Integrand size = 17, antiderivative size = 103 \[ \int \sin ^2(a+b x) \sin ^3(c+b x) \, dx=-\frac {\cos (2 a-3 c-b x)}{16 b}-\frac {3 \cos (2 a-c+b x)}{16 b}-\frac {3 \cos (c+b x)}{8 b}+\frac {\cos (2 a+c+3 b x)}{16 b}+\frac {\cos (3 c+3 b x)}{24 b}-\frac {\cos (2 a+3 c+5 b x)}{80 b} \] Output:
-1/16*cos(-b*x+2*a-3*c)/b-3/16*cos(b*x+2*a-c)/b-3/8*cos(b*x+c)/b+1/16*cos( 3*b*x+2*a+c)/b+1/24*cos(3*b*x+3*c)/b-1/80*cos(5*b*x+2*a+3*c)/b
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int \sin ^2(a+b x) \sin ^3(c+b x) \, dx=-\frac {15 \cos (2 a-3 c-b x)+45 \cos (2 a-c+b x)+90 \cos (c+b x)-10 \cos (3 (c+b x))-15 \cos (2 a+c+3 b x)+3 \cos (2 a+3 c+5 b x)}{240 b} \] Input:
Integrate[Sin[a + b*x]^2*Sin[c + b*x]^3,x]
Output:
-1/240*(15*Cos[2*a - 3*c - b*x] + 45*Cos[2*a - c + b*x] + 90*Cos[c + b*x] - 10*Cos[3*(c + b*x)] - 15*Cos[2*a + c + 3*b*x] + 3*Cos[2*a + 3*c + 5*b*x] )/b
Time = 0.27 (sec) , antiderivative size = 103, 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.
\(\displaystyle \int \sin ^2(a+b x) \sin ^3(b x+c) \, dx\) |
\(\Big \downarrow \) 5080 |
\(\displaystyle \int \left (-\frac {1}{16} \sin (2 a-b x-3 c)+\frac {3}{16} \sin (2 a+b x-c)-\frac {3}{16} \sin (2 a+3 b x+c)+\frac {1}{16} \sin (2 a+5 b x+3 c)+\frac {3}{8} \sin (b x+c)-\frac {1}{8} \sin (3 b x+3 c)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\cos (2 a-b x-3 c)}{16 b}-\frac {3 \cos (2 a+b x-c)}{16 b}+\frac {\cos (2 a+3 b x+c)}{16 b}-\frac {\cos (2 a+5 b x+3 c)}{80 b}-\frac {3 \cos (b x+c)}{8 b}+\frac {\cos (3 b x+3 c)}{24 b}\) |
Input:
Int[Sin[a + b*x]^2*Sin[c + b*x]^3,x]
Output:
-1/16*Cos[2*a - 3*c - b*x]/b - (3*Cos[2*a - c + b*x])/(16*b) - (3*Cos[c + b*x])/(8*b) + Cos[2*a + c + 3*b*x]/(16*b) + Cos[3*c + 3*b*x]/(24*b) - Cos[ 2*a + 3*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 = 3.82 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {\cos \left (-b x +2 a -3 c \right )}{16 b}-\frac {3 \cos \left (b x +2 a -c \right )}{16 b}-\frac {3 \cos \left (b x +c \right )}{8 b}+\frac {\cos \left (3 b x +2 a +c \right )}{16 b}+\frac {\cos \left (3 b x +3 c \right )}{24 b}-\frac {\cos \left (5 b x +2 a +3 c \right )}{80 b}\) | \(92\) |
risch | \(-\frac {\cos \left (-b x +2 a -3 c \right )}{16 b}-\frac {3 \cos \left (b x +2 a -c \right )}{16 b}-\frac {3 \cos \left (b x +c \right )}{8 b}+\frac {\cos \left (3 b x +2 a +c \right )}{16 b}+\frac {\cos \left (3 b x +3 c \right )}{24 b}-\frac {\cos \left (5 b x +2 a +3 c \right )}{80 b}\) | \(92\) |
parallelrisch | \(\frac {-\frac {16}{15}+\frac {16 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{6}}{15}-\frac {64 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{5} \tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{15}+\frac {16 \left (6 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}+1\right ) \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{4}}{15}+\frac {64 \left (-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{3}}{15}+\frac {16 \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}-9 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right ) \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2}}{15}+\frac {32 \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-3 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \tan \left (\frac {b x}{2}+\frac {c}{2}\right )}{15}+\frac {16 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}}{15}}{b \left (1+\tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{2}}\) | \(218\) |
orering | \(-\frac {259 \left (2 \sin \left (b x +a \right ) \sin \left (b x +c \right )^{3} b \cos \left (b x +a \right )+3 \sin \left (b x +a \right )^{2} \sin \left (b x +c \right )^{2} b \cos \left (b x +c \right )\right )}{225 b^{2}}-\frac {7 \left (-26 b^{3} \cos \left (b x +a \right ) \sin \left (b x +c \right )^{3} \sin \left (b x +a \right )+18 b^{3} \cos \left (b x +a \right )^{2} \sin \left (b x +c \right )^{2} \cos \left (b x +c \right )+36 \sin \left (b x +a \right ) \sin \left (b x +c \right ) b^{3} \cos \left (b x +a \right ) \cos \left (b x +c \right )^{2}-39 \sin \left (b x +a \right )^{2} \sin \left (b x +c \right )^{2} b^{3} \cos \left (b x +c \right )+6 \sin \left (b x +a \right )^{2} b^{3} \cos \left (b x +c \right )^{3}\right )}{45 b^{4}}-\frac {482 b^{5} \sin \left (b x +a \right ) \sin \left (b x +c \right )^{3} \cos \left (b x +a \right )+723 b^{5} \sin \left (b x +a \right )^{2} \sin \left (b x +c \right )^{2} \cos \left (b x +c \right )-1080 b^{5} \cos \left (b x +a \right ) \sin \left (b x +c \right ) \sin \left (b x +a \right ) \cos \left (b x +c \right )^{2}-540 b^{5} \cos \left (b x +a \right )^{2} \sin \left (b x +c \right )^{2} \cos \left (b x +c \right )+120 b^{5} \cos \left (b x +a \right )^{2} \cos \left (b x +c \right )^{3}-180 \sin \left (b x +a \right )^{2} b^{5} \cos \left (b x +c \right )^{3}}{225 b^{6}}\) | \(351\) |
Input:
int(sin(b*x+a)^2*sin(b*x+c)^3,x,method=_RETURNVERBOSE)
Output:
-1/16*cos(-b*x+2*a-3*c)/b-3/16*cos(b*x+2*a-c)/b-3/8*cos(b*x+c)/b+1/16*cos( 3*b*x+2*a+c)/b+1/24*cos(3*b*x+3*c)/b-1/80*cos(5*b*x+2*a+3*c)/b
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.15 \[ \int \sin ^2(a+b x) \sin ^3(c+b x) \, dx=-\frac {3 \, {\left (2 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{5} - 5 \, {\left (3 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{3} + 15 \, \cos \left (b x + c\right ) \cos \left (-a + c\right )^{2} + 6 \, {\left (\cos \left (b x + c\right )^{4} \cos \left (-a + c\right ) - 2 \, \cos \left (b x + c\right )^{2} \cos \left (-a + c\right ) + \cos \left (-a + c\right )\right )} \sin \left (b x + c\right ) \sin \left (-a + c\right )}{15 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(b*x+c)^3,x, algorithm="fricas")
Output:
-1/15*(3*(2*cos(-a + c)^2 - 1)*cos(b*x + c)^5 - 5*(3*cos(-a + c)^2 - 1)*co s(b*x + c)^3 + 15*cos(b*x + c)*cos(-a + c)^2 + 6*(cos(b*x + c)^4*cos(-a + c) - 2*cos(b*x + c)^2*cos(-a + c) + cos(-a + c))*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 = 2.12 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.70 \[ \int \sin ^2(a+b x) \sin ^3(c+b x) \, dx=\begin {cases} - \frac {3 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (b x + c \right )} \cos {\left (b x + c \right )}}{5 b} - \frac {2 \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (b x + c \right )}}{15 b} - \frac {2 \sin {\left (a + b x \right )} \sin ^{3}{\left (b x + c \right )} \cos {\left (a + b x \right )}}{5 b} - \frac {4 \sin {\left (a + b x \right )} \sin {\left (b x + c \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{5 b} - \frac {2 \sin ^{2}{\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (b x + c \right )}}{5 b} - \frac {8 \cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (b x + c \right )}}{15 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} \sin ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(b*x+a)**2*sin(b*x+c)**3,x)
Output:
Piecewise((-3*sin(a + b*x)**2*sin(b*x + c)**2*cos(b*x + c)/(5*b) - 2*sin(a + b*x)**2*cos(b*x + c)**3/(15*b) - 2*sin(a + b*x)*sin(b*x + c)**3*cos(a + b*x)/(5*b) - 4*sin(a + b*x)*sin(b*x + c)*cos(a + b*x)*cos(b*x + c)**2/(5* b) - 2*sin(b*x + c)**2*cos(a + b*x)**2*cos(b*x + c)/(5*b) - 8*cos(a + b*x) **2*cos(b*x + c)**3/(15*b), Ne(b, 0)), (x*sin(a)**2*sin(c)**3, True))
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \sin ^2(a+b x) \sin ^3(c+b x) \, dx=-\frac {3 \, \cos \left (5 \, b x + 2 \, a + 3 \, c\right ) - 15 \, \cos \left (3 \, b x + 2 \, a + c\right ) - 10 \, \cos \left (3 \, b x + 3 \, c\right ) + 45 \, \cos \left (b x + 2 \, a - c\right ) + 15 \, \cos \left (b x - 2 \, a + 3 \, c\right ) + 90 \, \cos \left (b x + c\right )}{240 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(b*x+c)^3,x, algorithm="maxima")
Output:
-1/240*(3*cos(5*b*x + 2*a + 3*c) - 15*cos(3*b*x + 2*a + c) - 10*cos(3*b*x + 3*c) + 45*cos(b*x + 2*a - c) + 15*cos(b*x - 2*a + 3*c) + 90*cos(b*x + c) )/b
Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \sin ^2(a+b x) \sin ^3(c+b x) \, dx=-\frac {\cos \left (5 \, b x + 2 \, a + 3 \, c\right )}{80 \, b} + \frac {\cos \left (3 \, b x + 2 \, a + c\right )}{16 \, b} + \frac {\cos \left (3 \, b x + 3 \, c\right )}{24 \, b} - \frac {3 \, \cos \left (b x + 2 \, a - c\right )}{16 \, b} - \frac {3 \, \cos \left (b x + c\right )}{8 \, b} - \frac {\cos \left (-b x + 2 \, a - 3 \, c\right )}{16 \, b} \] Input:
integrate(sin(b*x+a)^2*sin(b*x+c)^3,x, algorithm="giac")
Output:
-1/80*cos(5*b*x + 2*a + 3*c)/b + 1/16*cos(3*b*x + 2*a + c)/b + 1/24*cos(3* b*x + 3*c)/b - 3/16*cos(b*x + 2*a - c)/b - 3/8*cos(b*x + c)/b - 1/16*cos(- b*x + 2*a - 3*c)/b
Time = 1.84 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \sin ^2(a+b x) \sin ^3(c+b x) \, dx=-\frac {90\,\cos \left (c+b\,x\right )-15\,\cos \left (2\,a+c+3\,b\,x\right )+45\,\cos \left (2\,a-c+b\,x\right )+15\,\cos \left (3\,c-2\,a+b\,x\right )+3\,\cos \left (2\,a+3\,c+5\,b\,x\right )-10\,\cos \left (3\,c+3\,b\,x\right )}{240\,b} \] Input:
int(sin(a + b*x)^2*sin(c + b*x)^3,x)
Output:
-(90*cos(c + b*x) - 15*cos(2*a + c + 3*b*x) + 45*cos(2*a - c + b*x) + 15*c os(3*c - 2*a + b*x) + 3*cos(2*a + 3*c + 5*b*x) - 10*cos(3*c + 3*b*x))/(240 *b)
Time = 0.16 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.37 \[ \int \sin ^2(a+b x) \sin ^3(c+b x) \, dx=\frac {-12 \cos \left (b x +c \right ) \cos \left (b x +a \right )-9 \cos \left (b x +c \right ) \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )^{2}+2 \cos \left (b x +c \right ) \sin \left (b x +c \right )^{2}+6 \cos \left (b x +c \right ) \sin \left (b x +a \right )^{2}-8 \cos \left (b x +c \right )+6 \cos \left (b x +a \right ) \sin \left (b x +c \right )^{3} \sin \left (b x +a \right )-12 \cos \left (b x +a \right ) \sin \left (b x +c \right ) \sin \left (b x +a \right )-12 \sin \left (b x +c \right ) \sin \left (b x +a \right )-4}{15 b} \] Input:
int(sin(b*x+a)^2*sin(b*x+c)^3,x)
Output:
( - 12*cos(b*x + c)*cos(a + b*x) - 9*cos(b*x + c)*sin(b*x + c)**2*sin(a + b*x)**2 + 2*cos(b*x + c)*sin(b*x + c)**2 + 6*cos(b*x + c)*sin(a + b*x)**2 - 8*cos(b*x + c) + 6*cos(a + b*x)*sin(b*x + c)**3*sin(a + b*x) - 12*cos(a + b*x)*sin(b*x + c)*sin(a + b*x) - 12*sin(b*x + c)*sin(a + b*x) - 4)/(15*b )