Integrand size = 17, antiderivative size = 101 \[ \int \sin ^3(a+b x) \sin ^2(c+b x) \, dx=-\frac {3 \cos (a-2 c-b x)}{16 b}-\frac {3 \cos (a+b x)}{8 b}-\frac {\cos (3 a-2 c+b x)}{16 b}+\frac {\cos (3 a+3 b x)}{24 b}+\frac {\cos (a+2 c+3 b x)}{16 b}-\frac {\cos (3 a+2 c+5 b x)}{80 b} \] Output:
-3/16*cos(-b*x+a-2*c)/b-3/8*cos(b*x+a)/b-1/16*cos(b*x+3*a-2*c)/b+1/24*cos( 3*b*x+3*a)/b+1/16*cos(3*b*x+a+2*c)/b-1/80*cos(5*b*x+3*a+2*c)/b
Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \sin ^3(a+b x) \sin ^2(c+b x) \, dx=-\frac {45 \cos (a-2 c-b x)+90 \cos (a+b x)-10 \cos (3 (a+b x))+15 \cos (3 a-2 c+b x)-15 \cos (a+2 c+3 b x)+3 \cos (3 a+2 c+5 b x)}{240 b} \] Input:
Integrate[Sin[a + b*x]^3*Sin[c + b*x]^2,x]
Output:
-1/240*(45*Cos[a - 2*c - b*x] + 90*Cos[a + b*x] - 10*Cos[3*(a + b*x)] + 15 *Cos[3*a - 2*c + b*x] - 15*Cos[a + 2*c + 3*b*x] + 3*Cos[3*a + 2*c + 5*b*x] )/b
Time = 0.28 (sec) , antiderivative size = 101, 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.
\(\displaystyle \int \sin ^3(a+b x) \sin ^2(b x+c) \, dx\) |
\(\Big \downarrow \) 5080 |
\(\displaystyle \int \left (-\frac {3}{16} \sin (a-b x-2 c)+\frac {1}{16} \sin (3 a+b x-2 c)-\frac {3}{16} \sin (a+3 b x+2 c)+\frac {1}{16} \sin (3 a+5 b x+2 c)+\frac {3}{8} \sin (a+b x)-\frac {1}{8} \sin (3 a+3 b x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 \cos (a-b x-2 c)}{16 b}-\frac {\cos (3 a+b x-2 c)}{16 b}+\frac {\cos (a+3 b x+2 c)}{16 b}-\frac {\cos (3 a+5 b x+2 c)}{80 b}-\frac {3 \cos (a+b x)}{8 b}+\frac {\cos (3 a+3 b x)}{24 b}\) |
Input:
Int[Sin[a + b*x]^3*Sin[c + b*x]^2,x]
Output:
(-3*Cos[a - 2*c - b*x])/(16*b) - (3*Cos[a + b*x])/(8*b) - Cos[3*a - 2*c + b*x]/(16*b) + Cos[3*a + 3*b*x]/(24*b) + Cos[a + 2*c + 3*b*x]/(16*b) - Cos[ 3*a + 2*c + 5*b*x]/(80*b)
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]
Time = 3.81 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {3 \cos \left (-b x +a -2 c \right )}{16 b}-\frac {3 \cos \left (b x +a \right )}{8 b}-\frac {\cos \left (b x +3 a -2 c \right )}{16 b}+\frac {\cos \left (3 b x +3 a \right )}{24 b}+\frac {\cos \left (3 b x +a +2 c \right )}{16 b}-\frac {\cos \left (5 b x +3 a +2 c \right )}{80 b}\) | \(90\) |
risch | \(-\frac {3 \cos \left (-b x +a -2 c \right )}{16 b}-\frac {3 \cos \left (b x +a \right )}{8 b}-\frac {\cos \left (b x +3 a -2 c \right )}{16 b}+\frac {\cos \left (3 b x +3 a \right )}{24 b}+\frac {\cos \left (3 b x +a +2 c \right )}{16 b}-\frac {\cos \left (5 b x +3 a +2 c \right )}{80 b}\) | \(90\) |
parallelrisch | \(\frac {-16 \cos \left (2 a -2 c \right )-90 \cos \left (b x +a \right )-45 \cos \left (-b x +a -2 c \right )-15 \cos \left (b x +3 a -2 c \right )+10 \cos \left (3 b x +3 a \right )+15 \cos \left (3 b x +a +2 c \right )-3 \cos \left (5 b x +3 a +2 c \right )-128 \cos \left (a -c \right )+16}{240 b}\) | \(96\) |
orering | \(-\frac {259 \left (3 \sin \left (b x +a \right )^{2} \sin \left (b x +c \right )^{2} b \cos \left (b x +a \right )+2 \sin \left (b x +a \right )^{3} \sin \left (b x +c \right ) b \cos \left (b x +c \right )\right )}{225 b^{2}}-\frac {7 \left (6 b^{3} \cos \left (b x +a \right )^{3} \sin \left (b x +c \right )^{2}+36 \sin \left (b x +a \right ) \sin \left (b x +c \right ) b^{3} \cos \left (b x +a \right )^{2} \cos \left (b x +c \right )-39 \sin \left (b x +a \right )^{2} \sin \left (b x +c \right )^{2} b^{3} \cos \left (b x +a \right )+18 \sin \left (b x +a \right )^{2} b^{3} \cos \left (b x +c \right )^{2} \cos \left (b x +a \right )-26 \sin \left (b x +a \right )^{3} \sin \left (b x +c \right ) b^{3} \cos \left (b x +c \right )\right )}{45 b^{4}}-\frac {723 b^{5} \cos \left (b x +a \right ) \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )^{2}-1080 b^{5} \cos \left (b x +a \right )^{2} \sin \left (b x +c \right ) \sin \left (b x +a \right ) \cos \left (b x +c \right )-180 b^{5} \cos \left (b x +a \right )^{3} \sin \left (b x +c \right )^{2}+120 b^{5} \cos \left (b x +a \right )^{3} \cos \left (b x +c \right )^{2}-540 \sin \left (b x +a \right )^{2} b^{5} \cos \left (b x +c \right )^{2} \cos \left (b x +a \right )+482 \sin \left (b x +a \right )^{3} \sin \left (b x +c \right ) b^{5} \cos \left (b x +c \right )}{225 b^{6}}\) | \(351\) |
Input:
int(sin(b*x+a)^3*sin(b*x+c)^2,x,method=_RETURNVERBOSE)
Output:
-3/16*cos(-b*x+a-2*c)/b-3/8*cos(b*x+a)/b-1/16*cos(b*x+3*a-2*c)/b+1/24*cos( 3*b*x+3*a)/b+1/16*cos(3*b*x+a+2*c)/b-1/80*cos(5*b*x+3*a+2*c)/b
Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.48 \[ \int \sin ^3(a+b x) \sin ^2(c+b x) \, dx=-\frac {3 \, {\left (4 \, \cos \left (-a + c\right )^{3} - 3 \, \cos \left (-a + c\right )\right )} \cos \left (b x + c\right )^{5} - 5 \, {\left (5 \, \cos \left (-a + c\right )^{3} - 3 \, \cos \left (-a + c\right )\right )} \cos \left (b x + c\right )^{3} + 15 \, \cos \left (b x + c\right ) \cos \left (-a + c\right )^{3} + {\left (3 \, {\left (4 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{4} - {\left (19 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{2} + 7 \, \cos \left (-a + c\right )^{2} + 2\right )} \sin \left (b x + c\right ) \sin \left (-a + c\right )}{15 \, b} \] Input:
integrate(sin(b*x+a)^3*sin(b*x+c)^2,x, algorithm="fricas")
Output:
-1/15*(3*(4*cos(-a + c)^3 - 3*cos(-a + c))*cos(b*x + c)^5 - 5*(5*cos(-a + c)^3 - 3*cos(-a + c))*cos(b*x + c)^3 + 15*cos(b*x + c)*cos(-a + c)^3 + (3* (4*cos(-a + c)^2 - 1)*cos(b*x + c)^4 - (19*cos(-a + c)^2 - 1)*cos(b*x + c) ^2 + 7*cos(-a + c)^2 + 2)*sin(b*x + c)*sin(-a + c))/b
Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (87) = 174\).
Time = 2.16 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.73 \[ \int \sin ^3(a+b x) \sin ^2(c+b x) \, dx=\begin {cases} - \frac {2 \sin ^{3}{\left (a + b x \right )} \sin {\left (b x + c \right )} \cos {\left (b x + c \right )}}{5 b} - \frac {3 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (b x + c \right )} \cos {\left (a + b x \right )}}{5 b} - \frac {2 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{5 b} - \frac {4 \sin {\left (a + b x \right )} \sin {\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (b x + c \right )}}{5 b} - \frac {2 \sin ^{2}{\left (b x + c \right )} \cos ^{3}{\left (a + b x \right )}}{15 b} - \frac {8 \cos ^{3}{\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{15 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \sin ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(b*x+a)**3*sin(b*x+c)**2,x)
Output:
Piecewise((-2*sin(a + b*x)**3*sin(b*x + c)*cos(b*x + c)/(5*b) - 3*sin(a + b*x)**2*sin(b*x + c)**2*cos(a + b*x)/(5*b) - 2*sin(a + b*x)**2*cos(a + b*x )*cos(b*x + c)**2/(5*b) - 4*sin(a + b*x)*sin(b*x + c)*cos(a + b*x)**2*cos( b*x + c)/(5*b) - 2*sin(b*x + c)**2*cos(a + b*x)**3/(15*b) - 8*cos(a + b*x) **3*cos(b*x + c)**2/(15*b), Ne(b, 0)), (x*sin(a)**3*sin(c)**2, True))
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \sin ^3(a+b x) \sin ^2(c+b x) \, dx=-\frac {3 \, \cos \left (5 \, b x + 3 \, a + 2 \, c\right ) - 10 \, \cos \left (3 \, b x + 3 \, a\right ) - 15 \, \cos \left (3 \, b x + a + 2 \, c\right ) + 15 \, \cos \left (b x + 3 \, a - 2 \, c\right ) + 90 \, \cos \left (b x + a\right ) + 45 \, \cos \left (b x - a + 2 \, c\right )}{240 \, b} \] Input:
integrate(sin(b*x+a)^3*sin(b*x+c)^2,x, algorithm="maxima")
Output:
-1/240*(3*cos(5*b*x + 3*a + 2*c) - 10*cos(3*b*x + 3*a) - 15*cos(3*b*x + a + 2*c) + 15*cos(b*x + 3*a - 2*c) + 90*cos(b*x + a) + 45*cos(b*x - a + 2*c) )/b
Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88 \[ \int \sin ^3(a+b x) \sin ^2(c+b x) \, dx=-\frac {\cos \left (5 \, b x + 3 \, a + 2 \, c\right )}{80 \, b} + \frac {\cos \left (3 \, b x + 3 \, a\right )}{24 \, b} + \frac {\cos \left (3 \, b x + a + 2 \, c\right )}{16 \, b} - \frac {\cos \left (b x + 3 \, a - 2 \, c\right )}{16 \, b} - \frac {3 \, \cos \left (b x + a\right )}{8 \, b} - \frac {3 \, \cos \left (-b x + a - 2 \, c\right )}{16 \, b} \] Input:
integrate(sin(b*x+a)^3*sin(b*x+c)^2,x, algorithm="giac")
Output:
-1/80*cos(5*b*x + 3*a + 2*c)/b + 1/24*cos(3*b*x + 3*a)/b + 1/16*cos(3*b*x + a + 2*c)/b - 1/16*cos(b*x + 3*a - 2*c)/b - 3/8*cos(b*x + a)/b - 3/16*cos (-b*x + a - 2*c)/b
Time = 1.64 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \sin ^3(a+b x) \sin ^2(c+b x) \, dx=-\frac {90\,\cos \left (a+b\,x\right )-15\,\cos \left (a+2\,c+3\,b\,x\right )+45\,\cos \left (2\,c-a+b\,x\right )+15\,\cos \left (3\,a-2\,c+b\,x\right )+3\,\cos \left (3\,a+2\,c+5\,b\,x\right )-10\,\cos \left (3\,a+3\,b\,x\right )}{240\,b} \] Input:
int(sin(a + b*x)^3*sin(c + b*x)^2,x)
Output:
-(90*cos(a + b*x) - 15*cos(a + 2*c + 3*b*x) + 45*cos(2*c - a + b*x) + 15*c os(3*a - 2*c + b*x) + 3*cos(3*a + 2*c + 5*b*x) - 10*cos(3*a + 3*b*x))/(240 *b)
Time = 0.16 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.75 \[ \int \sin ^3(a+b x) \sin ^2(c+b x) \, dx=\frac {-12 \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +c \right ) \sin \left (b x +a \right )+6 \cos \left (b x +c \right ) \sin \left (b x +c \right ) \sin \left (b x +a \right )^{3}-12 \cos \left (b x +c \right ) \sin \left (b x +c \right ) \sin \left (b x +a \right )-9 \cos \left (b x +a \right ) \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )^{2}+6 \cos \left (b x +a \right ) \sin \left (b x +c \right )^{2}+2 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}-8 \cos \left (b x +a \right )-12 \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )^{2}+6 \sin \left (b x +c \right )^{2}+6 \sin \left (b x +a \right )^{2}+32}{15 b} \] Input:
int(sin(b*x+a)^3*sin(b*x+c)^2,x)
Output:
( - 12*cos(b*x + c)*cos(a + b*x)*sin(b*x + c)*sin(a + b*x) + 6*cos(b*x + c )*sin(b*x + c)*sin(a + b*x)**3 - 12*cos(b*x + c)*sin(b*x + c)*sin(a + b*x) - 9*cos(a + b*x)*sin(b*x + c)**2*sin(a + b*x)**2 + 6*cos(a + b*x)*sin(b*x + c)**2 + 2*cos(a + b*x)*sin(a + b*x)**2 - 8*cos(a + b*x) - 12*sin(b*x + c)**2*sin(a + b*x)**2 + 6*sin(b*x + c)**2 + 6*sin(a + b*x)**2 + 32)/(15*b)