Integrand size = 17, antiderivative size = 103 \[ \int \cos ^3(c+b x) \sin ^2(a+b x) \, dx=\frac {\sin (2 a-3 c-b x)}{16 b}-\frac {3 \sin (2 a-c+b x)}{16 b}+\frac {3 \sin (c+b x)}{8 b}-\frac {\sin (2 a+c+3 b x)}{16 b}+\frac {\sin (3 c+3 b x)}{24 b}-\frac {\sin (2 a+3 c+5 b x)}{80 b} \] Output:
1/16*sin(-b*x+2*a-3*c)/b-3/16*sin(b*x+2*a-c)/b+3/8*sin(b*x+c)/b-1/16*sin(3 *b*x+2*a+c)/b+1/24*sin(3*b*x+3*c)/b-1/80*sin(5*b*x+2*a+3*c)/b
Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+b x) \sin ^2(a+b x) \, dx=\frac {15 \sin (2 a-3 c-b x)-45 \sin (2 a-c+b x)+90 \sin (c+b x)+10 \sin (3 (c+b x))-15 \sin (2 a+c+3 b x)-3 \sin (2 a+3 c+5 b x)}{240 b} \] Input:
Integrate[Cos[c + b*x]^3*Sin[a + b*x]^2,x]
Output:
(15*Sin[2*a - 3*c - b*x] - 45*Sin[2*a - c + b*x] + 90*Sin[c + b*x] + 10*Si n[3*(c + b*x)] - 15*Sin[2*a + c + 3*b*x] - 3*Sin[2*a + 3*c + 5*b*x])/(240* 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 = {5085, 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.
\(\displaystyle \int \sin ^2(a+b x) \cos ^3(b x+c) \, dx\) |
\(\Big \downarrow \) 5085 |
\(\displaystyle \int \left (-\frac {1}{16} \cos (2 a-b x-3 c)-\frac {3}{16} \cos (2 a+b x-c)-\frac {3}{16} \cos (2 a+3 b x+c)-\frac {1}{16} \cos (2 a+5 b x+3 c)+\frac {3}{8} \cos (b x+c)+\frac {1}{8} \cos (3 b x+3 c)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sin (2 a-b x-3 c)}{16 b}-\frac {3 \sin (2 a+b x-c)}{16 b}-\frac {\sin (2 a+3 b x+c)}{16 b}-\frac {\sin (2 a+5 b x+3 c)}{80 b}+\frac {3 \sin (b x+c)}{8 b}+\frac {\sin (3 b x+3 c)}{24 b}\) |
Input:
Int[Cos[c + b*x]^3*Sin[a + b*x]^2,x]
Output:
Sin[2*a - 3*c - b*x]/(16*b) - (3*Sin[2*a - c + b*x])/(16*b) + (3*Sin[c + b *x])/(8*b) - Sin[2*a + c + 3*b*x]/(16*b) + Sin[3*c + 3*b*x]/(24*b) - Sin[2 *a + 3*c + 5*b*x]/(80*b)
Int[Cos[w_]^(q_.)*Sin[v_]^(p_.), x_Symbol] :> Int[ExpandTrigReduce[Sin[v]^p *Cos[w]^q, x], x] /; IGtQ[p, 0] && IGtQ[q, 0] && ((PolynomialQ[v, x] && Pol ynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w], x]))
Time = 3.73 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.89
method | result | size |
default | \(\frac {\sin \left (-b x +2 a -3 c \right )}{16 b}-\frac {3 \sin \left (b x +2 a -c \right )}{16 b}+\frac {3 \sin \left (b x +c \right )}{8 b}-\frac {\sin \left (3 b x +2 a +c \right )}{16 b}+\frac {\sin \left (3 b x +3 c \right )}{24 b}-\frac {\sin \left (5 b x +2 a +3 c \right )}{80 b}\) | \(92\) |
risch | \(\frac {\sin \left (-b x +2 a -3 c \right )}{16 b}-\frac {3 \sin \left (b x +2 a -c \right )}{16 b}+\frac {3 \sin \left (b x +c \right )}{8 b}-\frac {\sin \left (3 b x +2 a +c \right )}{16 b}+\frac {\sin \left (3 b x +3 c \right )}{24 b}-\frac {\sin \left (5 b x +2 a +3 c \right )}{80 b}\) | \(92\) |
parallelrisch | \(\frac {\left (80 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{3}+64 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+\left (-240 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{4}-80 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2}+64\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}+\left (240 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{5}-160 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{3}-144 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}+\left (-40 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{6}+200 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{4}+120 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2}-24\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )-40 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{5}+24 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )}{15 b \left (1+\tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{2}}\) | \(239\) |
orering | \(-\frac {259 \left (-3 \cos \left (b x +c \right )^{2} \sin \left (b x +a \right )^{2} b \sin \left (b x +c \right )+2 \cos \left (b x +c \right )^{3} \sin \left (b x +a \right ) b \cos \left (b x +a \right )\right )}{225 b^{2}}-\frac {7 \left (-6 b^{3} \sin \left (b x +c \right )^{3} \sin \left (b x +a \right )^{2}+36 \cos \left (b x +c \right ) \sin \left (b x +a \right ) b^{3} \sin \left (b x +c \right )^{2} \cos \left (b x +a \right )+39 \cos \left (b x +c \right )^{2} \sin \left (b x +a \right )^{2} b^{3} \sin \left (b x +c \right )-18 \cos \left (b x +c \right )^{2} b^{3} \cos \left (b x +a \right )^{2} \sin \left (b x +c \right )-26 \cos \left (b x +c \right )^{3} \sin \left (b x +a \right ) b^{3} \cos \left (b x +a \right )\right )}{45 b^{4}}-\frac {-723 b^{5} \sin \left (b x +c \right ) \sin \left (b x +a \right )^{2} \cos \left (b x +c \right )^{2}-1080 b^{5} \sin \left (b x +c \right )^{2} \sin \left (b x +a \right ) \cos \left (b x +c \right ) \cos \left (b x +a \right )+180 b^{5} \sin \left (b x +c \right )^{3} \sin \left (b x +a \right )^{2}-120 b^{5} \sin \left (b x +c \right )^{3} \cos \left (b x +a \right )^{2}+540 \cos \left (b x +c \right )^{2} b^{5} \cos \left (b x +a \right )^{2} \sin \left (b x +c \right )+482 \cos \left (b x +c \right )^{3} \sin \left (b x +a \right ) b^{5} \cos \left (b x +a \right )}{225 b^{6}}\) | \(351\) |
Input:
int(cos(b*x+c)^3*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/16*sin(-b*x+2*a-3*c)/b-3/16*sin(b*x+2*a-c)/b+3/8*sin(b*x+c)/b-1/16*sin(3 *b*x+2*a+c)/b+1/24*sin(3*b*x+3*c)/b-1/80*sin(5*b*x+2*a+3*c)/b
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \cos ^3(c+b x) \sin ^2(a+b x) \, dx=\frac {6 \, \cos \left (b x + c\right )^{5} \cos \left (-a + c\right ) \sin \left (-a + c\right ) - {\left (3 \, {\left (2 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{4} + {\left (3 \, \cos \left (-a + c\right )^{2} - 4\right )} \cos \left (b x + c\right )^{2} + 6 \, \cos \left (-a + c\right )^{2} - 8\right )} \sin \left (b x + c\right )}{15 \, b} \] Input:
integrate(cos(b*x+c)^3*sin(b*x+a)^2,x, algorithm="fricas")
Output:
1/15*(6*cos(b*x + c)^5*cos(-a + c)*sin(-a + c) - (3*(2*cos(-a + c)^2 - 1)* cos(b*x + c)^4 + (3*cos(-a + c)^2 - 4)*cos(b*x + c)^2 + 6*cos(-a + c)^2 - 8)*sin(b*x + c))/b
Time = 2.12 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.68 \[ \int \cos ^3(c+b x) \sin ^2(a+b x) \, dx=\begin {cases} \frac {2 \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (b x + c \right )}}{15 b} + \frac {3 \sin ^{2}{\left (a + b x \right )} \sin {\left (b x + c \right )} \cos ^{2}{\left (b x + c \right )}}{5 b} - \frac {4 \sin {\left (a + b x \right )} \sin ^{2}{\left (b x + c \right )} \cos {\left (a + b x \right )} \cos {\left (b x + c \right )}}{5 b} - \frac {2 \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (b x + c \right )}}{5 b} + \frac {8 \sin ^{3}{\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )}}{15 b} + \frac {2 \sin {\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{5 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+c)**3*sin(b*x+a)**2,x)
Output:
Piecewise((2*sin(a + b*x)**2*sin(b*x + c)**3/(15*b) + 3*sin(a + b*x)**2*si n(b*x + c)*cos(b*x + c)**2/(5*b) - 4*sin(a + b*x)*sin(b*x + c)**2*cos(a + b*x)*cos(b*x + c)/(5*b) - 2*sin(a + b*x)*cos(a + b*x)*cos(b*x + c)**3/(5*b ) + 8*sin(b*x + c)**3*cos(a + b*x)**2/(15*b) + 2*sin(b*x + c)*cos(a + b*x) **2*cos(b*x + c)**2/(5*b), Ne(b, 0)), (x*sin(a)**2*cos(c)**3, True))
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \cos ^3(c+b x) \sin ^2(a+b x) \, dx=-\frac {3 \, \sin \left (5 \, b x + 2 \, a + 3 \, c\right ) + 15 \, \sin \left (3 \, b x + 2 \, a + c\right ) - 10 \, \sin \left (3 \, b x + 3 \, c\right ) + 45 \, \sin \left (b x + 2 \, a - c\right ) + 15 \, \sin \left (b x - 2 \, a + 3 \, c\right ) - 90 \, \sin \left (b x + c\right )}{240 \, b} \] Input:
integrate(cos(b*x+c)^3*sin(b*x+a)^2,x, algorithm="maxima")
Output:
-1/240*(3*sin(5*b*x + 2*a + 3*c) + 15*sin(3*b*x + 2*a + c) - 10*sin(3*b*x + 3*c) + 45*sin(b*x + 2*a - c) + 15*sin(b*x - 2*a + 3*c) - 90*sin(b*x + c) )/b
Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \cos ^3(c+b x) \sin ^2(a+b x) \, dx=-\frac {\sin \left (5 \, b x + 2 \, a + 3 \, c\right )}{80 \, b} - \frac {\sin \left (3 \, b x + 2 \, a + c\right )}{16 \, b} + \frac {\sin \left (3 \, b x + 3 \, c\right )}{24 \, b} - \frac {3 \, \sin \left (b x + 2 \, a - c\right )}{16 \, b} + \frac {3 \, \sin \left (b x + c\right )}{8 \, b} + \frac {\sin \left (-b x + 2 \, a - 3 \, c\right )}{16 \, b} \] Input:
integrate(cos(b*x+c)^3*sin(b*x+a)^2,x, algorithm="giac")
Output:
-1/80*sin(5*b*x + 2*a + 3*c)/b - 1/16*sin(3*b*x + 2*a + c)/b + 1/24*sin(3* b*x + 3*c)/b - 3/16*sin(b*x + 2*a - c)/b + 3/8*sin(b*x + c)/b + 1/16*sin(- b*x + 2*a - 3*c)/b
Time = 1.75 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \cos ^3(c+b x) \sin ^2(a+b x) \, dx=-\frac {15\,\sin \left (2\,a+c+3\,b\,x\right )-90\,\sin \left (c+b\,x\right )+45\,\sin \left (2\,a-c+b\,x\right )+15\,\sin \left (3\,c-2\,a+b\,x\right )+3\,\sin \left (2\,a+3\,c+5\,b\,x\right )-10\,\sin \left (3\,c+3\,b\,x\right )}{240\,b} \] Input:
int(cos(c + b*x)^3*sin(a + b*x)^2,x)
Output:
-(15*sin(2*a + c + 3*b*x) - 90*sin(c + b*x) + 45*sin(2*a - c + b*x) + 15*s in(3*c - 2*a + b*x) + 3*sin(2*a + 3*c + 5*b*x) - 10*sin(3*c + 3*b*x))/(240 *b)
Time = 0.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.30 \[ \int \cos ^3(c+b x) \sin ^2(a+b x) \, dx=\frac {-6 \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )-6 \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +a \right )-30 \cos \left (b x +c \right ) \sin \left (b x +a \right )+30 \cos \left (b x +a \right ) \sin \left (b x +c \right )-9 \sin \left (b x +c \right )^{3} \sin \left (b x +a \right )^{2}+2 \sin \left (b x +c \right )^{3}+3 \sin \left (b x +c \right ) \sin \left (b x +a \right )^{2}+6 \sin \left (b x +c \right )}{15 b} \] Input:
int(cos(b*x+c)^3*sin(b*x+a)^2,x)
Output:
( - 6*cos(b*x + c)*cos(a + b*x)*sin(b*x + c)**2*sin(a + b*x) - 6*cos(b*x + c)*cos(a + b*x)*sin(a + b*x) - 30*cos(b*x + c)*sin(a + b*x) + 30*cos(a + b*x)*sin(b*x + c) - 9*sin(b*x + c)**3*sin(a + b*x)**2 + 2*sin(b*x + c)**3 + 3*sin(b*x + c)*sin(a + b*x)**2 + 6*sin(b*x + c))/(15*b)