubi steps

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

Mathematica [A]  time = 0.195882, size = 43, normalized size = 1. \[ -\frac{\cos (a+x (b-d)-c)}{2 (b-d)}-\frac{\cos (a+x (b+d)+c)}{2 (b+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 40, normalized size = 0.9 \begin{align*} -{\frac{\cos \left ( a-c+ \left ( b-d \right ) x \right ) }{2\,b-2\,d}}-{\frac{\cos \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(d*x+c)*sin(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.17808, size = 54, normalized size = 1.26 \begin{align*} -\frac{\cos \left (b x + d x + a + c\right )}{2 \,{\left (b + d\right )}} - \frac{\cos \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(d*x+c)*sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.473851, size = 100, normalized size = 2.33 \begin{align*} -\frac{b \cos \left (b x + a\right ) \cos \left (d x + c\right ) + d \sin \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(d*x+c)*sin(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.87205, size = 155, normalized size = 3.6 \begin{align*} \begin{cases} x \sin{\left (a \right )} \cos{\left (c \right )} & \text{for}\: b = 0 \wedge d = 0 \\\frac{x \sin{\left (a - d x \right )} \cos{\left (c + d x \right )}}{2} + \frac{x \sin{\left (c + d x \right )} \cos{\left (a - d x \right )}}{2} + \frac{\sin{\left (a - d x \right )} \sin{\left (c + d x \right )}}{2 d} & \text{for}\: b = - d \\\frac{x \sin{\left (a + d x \right )} \cos{\left (c + d x \right )}}{2} - \frac{x \sin{\left (c + d x \right )} \cos{\left (a + d x \right )}}{2} - \frac{\cos{\left (a + d x \right )} \cos{\left (c + d x \right )}}{2 d} & \text{for}\: b = d \\- \frac{b \cos{\left (a + b x \right )} \cos{\left (c + d x \right )}}{b^{2} - d^{2}} - \frac{d \sin{\left (a + b x \right )} \sin{\left (c + d 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(d*x+c)*sin(b*x+a),x)

[Out]

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

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Giac [A]  time = 1.13548, size = 54, normalized size = 1.26 \begin{align*} -\frac{\cos \left (b x + d x + a + c\right )}{2 \,{\left (b + d\right )}} - \frac{\cos \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(d*x+c)*sin(b*x+a),x, algorithm="giac")

[Out]

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