Integrand size = 17, antiderivative size = 101 \[ \int \cos ^3(a+b x) \cos ^2(c+b x) \, dx=-\frac {3 \sin (a-2 c-b x)}{16 b}+\frac {3 \sin (a+b x)}{8 b}+\frac {\sin (3 a-2 c+b x)}{16 b}+\frac {\sin (3 a+3 b x)}{24 b}+\frac {\sin (a+2 c+3 b x)}{16 b}+\frac {\sin (3 a+2 c+5 b x)}{80 b} \] Output:
-3/16*sin(-b*x+a-2*c)/b+3/8*sin(b*x+a)/b+1/16*sin(b*x+3*a-2*c)/b+1/24*sin( 3*b*x+3*a)/b+1/16*sin(3*b*x+a+2*c)/b+1/80*sin(5*b*x+3*a+2*c)/b
Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \cos ^3(a+b x) \cos ^2(c+b x) \, dx=\frac {-45 \sin (a-2 c-b x)+90 \sin (a+b x)+10 \sin (3 (a+b x))+15 \sin (3 a-2 c+b x)+15 \sin (a+2 c+3 b x)+3 \sin (3 a+2 c+5 b x)}{240 b} \] Input:
Integrate[Cos[a + b*x]^3*Cos[c + b*x]^2,x]
Output:
(-45*Sin[a - 2*c - b*x] + 90*Sin[a + b*x] + 10*Sin[3*(a + b*x)] + 15*Sin[3 *a - 2*c + b*x] + 15*Sin[a + 2*c + 3*b*x] + 3*Sin[3*a + 2*c + 5*b*x])/(240 *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 = {5081, 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.
\(\displaystyle \int \cos ^3(a+b x) \cos ^2(b x+c) \, dx\) |
\(\Big \downarrow \) 5081 |
\(\displaystyle \int \left (\frac {3}{16} \cos (a-b x-2 c)+\frac {1}{16} \cos (3 a+b x-2 c)+\frac {3}{16} \cos (a+3 b x+2 c)+\frac {1}{16} \cos (3 a+5 b x+2 c)+\frac {3}{8} \cos (a+b x)+\frac {1}{8} \cos (3 a+3 b x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 \sin (a-b x-2 c)}{16 b}+\frac {\sin (3 a+b x-2 c)}{16 b}+\frac {\sin (a+3 b x+2 c)}{16 b}+\frac {\sin (3 a+5 b x+2 c)}{80 b}+\frac {3 \sin (a+b x)}{8 b}+\frac {\sin (3 a+3 b x)}{24 b}\) |
Input:
Int[Cos[a + b*x]^3*Cos[c + b*x]^2,x]
Output:
(-3*Sin[a - 2*c - b*x])/(16*b) + (3*Sin[a + b*x])/(8*b) + Sin[3*a - 2*c + b*x]/(16*b) + Sin[3*a + 3*b*x]/(24*b) + Sin[a + 2*c + 3*b*x]/(16*b) + Sin[ 3*a + 2*c + 5*b*x]/(80*b)
Int[Cos[v_]^(p_.)*Cos[w_]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[Cos[v]^p *Cos[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 = 3.79 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {3 \sin \left (-b x +a -2 c \right )}{16 b}+\frac {3 \sin \left (b x +a \right )}{8 b}+\frac {\sin \left (b x +3 a -2 c \right )}{16 b}+\frac {\sin \left (3 b x +3 a \right )}{24 b}+\frac {\sin \left (3 b x +a +2 c \right )}{16 b}+\frac {\sin \left (5 b x +3 a +2 c \right )}{80 b}\) | \(90\) |
risch | \(-\frac {3 \sin \left (-b x +a -2 c \right )}{16 b}+\frac {3 \sin \left (b x +a \right )}{8 b}+\frac {\sin \left (b x +3 a -2 c \right )}{16 b}+\frac {\sin \left (3 b x +3 a \right )}{24 b}+\frac {\sin \left (3 b x +a +2 c \right )}{16 b}+\frac {\sin \left (5 b x +3 a +2 c \right )}{80 b}\) | \(90\) |
parallelrisch | \(\frac {15 \sin \left (b x +3 a -2 c \right )+10 \sin \left (3 b x +3 a \right )+3 \sin \left (5 b x +3 a +2 c \right )+112 \sin \left (2 a -2 c \right )-45 \sin \left (-b x +a -2 c \right )+15 \sin \left (3 b x +a +2 c \right )+90 \sin \left (b x +a \right )-80 \sin \left (a -c \right )}{240 b}\) | \(95\) |
orering | \(-\frac {259 \left (-3 \cos \left (b x +a \right )^{2} \cos \left (b x +c \right )^{2} b \sin \left (b x +a \right )-2 \cos \left (b x +a \right )^{3} \cos \left (b x +c \right ) b \sin \left (b x +c \right )\right )}{225 b^{2}}-\frac {7 \left (-6 b^{3} \sin \left (b x +a \right )^{3} \cos \left (b x +c \right )^{2}-36 \cos \left (b x +a \right ) \cos \left (b x +c \right ) b^{3} \sin \left (b x +a \right )^{2} \sin \left (b x +c \right )+39 \cos \left (b x +a \right )^{2} \cos \left (b x +c \right )^{2} b^{3} \sin \left (b x +a \right )-18 \cos \left (b x +a \right )^{2} b^{3} \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )+26 \cos \left (b x +a \right )^{3} \cos \left (b x +c \right ) b^{3} \sin \left (b x +c \right )\right )}{45 b^{4}}-\frac {-723 b^{5} \sin \left (b x +a \right ) \cos \left (b x +c \right )^{2} \cos \left (b x +a \right )^{2}+1080 b^{5} \sin \left (b x +a \right )^{2} \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +c \right )+180 b^{5} \sin \left (b x +a \right )^{3} \cos \left (b x +c \right )^{2}-120 b^{5} \sin \left (b x +a \right )^{3} \sin \left (b x +c \right )^{2}+540 \cos \left (b x +a \right )^{2} b^{5} \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )-482 \cos \left (b x +a \right )^{3} \cos \left (b x +c \right ) b^{5} \sin \left (b x +c \right )}{225 b^{6}}\) | \(351\) |
Input:
int(cos(b*x+a)^3*cos(b*x+c)^2,x,method=_RETURNVERBOSE)
Output:
-3/16*sin(-b*x+a-2*c)/b+3/8*sin(b*x+a)/b+1/16*sin(b*x+3*a-2*c)/b+1/24*sin( 3*b*x+3*a)/b+1/16*sin(3*b*x+a+2*c)/b+1/80*sin(5*b*x+3*a+2*c)/b
Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.37 \[ \int \cos ^3(a+b x) \cos ^2(c+b x) \, dx=\frac {{\left (3 \, {\left (4 \, \cos \left (-a + c\right )^{3} - 3 \, \cos \left (-a + c\right )\right )} \cos \left (b x + c\right )^{4} + {\left (\cos \left (-a + c\right )^{3} + 3 \, \cos \left (-a + c\right )\right )} \cos \left (b x + c\right )^{2} + 2 \, \cos \left (-a + c\right )^{3} + 6 \, \cos \left (-a + c\right )\right )} \sin \left (b x + c\right ) - {\left (3 \, {\left (4 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{5} - 5 \, {\left (\cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{3}\right )} \sin \left (-a + c\right )}{15 \, b} \] Input:
integrate(cos(b*x+a)^3*cos(b*x+c)^2,x, algorithm="fricas")
Output:
1/15*((3*(4*cos(-a + c)^3 - 3*cos(-a + c))*cos(b*x + c)^4 + (cos(-a + c)^3 + 3*cos(-a + c))*cos(b*x + c)^2 + 2*cos(-a + c)^3 + 6*cos(-a + c))*sin(b* x + c) - (3*(4*cos(-a + c)^2 - 1)*cos(b*x + c)^5 - 5*(cos(-a + c)^2 - 1)*c os(b*x + c)^3)*sin(-a + c))/b
Time = 2.23 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.71 \[ \int \cos ^3(a+b x) \cos ^2(c+b x) \, dx=\begin {cases} \frac {8 \sin ^{3}{\left (a + b x \right )} \sin ^{2}{\left (b x + c \right )}}{15 b} + \frac {2 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{15 b} + \frac {4 \sin ^{2}{\left (a + b x \right )} \sin {\left (b x + c \right )} \cos {\left (a + b x \right )} \cos {\left (b x + c \right )}}{5 b} + \frac {2 \sin {\left (a + b x \right )} \sin ^{2}{\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )}}{5 b} + \frac {3 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{5 b} + \frac {2 \sin {\left (b x + c \right )} \cos ^{3}{\left (a + b x \right )} \cos {\left (b x + c \right )}}{5 b} & \text {for}\: b \neq 0 \\x \cos ^{3}{\left (a \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+a)**3*cos(b*x+c)**2,x)
Output:
Piecewise((8*sin(a + b*x)**3*sin(b*x + c)**2/(15*b) + 2*sin(a + b*x)**3*co s(b*x + c)**2/(15*b) + 4*sin(a + b*x)**2*sin(b*x + c)*cos(a + b*x)*cos(b*x + c)/(5*b) + 2*sin(a + b*x)*sin(b*x + c)**2*cos(a + b*x)**2/(5*b) + 3*sin (a + b*x)*cos(a + b*x)**2*cos(b*x + c)**2/(5*b) + 2*sin(b*x + c)*cos(a + b *x)**3*cos(b*x + c)/(5*b), Ne(b, 0)), (x*cos(a)**3*cos(c)**2, True))
Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \cos ^3(a+b x) \cos ^2(c+b x) \, dx=\frac {3 \, \sin \left (5 \, b x + 3 \, a + 2 \, c\right ) + 10 \, \sin \left (3 \, b x + 3 \, a\right ) + 15 \, \sin \left (3 \, b x + a + 2 \, c\right ) + 15 \, \sin \left (b x + 3 \, a - 2 \, c\right ) + 90 \, \sin \left (b x + a\right ) + 45 \, \sin \left (b x - a + 2 \, c\right )}{240 \, b} \] Input:
integrate(cos(b*x+a)^3*cos(b*x+c)^2,x, algorithm="maxima")
Output:
1/240*(3*sin(5*b*x + 3*a + 2*c) + 10*sin(3*b*x + 3*a) + 15*sin(3*b*x + a + 2*c) + 15*sin(b*x + 3*a - 2*c) + 90*sin(b*x + a) + 45*sin(b*x - a + 2*c)) /b
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88 \[ \int \cos ^3(a+b x) \cos ^2(c+b x) \, dx=\frac {\sin \left (5 \, b x + 3 \, a + 2 \, c\right )}{80 \, b} + \frac {\sin \left (3 \, b x + 3 \, a\right )}{24 \, b} + \frac {\sin \left (3 \, b x + a + 2 \, c\right )}{16 \, b} + \frac {\sin \left (b x + 3 \, a - 2 \, c\right )}{16 \, b} + \frac {3 \, \sin \left (b x + a\right )}{8 \, b} - \frac {3 \, \sin \left (-b x + a - 2 \, c\right )}{16 \, b} \] Input:
integrate(cos(b*x+a)^3*cos(b*x+c)^2,x, algorithm="giac")
Output:
1/80*sin(5*b*x + 3*a + 2*c)/b + 1/24*sin(3*b*x + 3*a)/b + 1/16*sin(3*b*x + a + 2*c)/b + 1/16*sin(b*x + 3*a - 2*c)/b + 3/8*sin(b*x + a)/b - 3/16*sin( -b*x + a - 2*c)/b
Time = 1.46 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \cos ^3(a+b x) \cos ^2(c+b x) \, dx=\frac {15\,\sin \left (a+2\,c+3\,b\,x\right )+90\,\sin \left (a+b\,x\right )+45\,\sin \left (2\,c-a+b\,x\right )+15\,\sin \left (3\,a-2\,c+b\,x\right )+3\,\sin \left (3\,a+2\,c+5\,b\,x\right )+10\,\sin \left (3\,a+3\,b\,x\right )}{240\,b} \] Input:
int(cos(a + b*x)^3*cos(c + b*x)^2,x)
Output:
(15*sin(a + 2*c + 3*b*x) + 90*sin(a + b*x) + 45*sin(2*c - a + b*x) + 15*si n(3*a - 2*c + b*x) + 3*sin(3*a + 2*c + 5*b*x) + 10*sin(3*a + 3*b*x))/(240* b)
Time = 0.16 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.03 \[ \int \cos ^3(a+b x) \cos ^2(c+b x) \, dx=\frac {6 \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +c \right ) \sin \left (b x +a \right )^{2}+6 \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +c \right )-2 \cos \left (b x +c \right ) \sin \left (b x +c \right ) \sin \left (b x +a \right )^{2}+\cos \left (b x +c \right ) \sin \left (b x +c \right )+10 \cos \left (b x +c \right ) \sin \left (b x +a \right )+2 \cos \left (b x +a \right ) \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )-10 \cos \left (b x +a \right ) \sin \left (b x +c \right )-\cos \left (b x +a \right ) \sin \left (b x +a \right )+9 \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )^{3}-3 \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )-7 \sin \left (b x +a \right )^{3}+9 \sin \left (b x +a \right )}{15 b} \] Input:
int(cos(b*x+a)^3*cos(b*x+c)^2,x)
Output:
(6*cos(b*x + c)*cos(a + b*x)*sin(b*x + c)*sin(a + b*x)**2 + 6*cos(b*x + c) *cos(a + b*x)*sin(b*x + c) - 2*cos(b*x + c)*sin(b*x + c)*sin(a + b*x)**2 + cos(b*x + c)*sin(b*x + c) + 10*cos(b*x + c)*sin(a + b*x) + 2*cos(a + b*x) *sin(b*x + c)**2*sin(a + b*x) - 10*cos(a + b*x)*sin(b*x + c) - cos(a + b*x )*sin(a + b*x) + 9*sin(b*x + c)**2*sin(a + b*x)**3 - 3*sin(b*x + c)**2*sin (a + b*x) - 7*sin(a + b*x)**3 + 9*sin(a + b*x))/(15*b)