Integrand size = 17, antiderivative size = 103 \[ \int \cos ^2(a+b x) \cos ^3(c+b x) \, dx=-\frac {\sin (2 a-3 c-b x)}{16 b}+\frac {3 \sin (2 a-c+b x)}{16 b}+\frac {3 \sin (c+b x)}{8 b}+\frac {\sin (2 a+c+3 b x)}{16 b}+\frac {\sin (3 c+3 b x)}{24 b}+\frac {\sin (2 a+3 c+5 b x)}{80 b} \] 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 = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int \cos ^2(a+b x) \cos ^3(c+b x) \, dx=\frac {-15 \sin (2 a-3 c-b x)+45 \sin (2 a-c+b x)+90 \sin (c+b x)+10 \sin (3 (c+b x))+15 \sin (2 a+c+3 b x)+3 \sin (2 a+3 c+5 b x)}{240 b} \] Input:
Integrate[Cos[a + b*x]^2*Cos[c + b*x]^3,x]
Output:
(-15*Sin[2*a - 3*c - b*x] + 45*Sin[2*a - c + b*x] + 90*Sin[c + b*x] + 10*S in[3*(c + b*x)] + 15*Sin[2*a + c + 3*b*x] + 3*Sin[2*a + 3*c + 5*b*x])/(240 *b)
Time = 0.26 (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 = {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.
\(\displaystyle \int \cos ^2(a+b x) \cos ^3(b x+c) \, dx\) |
\(\Big \downarrow \) 5081 |
\(\displaystyle \int \left (\frac {1}{16} \cos (2 a-b x-3 c)+\frac {3}{16} \cos (2 a+b x-c)+\frac {3}{16} \cos (2 a+3 b x+c)+\frac {1}{16} \cos (2 a+5 b x+3 c)+\frac {3}{8} \cos (b x+c)+\frac {1}{8} \cos (3 b x+3 c)\right )dx\) |
\(\Big \downarrow \) 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[a + b*x]^2*Cos[c + b*x]^3,x]
Output:
-1/16*Sin[2*a - 3*c - b*x]/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[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 = 3.83 (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 | \(\frac {-20 \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{6} \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+\left (30 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+10\right ) \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{5}+\left (60 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-80 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{4}+\left (200 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-40\right ) \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{3}+\left (-40 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}+60 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )\right ) \tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2}+\left (26 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+24 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}+6\right ) \tan \left (\frac {b x}{2}+\frac {c}{2}\right )-4 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}+24 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{15 b \left (1+\tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{2}}\) | \(249\) |
orering | \(-\frac {259 \left (-2 \cos \left (b x +c \right )^{3} \sin \left (b x +a \right ) b \cos \left (b x +a \right )-3 \cos \left (b x +a \right )^{2} \cos \left (b x +c \right )^{2} b \sin \left (b x +c \right )\right )}{225 b^{2}}-\frac {7 \left (-36 \cos \left (b x +c \right ) \sin \left (b x +a \right ) b^{3} \sin \left (b x +c \right )^{2} \cos \left (b x +a \right )+39 \cos \left (b x +c \right )^{2} b^{3} \cos \left (b x +a \right )^{2} \sin \left (b x +c \right )-18 \cos \left (b x +c \right )^{2} \sin \left (b x +a \right )^{2} b^{3} \sin \left (b x +c \right )+26 \cos \left (b x +c \right )^{3} \sin \left (b x +a \right ) b^{3} \cos \left (b x +a \right )-6 \cos \left (b x +a \right )^{2} b^{3} \sin \left (b x +c \right )^{3}\right )}{45 b^{4}}-\frac {1080 b^{5} \sin \left (b x +c \right )^{2} \sin \left (b x +a \right ) \cos \left (b x +c \right ) \cos \left (b x +a \right )+180 b^{5} \sin \left (b x +c \right )^{3} \cos \left (b x +a \right )^{2}-120 b^{5} \sin \left (b x +c \right )^{3} \sin \left (b x +a \right )^{2}-723 \cos \left (b x +c \right )^{2} b^{5} \cos \left (b x +a \right )^{2} \sin \left (b x +c \right )-482 \cos \left (b x +c \right )^{3} \sin \left (b x +a \right ) b^{5} \cos \left (b x +a \right )+540 b^{5} \sin \left (b x +c \right ) \sin \left (b x +a \right )^{2} \cos \left (b x +c \right )^{2}}{225 b^{6}}\) | \(351\) |
Input:
int(cos(b*x+a)^2*cos(b*x+c)^3,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 \[ \int \cos ^2(a+b x) \cos ^3(c+b x) \, dx=-\frac {6 \, \cos \left (b x + c\right )^{5} \cos \left (-a + c\right ) \sin \left (-a + c\right ) - {\left (3 \, {\left (2 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{4} + {\left (3 \, \cos \left (-a + c\right )^{2} + 1\right )} \cos \left (b x + c\right )^{2} + 6 \, \cos \left (-a + c\right )^{2} + 2\right )} \sin \left (b x + c\right )}{15 \, b} \] Input:
integrate(cos(b*x+a)^2*cos(b*x+c)^3,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 + 1)*cos(b*x + c)^2 + 6*cos(-a + c)^2 + 2)*sin(b*x + c))/b
Time = 2.10 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.68 \[ \int \cos ^2(a+b x) \cos ^3(c+b x) \, dx=\begin {cases} \frac {8 \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (b x + c \right )}}{15 b} + \frac {2 \sin ^{2}{\left (a + b x \right )} \sin {\left (b x + c \right )} \cos ^{2}{\left (b x + c \right )}}{5 b} + \frac {4 \sin {\left (a + b x \right )} \sin ^{2}{\left (b x + c \right )} \cos {\left (a + b x \right )} \cos {\left (b x + c \right )}}{5 b} + \frac {2 \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (b x + c \right )}}{5 b} + \frac {2 \sin ^{3}{\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )}}{15 b} + \frac {3 \sin {\left (b x + c \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (b x + c \right )}}{5 b} & \text {for}\: b \neq 0 \\x \cos ^{2}{\left (a \right )} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+a)**2*cos(b*x+c)**3,x)
Output:
Piecewise((8*sin(a + b*x)**2*sin(b*x + c)**3/(15*b) + 2*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 ) + 2*sin(b*x + c)**3*cos(a + b*x)**2/(15*b) + 3*sin(b*x + c)*cos(a + b*x) **2*cos(b*x + c)**2/(5*b), Ne(b, 0)), (x*cos(a)**2*cos(c)**3, True))
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \cos ^2(a+b x) \cos ^3(c+b x) \, dx=\frac {3 \, \sin \left (5 \, b x + 2 \, a + 3 \, c\right ) + 15 \, \sin \left (3 \, b x + 2 \, a + c\right ) + 10 \, \sin \left (3 \, b x + 3 \, c\right ) + 45 \, \sin \left (b x + 2 \, a - c\right ) + 15 \, \sin \left (b x - 2 \, a + 3 \, c\right ) + 90 \, \sin \left (b x + c\right )}{240 \, b} \] Input:
integrate(cos(b*x+a)^2*cos(b*x+c)^3,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 = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \cos ^2(a+b x) \cos ^3(c+b x) \, dx=\frac {\sin \left (5 \, b x + 2 \, a + 3 \, c\right )}{80 \, b} + \frac {\sin \left (3 \, b x + 2 \, a + c\right )}{16 \, b} + \frac {\sin \left (3 \, b x + 3 \, c\right )}{24 \, b} + \frac {3 \, \sin \left (b x + 2 \, a - c\right )}{16 \, b} + \frac {3 \, \sin \left (b x + c\right )}{8 \, b} - \frac {\sin \left (-b x + 2 \, a - 3 \, c\right )}{16 \, b} \] Input:
integrate(cos(b*x+a)^2*cos(b*x+c)^3,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 = 1.66 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \cos ^2(a+b x) \cos ^3(c+b x) \, dx=\frac {15\,\sin \left (2\,a+c+3\,b\,x\right )+90\,\sin \left (c+b\,x\right )+45\,\sin \left (2\,a-c+b\,x\right )+15\,\sin \left (3\,c-2\,a+b\,x\right )+3\,\sin \left (2\,a+3\,c+5\,b\,x\right )+10\,\sin \left (3\,c+3\,b\,x\right )}{240\,b} \] Input:
int(cos(a + b*x)^2*cos(c + b*x)^3,x)
Output:
(15*sin(2*a + c + 3*b*x) + 90*sin(c + b*x) + 45*sin(2*a - c + b*x) + 15*si n(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 = 0.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.30 \[ \int \cos ^2(a+b x) \cos ^3(c+b x) \, dx=\frac {6 \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +c \right )^{2} \sin \left (b x +a \right )+6 \cos \left (b x +c \right ) \cos \left (b x +a \right ) \sin \left (b x +a \right )+30 \cos \left (b x +c \right ) \sin \left (b x +a \right )-30 \cos \left (b x +a \right ) \sin \left (b x +c \right )+9 \sin \left (b x +c \right )^{3} \sin \left (b x +a \right )^{2}-7 \sin \left (b x +c \right )^{3}-3 \sin \left (b x +c \right ) \sin \left (b x +a \right )^{2}+9 \sin \left (b x +c \right )}{15 b} \] Input:
int(cos(b*x+a)^2*cos(b*x+c)^3,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 - 7*sin(b*x + c)**3 - 3 *sin(b*x + c)*sin(a + b*x)**2 + 9*sin(b*x + c))/(15*b)