3.207 \(\int \sin ^3(a+b x) \sin ^2(c+d x) \, dx\)

Optimal. Leaf size=138 \[ \frac{3 \cos (a+x (b-2 d)-2 c)}{16 (b-2 d)}-\frac{\cos (3 a+x (3 b-2 d)-2 c)}{16 (3 b-2 d)}+\frac{3 \cos (a+x (b+2 d)+2 c)}{16 (b+2 d)}-\frac{\cos (3 a+x (3 b+2 d)+2 c)}{16 (3 b+2 d)}-\frac{3 \cos (a+b x)}{8 b}+\frac{\cos (3 a+3 b x)}{24 b} \]

[Out]

(-3*Cos[a + b*x])/(8*b) + Cos[3*a + 3*b*x]/(24*b) + (3*Cos[a - 2*c + (b - 2*d)*x])/(16*(b - 2*d)) - Cos[3*a -
2*c + (3*b - 2*d)*x]/(16*(3*b - 2*d)) + (3*Cos[a + 2*c + (b + 2*d)*x])/(16*(b + 2*d)) - Cos[3*a + 2*c + (3*b +
2*d)*x]/(16*(3*b + 2*d))

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Rubi [A]  time = 0.0984789, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {4569, 2638} \[ \frac{3 \cos (a+x (b-2 d)-2 c)}{16 (b-2 d)}-\frac{\cos (3 a+x (3 b-2 d)-2 c)}{16 (3 b-2 d)}+\frac{3 \cos (a+x (b+2 d)+2 c)}{16 (b+2 d)}-\frac{\cos (3 a+x (3 b+2 d)+2 c)}{16 (3 b+2 d)}-\frac{3 \cos (a+b x)}{8 b}+\frac{\cos (3 a+3 b x)}{24 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*Sin[c + d*x]^2,x]

[Out]

(-3*Cos[a + b*x])/(8*b) + Cos[3*a + 3*b*x]/(24*b) + (3*Cos[a - 2*c + (b - 2*d)*x])/(16*(b - 2*d)) - Cos[3*a -
2*c + (3*b - 2*d)*x]/(16*(3*b - 2*d)) + (3*Cos[a + 2*c + (b + 2*d)*x])/(16*(b + 2*d)) - Cos[3*a + 2*c + (3*b +
2*d)*x]/(16*(3*b + 2*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 ^3(a+b x) \sin ^2(c+d x) \, dx &=\int \left (\frac{3}{8} \sin (a+b x)-\frac{1}{8} \sin (3 a+3 b x)-\frac{3}{16} \sin (a-2 c+(b-2 d) x)+\frac{1}{16} \sin (3 a-2 c+(3 b-2 d) x)-\frac{3}{16} \sin (a+2 c+(b+2 d) x)+\frac{1}{16} \sin (3 a+2 c+(3 b+2 d) x)\right ) \, dx\\ &=\frac{1}{16} \int \sin (3 a-2 c+(3 b-2 d) x) \, dx+\frac{1}{16} \int \sin (3 a+2 c+(3 b+2 d) x) \, dx-\frac{1}{8} \int \sin (3 a+3 b x) \, dx-\frac{3}{16} \int \sin (a-2 c+(b-2 d) x) \, dx-\frac{3}{16} \int \sin (a+2 c+(b+2 d) x) \, dx+\frac{3}{8} \int \sin (a+b x) \, dx\\ &=-\frac{3 \cos (a+b x)}{8 b}+\frac{\cos (3 a+3 b x)}{24 b}+\frac{3 \cos (a-2 c+(b-2 d) x)}{16 (b-2 d)}-\frac{\cos (3 a-2 c+(3 b-2 d) x)}{16 (3 b-2 d)}+\frac{3 \cos (a+2 c+(b+2 d) x)}{16 (b+2 d)}-\frac{\cos (3 a+2 c+(3 b+2 d) x)}{16 (3 b+2 d)}\\ \end{align*}

Mathematica [A]  time = 1.67213, size = 153, normalized size = 1.11 \[ \frac{1}{48} \left (\frac{9 \cos (a+b x-2 c-2 d x)}{b-2 d}-\frac{3 \cos (3 a+3 b x-2 c-2 d x)}{3 b-2 d}+\frac{9 \cos (a+b x+2 c+2 d x)}{b+2 d}-\frac{3 \cos (3 a+3 b x+2 c+2 d x)}{3 b+2 d}+\frac{18 \sin (a) \sin (b x)}{b}-\frac{2 \sin (3 a) \sin (3 b x)}{b}-\frac{18 \cos (a) \cos (b x)}{b}+\frac{2 \cos (3 a) \cos (3 b x)}{b}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3*Sin[c + d*x]^2,x]

[Out]

((-18*Cos[a]*Cos[b*x])/b + (2*Cos[3*a]*Cos[3*b*x])/b + (9*Cos[a - 2*c + b*x - 2*d*x])/(b - 2*d) - (3*Cos[3*a -
2*c + 3*b*x - 2*d*x])/(3*b - 2*d) + (9*Cos[a + 2*c + b*x + 2*d*x])/(b + 2*d) - (3*Cos[3*a + 2*c + 3*b*x + 2*d
*x])/(3*b + 2*d) + (18*Sin[a]*Sin[b*x])/b - (2*Sin[3*a]*Sin[3*b*x])/b)/48

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Maple [A]  time = 0.022, size = 127, normalized size = 0.9 \begin{align*} -{\frac{3\,\cos \left ( bx+a \right ) }{8\,b}}+{\frac{\cos \left ( 3\,bx+3\,a \right ) }{24\,b}}+{\frac{3\,\cos \left ( a-2\,c+ \left ( b-2\,d \right ) x \right ) }{16\,b-32\,d}}-{\frac{\cos \left ( 3\,a-2\,c+ \left ( 3\,b-2\,d \right ) x \right ) }{48\,b-32\,d}}+{\frac{3\,\cos \left ( a+2\,c+ \left ( b+2\,d \right ) x \right ) }{16\,b+32\,d}}-{\frac{\cos \left ( 3\,a+2\,c+ \left ( 3\,b+2\,d \right ) x \right ) }{48\,b+32\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3*sin(d*x+c)^2,x)

[Out]

-3/8*cos(b*x+a)/b+1/24*cos(3*b*x+3*a)/b+3/16*cos(a-2*c+(b-2*d)*x)/(b-2*d)-1/16*cos(3*a-2*c+(3*b-2*d)*x)/(3*b-2
*d)+3/16*cos(a+2*c+(b+2*d)*x)/(b+2*d)-1/16*cos(3*a+2*c+(3*b+2*d)*x)/(3*b+2*d)

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Maxima [B]  time = 1.55865, size = 1836, normalized size = 13.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(d*x+c)^2,x, algorithm="maxima")

[Out]

-1/96*(3*(3*b^4*cos(2*c) - 2*b^3*d*cos(2*c) - 12*b^2*d^2*cos(2*c) + 8*b*d^3*cos(2*c))*cos((3*b + 2*d)*x + 3*a
+ 4*c) + 3*(3*b^4*cos(2*c) - 2*b^3*d*cos(2*c) - 12*b^2*d^2*cos(2*c) + 8*b*d^3*cos(2*c))*cos((3*b + 2*d)*x + 3*
a) + 3*(3*b^4*cos(2*c) + 2*b^3*d*cos(2*c) - 12*b^2*d^2*cos(2*c) - 8*b*d^3*cos(2*c))*cos(-(3*b - 2*d)*x - 3*a +
4*c) + 3*(3*b^4*cos(2*c) + 2*b^3*d*cos(2*c) - 12*b^2*d^2*cos(2*c) - 8*b*d^3*cos(2*c))*cos(-(3*b - 2*d)*x - 3*
a) - 9*(9*b^4*cos(2*c) - 18*b^3*d*cos(2*c) - 4*b^2*d^2*cos(2*c) + 8*b*d^3*cos(2*c))*cos((b + 2*d)*x + a + 4*c)
- 9*(9*b^4*cos(2*c) - 18*b^3*d*cos(2*c) - 4*b^2*d^2*cos(2*c) + 8*b*d^3*cos(2*c))*cos((b + 2*d)*x + a) - 9*(9*
b^4*cos(2*c) + 18*b^3*d*cos(2*c) - 4*b^2*d^2*cos(2*c) - 8*b*d^3*cos(2*c))*cos(-(b - 2*d)*x - a + 4*c) - 9*(9*b
^4*cos(2*c) + 18*b^3*d*cos(2*c) - 4*b^2*d^2*cos(2*c) - 8*b*d^3*cos(2*c))*cos(-(b - 2*d)*x - a) - 2*(9*b^4*cos(
2*c) - 40*b^2*d^2*cos(2*c) + 16*d^4*cos(2*c))*cos(3*b*x + 3*a + 2*c) - 2*(9*b^4*cos(2*c) - 40*b^2*d^2*cos(2*c)
+ 16*d^4*cos(2*c))*cos(3*b*x + 3*a - 2*c) + 18*(9*b^4*cos(2*c) - 40*b^2*d^2*cos(2*c) + 16*d^4*cos(2*c))*cos(b
*x + a + 2*c) + 18*(9*b^4*cos(2*c) - 40*b^2*d^2*cos(2*c) + 16*d^4*cos(2*c))*cos(b*x + a - 2*c) + 3*(3*b^4*sin(
2*c) - 2*b^3*d*sin(2*c) - 12*b^2*d^2*sin(2*c) + 8*b*d^3*sin(2*c))*sin((3*b + 2*d)*x + 3*a + 4*c) - 3*(3*b^4*si
n(2*c) - 2*b^3*d*sin(2*c) - 12*b^2*d^2*sin(2*c) + 8*b*d^3*sin(2*c))*sin((3*b + 2*d)*x + 3*a) + 3*(3*b^4*sin(2*
c) + 2*b^3*d*sin(2*c) - 12*b^2*d^2*sin(2*c) - 8*b*d^3*sin(2*c))*sin(-(3*b - 2*d)*x - 3*a + 4*c) - 3*(3*b^4*sin
(2*c) + 2*b^3*d*sin(2*c) - 12*b^2*d^2*sin(2*c) - 8*b*d^3*sin(2*c))*sin(-(3*b - 2*d)*x - 3*a) - 9*(9*b^4*sin(2*
c) - 18*b^3*d*sin(2*c) - 4*b^2*d^2*sin(2*c) + 8*b*d^3*sin(2*c))*sin((b + 2*d)*x + a + 4*c) + 9*(9*b^4*sin(2*c)
- 18*b^3*d*sin(2*c) - 4*b^2*d^2*sin(2*c) + 8*b*d^3*sin(2*c))*sin((b + 2*d)*x + a) - 9*(9*b^4*sin(2*c) + 18*b^
3*d*sin(2*c) - 4*b^2*d^2*sin(2*c) - 8*b*d^3*sin(2*c))*sin(-(b - 2*d)*x - a + 4*c) + 9*(9*b^4*sin(2*c) + 18*b^3
*d*sin(2*c) - 4*b^2*d^2*sin(2*c) - 8*b*d^3*sin(2*c))*sin(-(b - 2*d)*x - a) - 2*(9*b^4*sin(2*c) - 40*b^2*d^2*si
n(2*c) + 16*d^4*sin(2*c))*sin(3*b*x + 3*a + 2*c) + 2*(9*b^4*sin(2*c) - 40*b^2*d^2*sin(2*c) + 16*d^4*sin(2*c))*
sin(3*b*x + 3*a - 2*c) + 18*(9*b^4*sin(2*c) - 40*b^2*d^2*sin(2*c) + 16*d^4*sin(2*c))*sin(b*x + a + 2*c) - 18*(
9*b^4*sin(2*c) - 40*b^2*d^2*sin(2*c) + 16*d^4*sin(2*c))*sin(b*x + a - 2*c))/(9*b^5*cos(2*c)^2 + 9*b^5*sin(2*c)
^2 + 16*(b*cos(2*c)^2 + b*sin(2*c)^2)*d^4 - 40*(b^3*cos(2*c)^2 + b^3*sin(2*c)^2)*d^2)

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Fricas [A]  time = 0.533839, size = 427, normalized size = 3.09 \begin{align*} \frac{{\left (9 \, b^{4} - 38 \, b^{2} d^{2} + 8 \, d^{4}\right )} \cos \left (b x + a\right )^{3} + 6 \,{\left (7 \, b^{3} d - 4 \, b d^{3} -{\left (b^{3} d - 4 \, b d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \cos \left (d x + c\right ) \sin \left (b x + a\right ) \sin \left (d x + c\right ) - 9 \,{\left ({\left (b^{4} - 4 \, b^{2} d^{2}\right )} \cos \left (b x + a\right )^{3} -{\left (3 \, b^{4} - 4 \, b^{2} d^{2}\right )} \cos \left (b x + a\right )\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (9 \, b^{4} - 26 \, b^{2} d^{2} + 8 \, d^{4}\right )} \cos \left (b x + a\right )}{3 \,{\left (9 \, b^{5} - 40 \, b^{3} d^{2} + 16 \, b d^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(d*x+c)^2,x, algorithm="fricas")

[Out]

1/3*((9*b^4 - 38*b^2*d^2 + 8*d^4)*cos(b*x + a)^3 + 6*(7*b^3*d - 4*b*d^3 - (b^3*d - 4*b*d^3)*cos(b*x + a)^2)*co
s(d*x + c)*sin(b*x + a)*sin(d*x + c) - 9*((b^4 - 4*b^2*d^2)*cos(b*x + a)^3 - (3*b^4 - 4*b^2*d^2)*cos(b*x + a))
*cos(d*x + c)^2 - 3*(9*b^4 - 26*b^2*d^2 + 8*d^4)*cos(b*x + a))/(9*b^5 - 40*b^3*d^2 + 16*b*d^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**3*sin(d*x+c)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.11666, size = 167, normalized size = 1.21 \begin{align*} -\frac{\cos \left (3 \, b x + 2 \, d x + 3 \, a + 2 \, c\right )}{16 \,{\left (3 \, b + 2 \, d\right )}} - \frac{\cos \left (3 \, b x - 2 \, d x + 3 \, a - 2 \, c\right )}{16 \,{\left (3 \, b - 2 \, d\right )}} + \frac{\cos \left (3 \, b x + 3 \, a\right )}{24 \, b} + \frac{3 \, \cos \left (b x + 2 \, d x + a + 2 \, c\right )}{16 \,{\left (b + 2 \, d\right )}} + \frac{3 \, \cos \left (b x - 2 \, d x + a - 2 \, c\right )}{16 \,{\left (b - 2 \, d\right )}} - \frac{3 \, \cos \left (b x + a\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(d*x+c)^2,x, algorithm="giac")

[Out]

-1/16*cos(3*b*x + 2*d*x + 3*a + 2*c)/(3*b + 2*d) - 1/16*cos(3*b*x - 2*d*x + 3*a - 2*c)/(3*b - 2*d) + 1/24*cos(
3*b*x + 3*a)/b + 3/16*cos(b*x + 2*d*x + a + 2*c)/(b + 2*d) + 3/16*cos(b*x - 2*d*x + a - 2*c)/(b - 2*d) - 3/8*c
os(b*x + a)/b