Integrand size = 17, antiderivative size = 128 \[ \int \cos ^3(c+b x) \sin ^3(a+b x) \, dx=\frac {3 \cos (a-3 c-2 b x)}{64 b}+\frac {3 \cos (3 a-c+2 b x)}{64 b}-\frac {9 \cos (a+c+2 b x)}{64 b}+\frac {3 \cos (3 a+c+4 b x)}{128 b}-\frac {3 \cos (a+3 c+4 b x)}{128 b}+\frac {\cos (3 (a+c)+6 b x)}{192 b}+\frac {1}{32} x (9 \sin (a-c)-\sin (3 (a-c))) \] Output:
3/64*cos(-2*b*x+a-3*c)/b+3/64*cos(2*b*x+3*a-c)/b-9/64*cos(2*b*x+a+c)/b+3/1 28*cos(4*b*x+3*a+c)/b-3/128*cos(4*b*x+a+3*c)/b+1/192*cos(6*b*x+3*a+3*c)/b+ 1/32*x*(9*sin(a-c)-sin(3*a-3*c))
Time = 0.18 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80 \[ \int \cos ^3(c+b x) \sin ^3(a+b x) \, dx=\frac {18 \cos (a-3 c-2 b x)+18 \cos (3 a-c+2 b x)-54 \cos (a+c+2 b x)+2 \cos (3 (a+c+2 b x))+9 \cos (3 a+c+4 b x)-9 \cos (a+3 c+4 b x)+108 b x \sin (a-c)-12 b x \sin (3 (a-c))}{384 b} \] Input:
Integrate[Cos[c + b*x]^3*Sin[a + b*x]^3,x]
Output:
(18*Cos[a - 3*c - 2*b*x] + 18*Cos[3*a - c + 2*b*x] - 54*Cos[a + c + 2*b*x] + 2*Cos[3*(a + c + 2*b*x)] + 9*Cos[3*a + c + 4*b*x] - 9*Cos[a + 3*c + 4*b *x] + 108*b*x*Sin[a - c] - 12*b*x*Sin[3*(a - c)])/(384*b)
Time = 0.33 (sec) , antiderivative size = 128, 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.
\(\displaystyle \int \sin ^3(a+b x) \cos ^3(b x+c) \, dx\) |
\(\Big \downarrow \) 5085 |
\(\displaystyle \int \left (\frac {3}{32} \sin (a-2 b x-3 c)-\frac {3}{32} \sin (3 a+2 b x-c)+\frac {9}{32} \sin (a+2 b x+c)-\frac {3}{32} \sin (3 a+4 b x+c)+\frac {3}{32} \sin (a+4 b x+3 c)-\frac {1}{32} \sin (3 (a+c)+6 b x)-\frac {1}{32} \sin (3 a-3 c)+\frac {9}{32} \sin (a-c)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \cos (a-2 b x-3 c)}{64 b}+\frac {3 \cos (3 a+2 b x-c)}{64 b}-\frac {9 \cos (a+2 b x+c)}{64 b}+\frac {3 \cos (3 a+4 b x+c)}{128 b}-\frac {3 \cos (a+4 b x+3 c)}{128 b}+\frac {\cos (3 (a+c)+6 b x)}{192 b}+\frac {1}{32} x (9 \sin (a-c)-\sin (3 (a-c)))\) |
Input:
Int[Cos[c + b*x]^3*Sin[a + b*x]^3,x]
Output:
(3*Cos[a - 3*c - 2*b*x])/(64*b) + (3*Cos[3*a - c + 2*b*x])/(64*b) - (9*Cos [a + c + 2*b*x])/(64*b) + (3*Cos[3*a + c + 4*b*x])/(128*b) - (3*Cos[a + 3* c + 4*b*x])/(128*b) + Cos[3*(a + c) + 6*b*x]/(192*b) + (x*(9*Sin[a - c] - Sin[3*(a - c)]))/32
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 = 7.02 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.89
method | result | size |
default | \(\frac {9 x \sin \left (a -c \right )}{32}-\frac {x \sin \left (3 a -3 c \right )}{32}+\frac {3 \cos \left (-2 b x +a -3 c \right )}{64 b}-\frac {9 \cos \left (2 b x +a +c \right )}{64 b}+\frac {3 \cos \left (2 b x +3 a -c \right )}{64 b}-\frac {3 \cos \left (4 b x +a +3 c \right )}{128 b}+\frac {3 \cos \left (4 b x +3 a +c \right )}{128 b}+\frac {\cos \left (6 b x +3 a +3 c \right )}{192 b}\) | \(114\) |
risch | \(\frac {9 x \sin \left (a -c \right )}{32}-\frac {x \sin \left (3 a -3 c \right )}{32}+\frac {3 \cos \left (-2 b x +a -3 c \right )}{64 b}-\frac {9 \cos \left (2 b x +a +c \right )}{64 b}+\frac {3 \cos \left (2 b x +3 a -c \right )}{64 b}-\frac {3 \cos \left (4 b x +a +3 c \right )}{128 b}+\frac {3 \cos \left (4 b x +3 a +c \right )}{128 b}+\frac {\cos \left (6 b x +3 a +3 c \right )}{192 b}\) | \(114\) |
parallelrisch | \(\frac {-12 x b \sin \left (3 a -3 c \right )+108 x \sin \left (a -c \right ) b +18 \cos \left (2 b x +3 a -c \right )+9 \cos \left (4 b x +3 a +c \right )+2 \cos \left (6 b x +3 a +3 c \right )-29 \cos \left (3 a -3 c \right )+45 \cos \left (a -c \right )+18 \cos \left (-2 b x +a -3 c \right )-54 \cos \left (2 b x +a +c \right )-9 \cos \left (4 b x +a +3 c \right )}{384 b}\) | \(121\) |
orering | \(\text {Expression too large to display}\) | \(1006\) |
Input:
int(cos(b*x+c)^3*sin(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
9/32*x*sin(a-c)-1/32*x*sin(3*a-3*c)+3/64*cos(-2*b*x+a-3*c)/b-9/64*cos(2*b* x+a+c)/b+3/64*cos(2*b*x+3*a-c)/b-3/128*cos(4*b*x+a+3*c)/b+3/128*cos(4*b*x+ 3*a+c)/b+1/192*cos(6*b*x+3*a+3*c)/b
Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.22 \[ \int \cos ^3(c+b x) \sin ^3(a+b x) \, dx=\frac {8 \, {\left (4 \, \cos \left (-a + c\right )^{3} - 3 \, \cos \left (-a + c\right )\right )} \cos \left (b x + c\right )^{6} - 12 \, \cos \left (b x + c\right )^{4} \cos \left (-a + c\right )^{3} + {\left (8 \, {\left (4 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{5} + 2 \, {\left (2 \, \cos \left (-a + c\right )^{2} - 5\right )} \cos \left (b x + c\right )^{3} + 3 \, {\left (2 \, \cos \left (-a + c\right )^{2} - 5\right )} \cos \left (b x + c\right )\right )} \sin \left (b x + c\right ) \sin \left (-a + c\right ) + 3 \, {\left (2 \, b x \cos \left (-a + c\right )^{2} - 5 \, b x\right )} \sin \left (-a + c\right )}{48 \, b} \] Input:
integrate(cos(b*x+c)^3*sin(b*x+a)^3,x, algorithm="fricas")
Output:
1/48*(8*(4*cos(-a + c)^3 - 3*cos(-a + c))*cos(b*x + c)^6 - 12*cos(b*x + c) ^4*cos(-a + c)^3 + (8*(4*cos(-a + c)^2 - 1)*cos(b*x + c)^5 + 2*(2*cos(-a + c)^2 - 5)*cos(b*x + c)^3 + 3*(2*cos(-a + c)^2 - 5)*cos(b*x + c))*sin(b*x + c)*sin(-a + c) + 3*(2*b*x*cos(-a + c)^2 - 5*b*x)*sin(-a + c))/b
Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (117) = 234\).
Time = 5.12 (sec) , antiderivative size = 405, normalized size of antiderivative = 3.16 \[ \int \cos ^3(c+b x) \sin ^3(a+b x) \, dx=\begin {cases} \frac {3 x \sin ^{3}{\left (a + b x \right )} \sin ^{2}{\left (b x + c \right )} \cos {\left (b x + c \right )}}{16} + \frac {5 x \sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (b x + c \right )}}{16} - \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (b x + c \right )} \cos {\left (a + b x \right )}}{16} - \frac {9 x \sin ^{2}{\left (a + b x \right )} \sin {\left (b x + c \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{16} + \frac {9 x \sin {\left (a + b x \right )} \sin ^{2}{\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (b x + c \right )}}{16} + \frac {3 x \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (b x + c \right )}}{16} - \frac {5 x \sin ^{3}{\left (b x + c \right )} \cos ^{3}{\left (a + b x \right )}}{16} - \frac {3 x \sin {\left (b x + c \right )} \cos ^{3}{\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{16} - \frac {\sin ^{3}{\left (a + b x \right )} \sin ^{3}{\left (b x + c \right )}}{16 b} - \frac {11 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (b x + c \right )}}{16 b} - \frac {3 \sin {\left (a + b x \right )} \sin ^{3}{\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )}}{16 b} + \frac {3 \sin {\left (a + b x \right )} \sin {\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{4 b} - \frac {\sin ^{2}{\left (b x + c \right )} \cos ^{3}{\left (a + b x \right )} \cos {\left (b x + c \right )}}{2 b} - \frac {7 \cos ^{3}{\left (a + b x \right )} \cos ^{3}{\left (b x + c \right )}}{48 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+c)**3*sin(b*x+a)**3,x)
Output:
Piecewise((3*x*sin(a + b*x)**3*sin(b*x + c)**2*cos(b*x + c)/16 + 5*x*sin(a + b*x)**3*cos(b*x + c)**3/16 - 3*x*sin(a + b*x)**2*sin(b*x + c)**3*cos(a + b*x)/16 - 9*x*sin(a + b*x)**2*sin(b*x + c)*cos(a + b*x)*cos(b*x + c)**2/ 16 + 9*x*sin(a + b*x)*sin(b*x + c)**2*cos(a + b*x)**2*cos(b*x + c)/16 + 3* x*sin(a + b*x)*cos(a + b*x)**2*cos(b*x + c)**3/16 - 5*x*sin(b*x + c)**3*co s(a + b*x)**3/16 - 3*x*sin(b*x + c)*cos(a + b*x)**3*cos(b*x + c)**2/16 - s in(a + b*x)**3*sin(b*x + c)**3/(16*b) - 11*sin(a + b*x)**2*cos(a + b*x)*co s(b*x + c)**3/(16*b) - 3*sin(a + b*x)*sin(b*x + c)**3*cos(a + b*x)**2/(16* b) + 3*sin(a + b*x)*sin(b*x + c)*cos(a + b*x)**2*cos(b*x + c)**2/(4*b) - s in(b*x + c)**2*cos(a + b*x)**3*cos(b*x + c)/(2*b) - 7*cos(a + b*x)**3*cos( b*x + c)**3/(48*b), Ne(b, 0)), (x*sin(a)**3*cos(c)**3, True))
Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+b x) \sin ^3(a+b x) \, dx=-\frac {12 \, {\left (9 \, b \sin \left (-a + c\right ) - b \sin \left (-3 \, a + 3 \, c\right )\right )} x - 2 \, \cos \left (6 \, b x + 3 \, a + 3 \, c\right ) - 9 \, \cos \left (4 \, b x + 3 \, a + c\right ) + 9 \, \cos \left (4 \, b x + a + 3 \, c\right ) - 18 \, \cos \left (2 \, b x + 3 \, a - c\right ) + 54 \, \cos \left (2 \, b x + a + c\right ) - 18 \, \cos \left (2 \, b x - a + 3 \, c\right )}{384 \, b} \] Input:
integrate(cos(b*x+c)^3*sin(b*x+a)^3,x, algorithm="maxima")
Output:
-1/384*(12*(9*b*sin(-a + c) - b*sin(-3*a + 3*c))*x - 2*cos(6*b*x + 3*a + 3 *c) - 9*cos(4*b*x + 3*a + c) + 9*cos(4*b*x + a + 3*c) - 18*cos(2*b*x + 3*a - c) + 54*cos(2*b*x + a + c) - 18*cos(2*b*x - a + 3*c))/b
Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88 \[ \int \cos ^3(c+b x) \sin ^3(a+b x) \, dx=-\frac {1}{32} \, x \sin \left (3 \, a - 3 \, c\right ) + \frac {9}{32} \, x \sin \left (a - c\right ) + \frac {\cos \left (6 \, b x + 3 \, a + 3 \, c\right )}{192 \, b} + \frac {3 \, \cos \left (4 \, b x + 3 \, a + c\right )}{128 \, b} - \frac {3 \, \cos \left (4 \, b x + a + 3 \, c\right )}{128 \, b} + \frac {3 \, \cos \left (2 \, b x + 3 \, a - c\right )}{64 \, b} - \frac {9 \, \cos \left (2 \, b x + a + c\right )}{64 \, b} + \frac {3 \, \cos \left (-2 \, b x + a - 3 \, c\right )}{64 \, b} \] Input:
integrate(cos(b*x+c)^3*sin(b*x+a)^3,x, algorithm="giac")
Output:
-1/32*x*sin(3*a - 3*c) + 9/32*x*sin(a - c) + 1/192*cos(6*b*x + 3*a + 3*c)/ b + 3/128*cos(4*b*x + 3*a + c)/b - 3/128*cos(4*b*x + a + 3*c)/b + 3/64*cos (2*b*x + 3*a - c)/b - 9/64*cos(2*b*x + a + c)/b + 3/64*cos(-2*b*x + a - 3* c)/b
Time = 22.66 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.81 \[ \int \cos ^3(c+b x) \sin ^3(a+b x) \, dx=\frac {\frac {9\,\cos \left (3\,a+c+4\,b\,x\right )}{8}-\frac {9\,\cos \left (a+3\,c+4\,b\,x\right )}{8}+\frac {9\,\cos \left (3\,c-a+2\,b\,x\right )}{4}+\frac {9\,\cos \left (3\,a-c+2\,b\,x\right )}{4}+\frac {\cos \left (3\,a+3\,c+6\,b\,x\right )}{4}-\frac {27\,\cos \left (a+c+2\,b\,x\right )}{4}+\frac {27\,b\,x\,\sin \left (a-c\right )}{2}-\frac {3\,b\,x\,\sin \left (3\,a-3\,c\right )}{2}}{48\,b} \] Input:
int(cos(c + b*x)^3*sin(a + b*x)^3,x)
Output:
((9*cos(3*a + c + 4*b*x))/8 - (9*cos(a + 3*c + 4*b*x))/8 + (9*cos(3*c - a + 2*b*x))/4 + (9*cos(3*a - c + 2*b*x))/4 + cos(3*a + 3*c + 6*b*x)/4 - (27* cos(a + c + 2*b*x))/4 + (27*b*x*sin(a - c))/2 - (3*b*x*sin(3*a - 3*c))/2)/ (48*b)
\[ \int \cos ^3(c+b x) \sin ^3(a+b x) \, dx=\int \cos \left (b x +c \right )^{3} \sin \left (b x +a \right )^{3}d x \] Input:
int(cos(b*x+c)^3*sin(b*x+a)^3,x)
Output:
int(cos(b*x + c)**3*sin(a + b*x)**3,x)