\(\int \sin ^3(a+b x) \sin ^2(c+d x) \, dx\) [124]

Optimal result

Integrand size = 17, antiderivative size = 138 \[ \int \sin ^3(a+b x) \sin ^2(c+d x) \, dx=-\frac {3 \cos (a+b x)}{8 b}+\frac {\cos (3 a+3 b x)}{24 b}+\frac {3 \cos (a-2 c+(b-2 d) x)}{16 (b-2 d)}-\frac {\cos (3 a-2 c+(3 b-2 d) x)}{16 (3 b-2 d)}+\frac {3 \cos (a+2 c+(b+2 d) x)}{16 (b+2 d)}-\frac {\cos (3 a+2 c+(3 b+2 d) x)}{16 (3 b+2 d)} \] Output:

-3/8*cos(b*x+a)/b+1/24*cos(3*b*x+3*a)/b+3*cos(a-2*c+(b-2*d)*x)/(16*b-32*d) 
-cos(3*a-2*c+(3*b-2*d)*x)/(48*b-32*d)+3*cos(a+2*c+(b+2*d)*x)/(16*b+32*d)-c 
os(3*a+2*c+(3*b+2*d)*x)/(48*b+32*d)
 

Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11 \[ \int \sin ^3(a+b x) \sin ^2(c+d x) \, dx=\frac {1}{48} \left (-\frac {18 \cos (a) \cos (b x)}{b}+\frac {2 \cos (3 a) \cos (3 b x)}{b}+\frac {9 \cos (a-2 c+b x-2 d x)}{b-2 d}-\frac {3 \cos (3 a-2 c+3 b x-2 d x)}{3 b-2 d}+\frac {9 \cos (a+2 c+b x+2 d x)}{b+2 d}-\frac {3 \cos (3 a+2 c+3 b x+2 d x)}{3 b+2 d}+\frac {18 \sin (a) \sin (b x)}{b}-\frac {2 \sin (3 a) \sin (3 b x)}{b}\right ) \] Input:

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

Output:

((-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
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {5080, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5080

\(\displaystyle \int \left (-\frac {3}{16} \sin (a+x (b-2 d)-2 c)+\frac {1}{16} \sin (3 a+x (3 b-2 d)-2 c)-\frac {3}{16} \sin (a+x (b+2 d)+2 c)+\frac {1}{16} \sin (3 a+x (3 b+2 d)+2 c)+\frac {3}{8} \sin (a+b x)-\frac {1}{8} \sin (3 a+3 b x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \cos (a+x (b-2 d)-2 c)}{16 (b-2 d)}-\frac {\cos (3 a+x (3 b-2 d)-2 c)}{16 (3 b-2 d)}+\frac {3 \cos (a+x (b+2 d)+2 c)}{16 (b+2 d)}-\frac {\cos (3 a+x (3 b+2 d)+2 c)}{16 (3 b+2 d)}-\frac {3 \cos (a+b x)}{8 b}+\frac {\cos (3 a+3 b x)}{24 b}\)

Input:

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

Output:

(-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))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5080

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]) || (Binomial 
Q[{v, w}, x] && IndependentQ[Cancel[v/w], x])) && IGtQ[p, 0] && IGtQ[q, 0]
 

Maple [A] (verified)

Time = 13.20 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {3 \cos \left (b x +a \right )}{8 b}+\frac {\cos \left (3 b x +3 a \right )}{24 b}+\frac {3 \cos \left (a -2 c +\left (b -2 d \right ) x \right )}{16 \left (b -2 d \right )}+\frac {3 \cos \left (a +2 c +\left (b +2 d \right ) x \right )}{16 \left (b +2 d \right )}-\frac {\cos \left (3 a -2 c +\left (3 b -2 d \right ) x \right )}{16 \left (3 b -2 d \right )}-\frac {\cos \left (3 a +2 c +\left (3 b +2 d \right ) x \right )}{16 \left (3 b +2 d \right )}\)	\(127\)
parallelrisch	\(\frac {-9 \left (b +\frac {2 d}{3}\right ) \left (b +2 d \right ) b \left (b -2 d \right ) \cos \left (3 a -2 c +\left (3 b -2 d \right ) x \right )-9 \left (b +2 d \right ) \left (b -\frac {2 d}{3}\right ) b \left (b -2 d \right ) \cos \left (3 a +2 c +\left (3 b +2 d \right ) x \right )+81 \left (b +\frac {2 d}{3}\right ) \left (b +2 d \right ) \left (b -\frac {2 d}{3}\right ) b \cos \left (a -2 c +\left (b -2 d \right ) x \right )+81 \left (b +\frac {2 d}{3}\right ) \left (b -\frac {2 d}{3}\right ) b \left (b -2 d \right ) \cos \left (a +2 c +\left (b +2 d \right ) x \right )+\left (18 b^{4}-80 b^{2} d^{2}+32 d^{4}\right ) \cos \left (3 b x +3 a \right )+\left (-162 b^{4}+720 b^{2} d^{2}-288 d^{4}\right ) \cos \left (b x +a \right )+640 b^{2} d^{2}-256 d^{4}}{432 b^{5}-1920 b^{3} d^{2}+768 b \,d^{4}}\)	\(225\)
risch	\(-\frac {3 \cos \left (b x +a \right )}{8 b}+\frac {27 \cos \left (b x -2 d x +a -2 c \right ) b^{3}}{16 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}+\frac {27 \cos \left (b x -2 d x +a -2 c \right ) b^{2} d}{8 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}-\frac {3 \cos \left (b x -2 d x +a -2 c \right ) b \,d^{2}}{4 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}-\frac {3 \cos \left (b x -2 d x +a -2 c \right ) d^{3}}{2 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}+\frac {27 \cos \left (b x +2 d x +a +2 c \right ) b^{3}}{16 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}-\frac {27 \cos \left (b x +2 d x +a +2 c \right ) b^{2} d}{8 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}-\frac {3 \cos \left (b x +2 d x +a +2 c \right ) b \,d^{2}}{4 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}+\frac {3 \cos \left (b x +2 d x +a +2 c \right ) d^{3}}{2 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}-\frac {3 \cos \left (3 b x -2 d x +3 a -2 c \right ) b^{3}}{16 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}-\frac {\cos \left (3 b x -2 d x +3 a -2 c \right ) b^{2} d}{8 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}+\frac {3 \cos \left (3 b x -2 d x +3 a -2 c \right ) b \,d^{2}}{4 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}+\frac {\cos \left (3 b x -2 d x +3 a -2 c \right ) d^{3}}{2 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}-\frac {3 \cos \left (3 b x +2 d x +3 a +2 c \right ) b^{3}}{16 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}+\frac {\cos \left (3 b x +2 d x +3 a +2 c \right ) b^{2} d}{8 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}+\frac {3 \cos \left (3 b x +2 d x +3 a +2 c \right ) b \,d^{2}}{4 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}-\frac {\cos \left (3 b x +2 d x +3 a +2 c \right ) d^{3}}{2 \left (b +2 d \right ) \left (3 b +2 d \right ) \left (3 b -2 d \right ) \left (b -2 d \right )}+\frac {\cos \left (3 b x +3 a \right )}{24 b}\)	\(859\)
orering	\(\text {Expression too large to display}\)	\(3070\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.37 \[ \int \sin ^3(a+b x) \sin ^2(c+d x) \, dx=\frac {{\left (9 \, b^{4} - 38 \, b^{2} d^{2} + 8 \, d^{4}\right )} \cos \left (b x + a\right )^{3} + 6 \, {\left (7 \, b^{3} d - 4 \, b d^{3} - {\left (b^{3} d - 4 \, b d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \cos \left (d x + c\right ) \sin \left (b x + a\right ) \sin \left (d x + c\right ) - 9 \, {\left ({\left (b^{4} - 4 \, b^{2} d^{2}\right )} \cos \left (b x + a\right )^{3} - {\left (3 \, b^{4} - 4 \, b^{2} d^{2}\right )} \cos \left (b x + a\right )\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (9 \, b^{4} - 26 \, b^{2} d^{2} + 8 \, d^{4}\right )} \cos \left (b x + a\right )}{3 \, {\left (9 \, b^{5} - 40 \, b^{3} d^{2} + 16 \, b d^{4}\right )}} \] Input:

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

Output:

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)*cos(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))*c 
os(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)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2030 vs. \(2 (116) = 232\).

Time = 5.81 (sec) , antiderivative size = 2030, normalized size of antiderivative = 14.71 \[ \int \sin ^3(a+b x) \sin ^2(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((x*sin(a)**3*sin(c)**2, Eq(b, 0) & Eq(d, 0)), ((x*sin(c + d*x)** 
2/2 + x*cos(c + d*x)**2/2 - sin(c + d*x)*cos(c + d*x)/(2*d))*sin(a)**3, Eq 
(b, 0)), (3*x*sin(a - 2*d*x)**3*sin(c + d*x)**2/16 - 3*x*sin(a - 2*d*x)**3 
*cos(c + d*x)**2/16 - 3*x*sin(a - 2*d*x)**2*sin(c + d*x)*cos(a - 2*d*x)*co 
s(c + d*x)/8 + 3*x*sin(a - 2*d*x)*sin(c + d*x)**2*cos(a - 2*d*x)**2/16 - 3 
*x*sin(a - 2*d*x)*cos(a - 2*d*x)**2*cos(c + d*x)**2/16 - 3*x*sin(c + d*x)* 
cos(a - 2*d*x)**3*cos(c + d*x)/8 - 13*sin(a - 2*d*x)**3*sin(c + d*x)*cos(c 
 + d*x)/(16*d) + sin(a - 2*d*x)**2*cos(a - 2*d*x)*cos(c + d*x)**2/(2*d) - 
7*sin(a - 2*d*x)*sin(c + d*x)*cos(a - 2*d*x)**2*cos(c + d*x)/(8*d) - 17*si 
n(c + d*x)**2*cos(a - 2*d*x)**3/(96*d) + 49*cos(a - 2*d*x)**3*cos(c + d*x) 
**2/(96*d), Eq(b, -2*d)), (x*sin(a - 2*d*x/3)**3*sin(c + d*x)**2/16 - x*si 
n(a - 2*d*x/3)**3*cos(c + d*x)**2/16 - 3*x*sin(a - 2*d*x/3)**2*sin(c + d*x 
)*cos(a - 2*d*x/3)*cos(c + d*x)/8 - 3*x*sin(a - 2*d*x/3)*sin(c + d*x)**2*c 
os(a - 2*d*x/3)**2/16 + 3*x*sin(a - 2*d*x/3)*cos(a - 2*d*x/3)**2*cos(c + d 
*x)**2/16 + x*sin(c + d*x)*cos(a - 2*d*x/3)**3*cos(c + d*x)/8 - 15*sin(a - 
 2*d*x/3)**3*sin(c + d*x)*cos(c + d*x)/(16*d) + 3*sin(a - 2*d*x/3)**2*cos( 
a - 2*d*x/3)*cos(c + d*x)**2/(2*d) + 9*sin(a - 2*d*x/3)*sin(c + d*x)*cos(a 
 - 2*d*x/3)**2*cos(c + d*x)/(8*d) + 21*sin(c + d*x)**2*cos(a - 2*d*x/3)**3 
/(32*d) + 11*cos(a - 2*d*x/3)**3*cos(c + d*x)**2/(32*d), Eq(b, -2*d/3)), ( 
x*sin(a + 2*d*x/3)**3*sin(c + d*x)**2/16 - x*sin(a + 2*d*x/3)**3*cos(c ...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1360 vs. \(2 (126) = 252\).

Time = 0.10 (sec) , antiderivative size = 1360, normalized size of antiderivative = 9.86 \[ \int \sin ^3(a+b x) \sin ^2(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

-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*c 
os(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*co 
s(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*co 
s(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) - 4 
0*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*co 
s(2*c) - 40*b^2*d^2*cos(2*c) + 16*d^4*cos(2*c))*cos(3*b*x + 3*a - 2*c) + 1 
8*(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*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) + 3*(3*b^4*sin(2*c) + 2*b^3*d*sin(2*c) - 12*b^2*d^2*sin(2*c) - 8...
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \sin ^3(a+b x) \sin ^2(c+d x) \, dx=-\frac {\cos \left (3 \, b x + 2 \, d x + 3 \, a + 2 \, c\right )}{16 \, {\left (3 \, b + 2 \, d\right )}} - \frac {\cos \left (3 \, b x - 2 \, d x + 3 \, a - 2 \, c\right )}{16 \, {\left (3 \, b - 2 \, d\right )}} + \frac {\cos \left (3 \, b x + 3 \, a\right )}{24 \, b} + \frac {3 \, \cos \left (b x + 2 \, d x + a + 2 \, c\right )}{16 \, {\left (b + 2 \, d\right )}} + \frac {3 \, \cos \left (b x - 2 \, d x + a - 2 \, c\right )}{16 \, {\left (b - 2 \, d\right )}} - \frac {3 \, \cos \left (b x + a\right )}{8 \, b} \] Input:

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

Output:

-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*cos 
(b*x + a)/b
 

Mupad [B] (verification not implemented)

Time = 19.16 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.17 \[ \int \sin ^3(a+b x) \sin ^2(c+d x) \, dx=\frac {81\,b^4\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )+81\,b^4\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )-162\,b^4\,\cos \left (a+b\,x\right )-288\,d^4\,\cos \left (a+b\,x\right )-9\,b^4\,\cos \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )-9\,b^4\,\cos \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )+18\,b^4\,\cos \left (3\,a+3\,b\,x\right )+32\,d^4\,\cos \left (3\,a+3\,b\,x\right )+24\,b\,d^3\,\cos \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )-24\,b\,d^3\,\cos \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )-6\,b^3\,d\,\cos \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )+6\,b^3\,d\,\cos \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )-36\,b^2\,d^2\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )-36\,b^2\,d^2\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )+720\,b^2\,d^2\,\cos \left (a+b\,x\right )+36\,b^2\,d^2\,\cos \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )+36\,b^2\,d^2\,\cos \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )-80\,b^2\,d^2\,\cos \left (3\,a+3\,b\,x\right )-72\,b\,d^3\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )+72\,b\,d^3\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )+162\,b^3\,d\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )-162\,b^3\,d\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )}{48\,\left (9\,b^5-40\,b^3\,d^2+16\,b\,d^4\right )} \] Input:

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

Output:

(81*b^4*cos(a - 2*c + b*x - 2*d*x) + 81*b^4*cos(a + 2*c + b*x + 2*d*x) - 1 
62*b^4*cos(a + b*x) - 288*d^4*cos(a + b*x) - 9*b^4*cos(3*a - 2*c + 3*b*x - 
 2*d*x) - 9*b^4*cos(3*a + 2*c + 3*b*x + 2*d*x) + 18*b^4*cos(3*a + 3*b*x) + 
 32*d^4*cos(3*a + 3*b*x) + 24*b*d^3*cos(3*a - 2*c + 3*b*x - 2*d*x) - 24*b* 
d^3*cos(3*a + 2*c + 3*b*x + 2*d*x) - 6*b^3*d*cos(3*a - 2*c + 3*b*x - 2*d*x 
) + 6*b^3*d*cos(3*a + 2*c + 3*b*x + 2*d*x) - 36*b^2*d^2*cos(a - 2*c + b*x 
- 2*d*x) - 36*b^2*d^2*cos(a + 2*c + b*x + 2*d*x) + 720*b^2*d^2*cos(a + b*x 
) + 36*b^2*d^2*cos(3*a - 2*c + 3*b*x - 2*d*x) + 36*b^2*d^2*cos(3*a + 2*c + 
 3*b*x + 2*d*x) - 80*b^2*d^2*cos(3*a + 3*b*x) - 72*b*d^3*cos(a - 2*c + b*x 
 - 2*d*x) + 72*b*d^3*cos(a + 2*c + b*x + 2*d*x) + 162*b^3*d*cos(a - 2*c + 
b*x - 2*d*x) - 162*b^3*d*cos(a + 2*c + b*x + 2*d*x))/(48*(16*b*d^4 + 9*b^5 
 - 40*b^3*d^2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.87 \[ \int \sin ^3(a+b x) \sin ^2(c+d x) \, dx=\frac {-9 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} \sin \left (d x +c \right )^{2} b^{4}+36 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} \sin \left (d x +c \right )^{2} b^{2} d^{2}+2 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b^{2} d^{2}-8 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} d^{4}-18 \cos \left (b x +a \right ) \sin \left (d x +c \right )^{2} b^{4}+40 \cos \left (b x +a \right ) b^{2} d^{2}-16 \cos \left (b x +a \right ) d^{4}+6 \cos \left (d x +c \right ) \sin \left (b x +a \right )^{3} \sin \left (d x +c \right ) b^{3} d -24 \cos \left (d x +c \right ) \sin \left (b x +a \right )^{3} \sin \left (d x +c \right ) b \,d^{3}+36 \cos \left (d x +c \right ) \sin \left (b x +a \right ) \sin \left (d x +c \right ) b^{3} d -16 b^{2} d^{2}+16 d^{4}}{3 b \left (9 b^{4}-40 b^{2} d^{2}+16 d^{4}\right )} \] Input:

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

Output:

( - 9*cos(a + b*x)*sin(a + b*x)**2*sin(c + d*x)**2*b**4 + 36*cos(a + b*x)* 
sin(a + b*x)**2*sin(c + d*x)**2*b**2*d**2 + 2*cos(a + b*x)*sin(a + b*x)**2 
*b**2*d**2 - 8*cos(a + b*x)*sin(a + b*x)**2*d**4 - 18*cos(a + b*x)*sin(c + 
 d*x)**2*b**4 + 40*cos(a + b*x)*b**2*d**2 - 16*cos(a + b*x)*d**4 + 6*cos(c 
 + d*x)*sin(a + b*x)**3*sin(c + d*x)*b**3*d - 24*cos(c + d*x)*sin(a + b*x) 
**3*sin(c + d*x)*b*d**3 + 36*cos(c + d*x)*sin(a + b*x)*sin(c + d*x)*b**3*d 
 - 16*b**2*d**2 + 16*d**4)/(3*b*(9*b**4 - 40*b**2*d**2 + 16*d**4))