3.136 \(\int \sin (c-b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=27 \[ \frac{\sin (a+2 b x-c)}{4 b}-\frac{1}{2} x \cos (a+c) \]

[Out]

-(x*Cos[a + c])/2 + Sin[a - c + 2*b*x]/(4*b)

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Rubi [A]  time = 0.025399, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4569, 2637} \[ \frac{\sin (a+2 b x-c)}{4 b}-\frac{1}{2} x \cos (a+c) \]

Antiderivative was successfully verified.

[In]

Int[Sin[c - b*x]*Sin[a + b*x],x]

[Out]

-(x*Cos[a + c])/2 + Sin[a - c + 2*b*x]/(4*b)

Rule 4569

Int[Sin[v_]^(p_.)*Sin[w_]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[Sin[v]^p*Sin[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 \sin (c-b x) \sin (a+b x) \, dx &=\int \left (-\frac{1}{2} \cos (a+c)+\frac{1}{2} \cos (a-c+2 b x)\right ) \, dx\\ &=-\frac{1}{2} x \cos (a+c)+\frac{1}{2} \int \cos (a-c+2 b x) \, dx\\ &=-\frac{1}{2} x \cos (a+c)+\frac{\sin (a-c+2 b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0338703, size = 26, normalized size = 0.96 \[ \frac{\sin (a+2 b x-c)-2 b x \cos (a+c)}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c - b*x]*Sin[a + b*x],x]

[Out]

(-2*b*x*Cos[a + c] + Sin[a - c + 2*b*x])/(4*b)

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Maple [A]  time = 0.01, size = 24, normalized size = 0.9 \begin{align*} -{\frac{x\cos \left ( a+c \right ) }{2}}+{\frac{\sin \left ( 2\,bx+a-c \right ) }{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-sin(b*x-c)*sin(b*x+a),x)

[Out]

-1/2*x*cos(a+c)+1/4*sin(2*b*x+a-c)/b

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Maxima [A]  time = 1.00554, size = 31, normalized size = 1.15 \begin{align*} -\frac{1}{2} \, x \cos \left (a + c\right ) + \frac{\sin \left (2 \, b x + a - c\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-sin(b*x-c)*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/2*x*cos(a + c) + 1/4*sin(2*b*x + a - c)/b

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Fricas [A]  time = 2.4156, size = 124, normalized size = 4.59 \begin{align*} -\frac{b x \cos \left (a + c\right ) - \cos \left (b x + a\right ) \cos \left (a + c\right ) \sin \left (b x + a\right ) + \cos \left (b x + a\right )^{2} \sin \left (a + c\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-sin(b*x-c)*sin(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.21596, size = 61, normalized size = 2.26 \begin{align*} - \begin{cases} \frac{x \sin{\left (a + b x \right )} \sin{\left (b x - c \right )}}{2} + \frac{x \cos{\left (a + b x \right )} \cos{\left (b x - c \right )}}{2} - \frac{\sin{\left (b x - c \right )} \cos{\left (a + b x \right )}}{2 b} & \text{for}\: b \neq 0 \\- x \sin{\left (a \right )} \sin{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-sin(b*x-c)*sin(b*x+a),x)

[Out]

-Piecewise((x*sin(a + b*x)*sin(b*x - c)/2 + x*cos(a + b*x)*cos(b*x - c)/2 - sin(b*x - c)*cos(a + b*x)/(2*b), N
e(b, 0)), (-x*sin(a)*sin(c), True))

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Giac [A]  time = 1.11759, size = 31, normalized size = 1.15 \begin{align*} -\frac{1}{2} \, x \cos \left (a + c\right ) + \frac{\sin \left (2 \, b x + a - c\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-sin(b*x-c)*sin(b*x+a),x, algorithm="giac")

[Out]

-1/2*x*cos(a + c) + 1/4*sin(2*b*x + a - c)/b