3.202 \(\int \sin ^2(a+b x) \sin (c+d x) \, dx\)

Optimal. Leaf size=68 \[ -\frac{\cos (2 a+x (2 b-d)-c)}{4 (2 b-d)}+\frac{\cos (2 a+x (2 b+d)+c)}{4 (2 b+d)}-\frac{\cos (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0536243, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4569, 2638} \[ -\frac{\cos (2 a+x (2 b-d)-c)}{4 (2 b-d)}+\frac{\cos (2 a+x (2 b+d)+c)}{4 (2 b+d)}-\frac{\cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2*Sin[c + d*x],x]

[Out]

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

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 2638

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

Rubi steps

\begin{align*} \int \sin ^2(a+b x) \sin (c+d x) \, dx &=\int \left (\frac{1}{4} \sin (2 a-c+(2 b-d) x)+\frac{1}{2} \sin (c+d x)-\frac{1}{4} \sin (2 a+c+(2 b+d) x)\right ) \, dx\\ &=\frac{1}{4} \int \sin (2 a-c+(2 b-d) x) \, dx-\frac{1}{4} \int \sin (2 a+c+(2 b+d) x) \, dx+\frac{1}{2} \int \sin (c+d x) \, dx\\ &=-\frac{\cos (2 a-c+(2 b-d) x)}{4 (2 b-d)}-\frac{\cos (c+d x)}{2 d}+\frac{\cos (2 a+c+(2 b+d) x)}{4 (2 b+d)}\\ \end{align*}

Mathematica [A]  time = 0.35091, size = 80, normalized size = 1.18 \[ -\frac{\cos (2 a+2 b x-c-d x)}{4 (2 b-d)}+\frac{\cos (2 a+x (2 b+d)+c)}{4 (2 b+d)}+\frac{1}{2} \left (\frac{\sin (c) \sin (d x)}{d}-\frac{\cos (c) \cos (d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2*Sin[c + d*x],x]

[Out]

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

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Maple [A]  time = 0.017, size = 63, normalized size = 0.9 \begin{align*} -{\frac{\cos \left ( 2\,a-c+ \left ( 2\,b-d \right ) x \right ) }{8\,b-4\,d}}-{\frac{\cos \left ( dx+c \right ) }{2\,d}}+{\frac{\cos \left ( 2\,a+c+ \left ( 2\,b+d \right ) x \right ) }{8\,b+4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^2*sin(d*x+c),x)

[Out]

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

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Maxima [B]  time = 1.149, size = 501, normalized size = 7.37 \begin{align*} -\frac{{\left (2 \, b d \cos \left (c\right ) - d^{2} \cos \left (c\right )\right )} \cos \left ({\left (2 \, b + d\right )} x + 2 \, a + 2 \, c\right ) +{\left (2 \, b d \cos \left (c\right ) - d^{2} \cos \left (c\right )\right )} \cos \left ({\left (2 \, b + d\right )} x + 2 \, a\right ) -{\left (2 \, b d \cos \left (c\right ) + d^{2} \cos \left (c\right )\right )} \cos \left (-{\left (2 \, b - d\right )} x - 2 \, a + 2 \, c\right ) -{\left (2 \, b d \cos \left (c\right ) + d^{2} \cos \left (c\right )\right )} \cos \left (-{\left (2 \, b - d\right )} x - 2 \, a\right ) - 2 \,{\left (4 \, b^{2} \cos \left (c\right ) - d^{2} \cos \left (c\right )\right )} \cos \left (d x + 2 \, c\right ) - 2 \,{\left (4 \, b^{2} \cos \left (c\right ) - d^{2} \cos \left (c\right )\right )} \cos \left (d x\right ) +{\left (2 \, b d \sin \left (c\right ) - d^{2} \sin \left (c\right )\right )} \sin \left ({\left (2 \, b + d\right )} x + 2 \, a + 2 \, c\right ) -{\left (2 \, b d \sin \left (c\right ) - d^{2} \sin \left (c\right )\right )} \sin \left ({\left (2 \, b + d\right )} x + 2 \, a\right ) -{\left (2 \, b d \sin \left (c\right ) + d^{2} \sin \left (c\right )\right )} \sin \left (-{\left (2 \, b - d\right )} x - 2 \, a + 2 \, c\right ) +{\left (2 \, b d \sin \left (c\right ) + d^{2} \sin \left (c\right )\right )} \sin \left (-{\left (2 \, b - d\right )} x - 2 \, a\right ) - 2 \,{\left (4 \, b^{2} \sin \left (c\right ) - d^{2} \sin \left (c\right )\right )} \sin \left (d x + 2 \, c\right ) + 2 \,{\left (4 \, b^{2} \sin \left (c\right ) - d^{2} \sin \left (c\right )\right )} \sin \left (d x\right )}{8 \,{\left ({\left (\cos \left (c\right )^{2} + \sin \left (c\right )^{2}\right )} d^{3} - 4 \,{\left (b^{2} \cos \left (c\right )^{2} + b^{2} \sin \left (c\right )^{2}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^2*sin(d*x+c),x, algorithm="maxima")

[Out]

-1/8*((2*b*d*cos(c) - d^2*cos(c))*cos((2*b + d)*x + 2*a + 2*c) + (2*b*d*cos(c) - d^2*cos(c))*cos((2*b + d)*x +
2*a) - (2*b*d*cos(c) + d^2*cos(c))*cos(-(2*b - d)*x - 2*a + 2*c) - (2*b*d*cos(c) + d^2*cos(c))*cos(-(2*b - d)
*x - 2*a) - 2*(4*b^2*cos(c) - d^2*cos(c))*cos(d*x + 2*c) - 2*(4*b^2*cos(c) - d^2*cos(c))*cos(d*x) + (2*b*d*sin
(c) - d^2*sin(c))*sin((2*b + d)*x + 2*a + 2*c) - (2*b*d*sin(c) - d^2*sin(c))*sin((2*b + d)*x + 2*a) - (2*b*d*s
in(c) + d^2*sin(c))*sin(-(2*b - d)*x - 2*a + 2*c) + (2*b*d*sin(c) + d^2*sin(c))*sin(-(2*b - d)*x - 2*a) - 2*(4
*b^2*sin(c) - d^2*sin(c))*sin(d*x + 2*c) + 2*(4*b^2*sin(c) - d^2*sin(c))*sin(d*x))/((cos(c)^2 + sin(c)^2)*d^3
- 4*(b^2*cos(c)^2 + b^2*sin(c)^2)*d)

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Fricas [A]  time = 0.477438, size = 155, normalized size = 2.28 \begin{align*} -\frac{2 \, b d \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (d x + c\right ) +{\left (d^{2} \cos \left (b x + a\right )^{2} + 2 \, b^{2} - d^{2}\right )} \cos \left (d x + c\right )}{4 \, b^{2} d - d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^2*sin(d*x+c),x, algorithm="fricas")

[Out]

-(2*b*d*cos(b*x + a)*sin(b*x + a)*sin(d*x + c) + (d^2*cos(b*x + a)^2 + 2*b^2 - d^2)*cos(d*x + c))/(4*b^2*d - d
^3)

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Sympy [A]  time = 18.2682, size = 410, normalized size = 6.03 \begin{align*} \begin{cases} x \sin ^{2}{\left (a \right )} \sin{\left (c \right )} & \text{for}\: b = 0 \wedge d = 0 \\\frac{x \sin ^{2}{\left (a - \frac{d x}{2} \right )} \sin{\left (c + d x \right )}}{4} - \frac{x \sin{\left (a - \frac{d x}{2} \right )} \cos{\left (a - \frac{d x}{2} \right )} \cos{\left (c + d x \right )}}{2} - \frac{x \sin{\left (c + d x \right )} \cos ^{2}{\left (a - \frac{d x}{2} \right )}}{4} - \frac{\sin ^{2}{\left (a - \frac{d x}{2} \right )} \cos{\left (c + d x \right )}}{d} - \frac{\sin{\left (a - \frac{d x}{2} \right )} \sin{\left (c + d x \right )} \cos{\left (a - \frac{d x}{2} \right )}}{2 d} & \text{for}\: b = - \frac{d}{2} \\\frac{x \sin ^{2}{\left (a + \frac{d x}{2} \right )} \sin{\left (c + d x \right )}}{4} + \frac{x \sin{\left (a + \frac{d x}{2} \right )} \cos{\left (a + \frac{d x}{2} \right )} \cos{\left (c + d x \right )}}{2} - \frac{x \sin{\left (c + d x \right )} \cos ^{2}{\left (a + \frac{d x}{2} \right )}}{4} - \frac{3 \sin{\left (a + \frac{d x}{2} \right )} \sin{\left (c + d x \right )} \cos{\left (a + \frac{d x}{2} \right )}}{2 d} - \frac{\cos ^{2}{\left (a + \frac{d x}{2} \right )} \cos{\left (c + d x \right )}}{d} & \text{for}\: b = \frac{d}{2} \\\left (\frac{x \sin ^{2}{\left (a + b x \right )}}{2} + \frac{x \cos ^{2}{\left (a + b x \right )}}{2} - \frac{\sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b}\right ) \sin{\left (c \right )} & \text{for}\: d = 0 \\- \frac{2 b^{2} \sin ^{2}{\left (a + b x \right )} \cos{\left (c + d x \right )}}{4 b^{2} d - d^{3}} - \frac{2 b^{2} \cos ^{2}{\left (a + b x \right )} \cos{\left (c + d x \right )}}{4 b^{2} d - d^{3}} - \frac{2 b d \sin{\left (a + b x \right )} \sin{\left (c + d x \right )} \cos{\left (a + b x \right )}}{4 b^{2} d - d^{3}} + \frac{d^{2} \sin ^{2}{\left (a + b x \right )} \cos{\left (c + d x \right )}}{4 b^{2} d - d^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**2*sin(d*x+c),x)

[Out]

Piecewise((x*sin(a)**2*sin(c), Eq(b, 0) & Eq(d, 0)), (x*sin(a - d*x/2)**2*sin(c + d*x)/4 - x*sin(a - d*x/2)*co
s(a - d*x/2)*cos(c + d*x)/2 - x*sin(c + d*x)*cos(a - d*x/2)**2/4 - sin(a - d*x/2)**2*cos(c + d*x)/d - sin(a -
d*x/2)*sin(c + d*x)*cos(a - d*x/2)/(2*d), Eq(b, -d/2)), (x*sin(a + d*x/2)**2*sin(c + d*x)/4 + x*sin(a + d*x/2)
*cos(a + d*x/2)*cos(c + d*x)/2 - x*sin(c + d*x)*cos(a + d*x/2)**2/4 - 3*sin(a + d*x/2)*sin(c + d*x)*cos(a + d*
x/2)/(2*d) - cos(a + d*x/2)**2*cos(c + d*x)/d, Eq(b, d/2)), ((x*sin(a + b*x)**2/2 + x*cos(a + b*x)**2/2 - sin(
a + b*x)*cos(a + b*x)/(2*b))*sin(c), Eq(d, 0)), (-2*b**2*sin(a + b*x)**2*cos(c + d*x)/(4*b**2*d - d**3) - 2*b*
*2*cos(a + b*x)**2*cos(c + d*x)/(4*b**2*d - d**3) - 2*b*d*sin(a + b*x)*sin(c + d*x)*cos(a + b*x)/(4*b**2*d - d
**3) + d**2*sin(a + b*x)**2*cos(c + d*x)/(4*b**2*d - d**3), True))

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Giac [A]  time = 1.13754, size = 82, normalized size = 1.21 \begin{align*} \frac{\cos \left (2 \, b x + d x + 2 \, a + c\right )}{4 \,{\left (2 \, b + d\right )}} - \frac{\cos \left (2 \, b x - d x + 2 \, a - c\right )}{4 \,{\left (2 \, b - d\right )}} - \frac{\cos \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^2*sin(d*x+c),x, algorithm="giac")

[Out]

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