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

Optimal. Leaf size=144 \[ \frac{\cos (2 a+x (2 b-3 d)-3 c)}{16 (2 b-3 d)}-\frac{3 \cos (2 a+x (2 b-d)-c)}{16 (2 b-d)}+\frac{3 \cos (2 a+x (2 b+d)+c)}{16 (2 b+d)}-\frac{\cos (2 a+x (2 b+3 d)+3 c)}{16 (2 b+3 d)}-\frac{3 \cos (c+d x)}{8 d}+\frac{\cos (3 c+3 d x)}{24 d} \]

[Out]

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

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Rubi [A]  time = 0.0972936, antiderivative size = 144, 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} \[ \frac{\cos (2 a+x (2 b-3 d)-3 c)}{16 (2 b-3 d)}-\frac{3 \cos (2 a+x (2 b-d)-c)}{16 (2 b-d)}+\frac{3 \cos (2 a+x (2 b+d)+c)}{16 (2 b+d)}-\frac{\cos (2 a+x (2 b+3 d)+3 c)}{16 (2 b+3 d)}-\frac{3 \cos (c+d x)}{8 d}+\frac{\cos (3 c+3 d x)}{24 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2*Sin[c + d*x]^3,x]

[Out]

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

Mathematica [A]  time = 1.64462, size = 158, normalized size = 1.1 \[ \frac{1}{48} \left (\frac{3 \cos (2 a+2 b x-3 c-3 d x)}{2 b-3 d}-\frac{9 \cos (2 a+2 b x-c-d x)}{2 b-d}+\frac{9 \cos (2 a+2 b x+c+d x)}{2 b+d}-\frac{3 \cos (2 a+2 b x+3 c+3 d x)}{2 b+3 d}+\frac{18 \sin (c) \sin (d x)}{d}-\frac{2 \sin (3 c) \sin (3 d x)}{d}-\frac{18 \cos (c) \cos (d x)}{d}+\frac{2 \cos (3 c) \cos (3 d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 133, normalized size = 0.9 \begin{align*}{\frac{\cos \left ( 2\,a-3\,c+ \left ( 2\,b-3\,d \right ) x \right ) }{32\,b-48\,d}}-{\frac{3\,\cos \left ( 2\,a-c+ \left ( 2\,b-d \right ) x \right ) }{32\,b-16\,d}}-{\frac{3\,\cos \left ( dx+c \right ) }{8\,d}}+{\frac{\cos \left ( 3\,dx+3\,c \right ) }{24\,d}}+{\frac{3\,\cos \left ( 2\,a+c+ \left ( 2\,b+d \right ) x \right ) }{32\,b+16\,d}}-{\frac{\cos \left ( 2\,a+3\,c+ \left ( 2\,b+3\,d \right ) x \right ) }{32\,b+48\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.46137, size = 1839, normalized size = 12.77 \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)^2*sin(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/96*(3*(8*b^3*d*cos(3*c) - 12*b^2*d^2*cos(3*c) - 2*b*d^3*cos(3*c) + 3*d^4*cos(3*c))*cos((2*b + 3*d)*x + 2*a
+ 6*c) + 3*(8*b^3*d*cos(3*c) - 12*b^2*d^2*cos(3*c) - 2*b*d^3*cos(3*c) + 3*d^4*cos(3*c))*cos((2*b + 3*d)*x + 2*
a) - 9*(8*b^3*d*cos(3*c) - 4*b^2*d^2*cos(3*c) - 18*b*d^3*cos(3*c) + 9*d^4*cos(3*c))*cos((2*b + d)*x + 2*a + 4*
c) - 9*(8*b^3*d*cos(3*c) - 4*b^2*d^2*cos(3*c) - 18*b*d^3*cos(3*c) + 9*d^4*cos(3*c))*cos((2*b + d)*x + 2*a - 2*
c) + 9*(8*b^3*d*cos(3*c) + 4*b^2*d^2*cos(3*c) - 18*b*d^3*cos(3*c) - 9*d^4*cos(3*c))*cos(-(2*b - d)*x - 2*a + 4
*c) + 9*(8*b^3*d*cos(3*c) + 4*b^2*d^2*cos(3*c) - 18*b*d^3*cos(3*c) - 9*d^4*cos(3*c))*cos(-(2*b - d)*x - 2*a -
2*c) - 3*(8*b^3*d*cos(3*c) + 12*b^2*d^2*cos(3*c) - 2*b*d^3*cos(3*c) - 3*d^4*cos(3*c))*cos(-(2*b - 3*d)*x - 2*a
+ 6*c) - 3*(8*b^3*d*cos(3*c) + 12*b^2*d^2*cos(3*c) - 2*b*d^3*cos(3*c) - 3*d^4*cos(3*c))*cos(-(2*b - 3*d)*x -
2*a) - 2*(16*b^4*cos(3*c) - 40*b^2*d^2*cos(3*c) + 9*d^4*cos(3*c))*cos(3*d*x) - 2*(16*b^4*cos(3*c) - 40*b^2*d^2
*cos(3*c) + 9*d^4*cos(3*c))*cos(3*d*x + 6*c) + 18*(16*b^4*cos(3*c) - 40*b^2*d^2*cos(3*c) + 9*d^4*cos(3*c))*cos
(d*x + 4*c) + 18*(16*b^4*cos(3*c) - 40*b^2*d^2*cos(3*c) + 9*d^4*cos(3*c))*cos(d*x - 2*c) + 3*(8*b^3*d*sin(3*c)
- 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) + 3*d^4*sin(3*c))*sin((2*b + 3*d)*x + 2*a + 6*c) - 3*(8*b^3*d*sin(3*
c) - 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) + 3*d^4*sin(3*c))*sin((2*b + 3*d)*x + 2*a) - 9*(8*b^3*d*sin(3*c) -
4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) + 9*d^4*sin(3*c))*sin((2*b + d)*x + 2*a + 4*c) + 9*(8*b^3*d*sin(3*c) -
4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) + 9*d^4*sin(3*c))*sin((2*b + d)*x + 2*a - 2*c) + 9*(8*b^3*d*sin(3*c) +
4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) - 9*d^4*sin(3*c))*sin(-(2*b - d)*x - 2*a + 4*c) - 9*(8*b^3*d*sin(3*c)
+ 4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) - 9*d^4*sin(3*c))*sin(-(2*b - d)*x - 2*a - 2*c) - 3*(8*b^3*d*sin(3*c)
+ 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) - 3*d^4*sin(3*c))*sin(-(2*b - 3*d)*x - 2*a + 6*c) + 3*(8*b^3*d*sin(3
*c) + 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) - 3*d^4*sin(3*c))*sin(-(2*b - 3*d)*x - 2*a) + 2*(16*b^4*sin(3*c)
- 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*sin(3*d*x) - 2*(16*b^4*sin(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c)
)*sin(3*d*x + 6*c) + 18*(16*b^4*sin(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*sin(d*x + 4*c) - 18*(16*b^4*s
in(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*sin(d*x - 2*c))/(9*(cos(3*c)^2 + sin(3*c)^2)*d^5 - 40*(b^2*cos
(3*c)^2 + b^2*sin(3*c)^2)*d^3 + 16*(b^4*cos(3*c)^2 + b^4*sin(3*c)^2)*d)

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Fricas [A]  time = 0.545475, size = 429, normalized size = 2.98 \begin{align*} \frac{{\left (8 \, b^{4} - 38 \, b^{2} d^{2} + 9 \, d^{4} + 9 \,{\left (4 \, b^{2} d^{2} - d^{4}\right )} \cos \left (b x + a\right )^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left ({\left (4 \, b^{3} d - b d^{3}\right )} \cos \left (b x + a\right ) \cos \left (d x + c\right )^{2} -{\left (4 \, b^{3} d - 7 \, b d^{3}\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right ) \sin \left (d x + c\right ) - 3 \,{\left (8 \, b^{4} - 26 \, b^{2} d^{2} + 9 \, d^{4} + 3 \,{\left (4 \, b^{2} d^{2} - 3 \, d^{4}\right )} \cos \left (b x + a\right )^{2}\right )} \cos \left (d x + c\right )}{3 \,{\left (16 \, b^{4} d - 40 \, b^{2} d^{3} + 9 \, d^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**2*sin(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.16644, size = 174, normalized size = 1.21 \begin{align*} -\frac{\cos \left (2 \, b x + 3 \, d x + 2 \, a + 3 \, c\right )}{16 \,{\left (2 \, b + 3 \, d\right )}} + \frac{3 \, \cos \left (2 \, b x + d x + 2 \, a + c\right )}{16 \,{\left (2 \, b + d\right )}} - \frac{3 \, \cos \left (2 \, b x - d x + 2 \, a - c\right )}{16 \,{\left (2 \, b - d\right )}} + \frac{\cos \left (2 \, b x - 3 \, d x + 2 \, a - 3 \, c\right )}{16 \,{\left (2 \, b - 3 \, d\right )}} + \frac{\cos \left (3 \, d x + 3 \, c\right )}{24 \, d} - \frac{3 \, \cos \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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