Integrand size = 17, antiderivative size = 138 \[ \int \cos ^3(a+b x) \cos ^2(c+d x) \, dx=\frac {3 \sin (a+b x)}{8 b}+\frac {\sin (3 a+3 b x)}{24 b}+\frac {3 \sin (a-2 c+(b-2 d) x)}{16 (b-2 d)}+\frac {\sin (3 a-2 c+(3 b-2 d) x)}{16 (3 b-2 d)}+\frac {3 \sin (a+2 c+(b+2 d) x)}{16 (b+2 d)}+\frac {\sin (3 a+2 c+(3 b+2 d) x)}{16 (3 b+2 d)} \] Output:
3/8*sin(b*x+a)/b+1/24*sin(3*b*x+3*a)/b+3*sin(a-2*c+(b-2*d)*x)/(16*b-32*d)+ sin(3*a-2*c+(3*b-2*d)*x)/(48*b-32*d)+3*sin(a+2*c+(b+2*d)*x)/(16*b+32*d)+si n(3*a+2*c+(3*b+2*d)*x)/(48*b+32*d)
Time = 1.14 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11 \[ \int \cos ^3(a+b x) \cos ^2(c+d x) \, dx=\frac {1}{48} \left (\frac {18 \cos (b x) \sin (a)}{b}+\frac {2 \cos (3 b x) \sin (3 a)}{b}+\frac {18 \cos (a) \sin (b x)}{b}+\frac {2 \cos (3 a) \sin (3 b x)}{b}+\frac {9 \sin (a-2 c+b x-2 d x)}{b-2 d}+\frac {3 \sin (3 a-2 c+3 b x-2 d x)}{3 b-2 d}+\frac {9 \sin (a+2 c+b x+2 d x)}{b+2 d}+\frac {3 \sin (3 a+2 c+3 b x+2 d x)}{3 b+2 d}\right ) \] Input:
Integrate[Cos[a + b*x]^3*Cos[c + d*x]^2,x]
Output:
((18*Cos[b*x]*Sin[a])/b + (2*Cos[3*b*x]*Sin[3*a])/b + (18*Cos[a]*Sin[b*x]) /b + (2*Cos[3*a]*Sin[3*b*x])/b + (9*Sin[a - 2*c + b*x - 2*d*x])/(b - 2*d) + (3*Sin[3*a - 2*c + 3*b*x - 2*d*x])/(3*b - 2*d) + (9*Sin[a + 2*c + b*x + 2*d*x])/(b + 2*d) + (3*Sin[3*a + 2*c + 3*b*x + 2*d*x])/(3*b + 2*d))/48
Time = 0.32 (sec) , antiderivative size = 138, 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.
\(\displaystyle \int \cos ^3(a+b x) \cos ^2(c+d x) \, dx\) |
\(\Big \downarrow \) 5081 |
\(\displaystyle \int \left (\frac {3}{16} \cos (a+x (b-2 d)-2 c)+\frac {1}{16} \cos (3 a+x (3 b-2 d)-2 c)+\frac {3}{16} \cos (a+x (b+2 d)+2 c)+\frac {1}{16} \cos (3 a+x (3 b+2 d)+2 c)+\frac {3}{8} \cos (a+b x)+\frac {1}{8} \cos (3 a+3 b x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \sin (a+x (b-2 d)-2 c)}{16 (b-2 d)}+\frac {\sin (3 a+x (3 b-2 d)-2 c)}{16 (3 b-2 d)}+\frac {3 \sin (a+x (b+2 d)+2 c)}{16 (b+2 d)}+\frac {\sin (3 a+x (3 b+2 d)+2 c)}{16 (3 b+2 d)}+\frac {3 \sin (a+b x)}{8 b}+\frac {\sin (3 a+3 b x)}{24 b}\) |
Input:
Int[Cos[a + b*x]^3*Cos[c + d*x]^2,x]
Output:
(3*Sin[a + b*x])/(8*b) + Sin[3*a + 3*b*x]/(24*b) + (3*Sin[a - 2*c + (b - 2 *d)*x])/(16*(b - 2*d)) + Sin[3*a - 2*c + (3*b - 2*d)*x]/(16*(3*b - 2*d)) + (3*Sin[a + 2*c + (b + 2*d)*x])/(16*(b + 2*d)) + Sin[3*a + 2*c + (3*b + 2* d)*x]/(16*(3*b + 2*d))
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 = 13.52 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {3 \sin \left (b x +a \right )}{8 b}+\frac {\sin \left (3 b x +3 a \right )}{24 b}+\frac {3 \sin \left (a -2 c +\left (b -2 d \right ) x \right )}{16 \left (b -2 d \right )}+\frac {3 \sin \left (a +2 c +\left (b +2 d \right ) x \right )}{16 \left (b +2 d \right )}+\frac {\sin \left (3 a -2 c +\left (3 b -2 d \right ) x \right )}{48 b -32 d}+\frac {\sin \left (3 a +2 c +\left (3 b +2 d \right ) x \right )}{48 b +32 d}\) | \(127\) |
parallelrisch | \(\frac {9 \left (b +2 d \right ) \left (b -2 d \right ) \left (b +\frac {2 d}{3}\right ) b \sin \left (3 a -2 c +\left (3 b -2 d \right ) x \right )+9 \left (\left (b^{3}-4 b \,d^{2}\right ) \sin \left (3 a +2 c +\left (3 b +2 d \right ) x \right )+18 \left (b +\frac {2 d}{3}\right ) \left (\frac {b \left (b +2 d \right ) \sin \left (a -2 c +\left (b -2 d \right ) x \right )}{2}+\left (\frac {b \sin \left (a +2 c +\left (b +2 d \right ) x \right )}{2}+\left (b +2 d \right ) \left (\sin \left (b x +a \right )+\frac {\sin \left (3 b x +3 a \right )}{9}\right )\right ) \left (b -2 d \right )\right )\right ) \left (b -\frac {2 d}{3}\right )}{432 b^{5}-1920 b^{3} d^{2}+768 b \,d^{4}}\) | \(172\) |
risch | \(\frac {3 \sin \left (b x +a \right )}{8 b}+\frac {27 \sin \left (b x -2 d x +a -2 c \right ) b^{3}}{16 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}+\frac {27 \sin \left (b x -2 d x +a -2 c \right ) b^{2} d}{8 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}-\frac {3 \sin \left (b x -2 d x +a -2 c \right ) b \,d^{2}}{4 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}-\frac {3 \sin \left (b x -2 d x +a -2 c \right ) d^{3}}{2 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}+\frac {27 \sin \left (b x +2 d x +a +2 c \right ) b^{3}}{16 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}-\frac {27 \sin \left (b x +2 d x +a +2 c \right ) b^{2} d}{8 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}-\frac {3 \sin \left (b x +2 d x +a +2 c \right ) b \,d^{2}}{4 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}+\frac {3 \sin \left (b x +2 d x +a +2 c \right ) d^{3}}{2 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}+\frac {3 \sin \left (3 b x -2 d x +3 a -2 c \right ) b^{3}}{16 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}+\frac {\sin \left (3 b x -2 d x +3 a -2 c \right ) b^{2} d}{8 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}-\frac {3 \sin \left (3 b x -2 d x +3 a -2 c \right ) b \,d^{2}}{4 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}-\frac {\sin \left (3 b x -2 d x +3 a -2 c \right ) d^{3}}{2 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}+\frac {3 \sin \left (3 b x +2 d x +3 a +2 c \right ) b^{3}}{16 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}-\frac {\sin \left (3 b x +2 d x +3 a +2 c \right ) b^{2} d}{8 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}-\frac {3 \sin \left (3 b x +2 d x +3 a +2 c \right ) b \,d^{2}}{4 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}+\frac {\sin \left (3 b x +2 d x +3 a +2 c \right ) d^{3}}{2 \left (b -2 d \right ) \left (3 b -2 d \right ) \left (3 b +2 d \right ) \left (b +2 d \right )}+\frac {\sin \left (3 b x +3 a \right )}{24 b}\) | \(859\) |
orering | \(\text {Expression too large to display}\) | \(3070\) |
Input:
int(cos(b*x+a)^3*cos(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
3/8*sin(b*x+a)/b+1/24*sin(3*b*x+3*a)/b+3/16/(b-2*d)*sin(a-2*c+(b-2*d)*x)+3 /16/(b+2*d)*sin(a+2*c+(b+2*d)*x)+1/16/(3*b-2*d)*sin(3*a-2*c+(3*b-2*d)*x)+1 /16/(3*b+2*d)*sin(3*a+2*c+(3*b+2*d)*x)
Time = 0.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.12 \[ \int \cos ^3(a+b x) \cos ^2(c+d x) \, dx=-\frac {6 \, {\left (6 \, b^{3} d \cos \left (b x + a\right ) + {\left (b^{3} d - 4 \, b d^{3}\right )} \cos \left (b x + a\right )^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (40 \, b^{2} d^{2} - 16 \, d^{4} + 2 \, {\left (b^{2} d^{2} - 4 \, d^{4}\right )} \cos \left (b x + a\right )^{2} - 9 \, {\left (2 \, b^{4} + {\left (b^{4} - 4 \, b^{2} d^{2}\right )} \cos \left (b x + a\right )^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (b x + a\right )}{3 \, {\left (9 \, b^{5} - 40 \, b^{3} d^{2} + 16 \, b d^{4}\right )}} \] Input:
integrate(cos(b*x+a)^3*cos(d*x+c)^2,x, algorithm="fricas")
Output:
-1/3*(6*(6*b^3*d*cos(b*x + a) + (b^3*d - 4*b*d^3)*cos(b*x + a)^3)*cos(d*x + c)*sin(d*x + c) + (40*b^2*d^2 - 16*d^4 + 2*(b^2*d^2 - 4*d^4)*cos(b*x + a )^2 - 9*(2*b^4 + (b^4 - 4*b^2*d^2)*cos(b*x + a)^2)*cos(d*x + c)^2)*sin(b*x + a))/(9*b^5 - 40*b^3*d^2 + 16*b*d^4)
Leaf count of result is larger than twice the leaf count of optimal. 2020 vs. \(2 (116) = 232\).
Time = 5.68 (sec) , antiderivative size = 2020, normalized size of antiderivative = 14.64 \[ \int \cos ^3(a+b x) \cos ^2(c+d x) \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)**3*cos(d*x+c)**2,x)
Output:
Piecewise((x*cos(a)**3*cos(c)**2, Eq(b, 0) & Eq(d, 0)), ((x*sin(c + d*x)** 2/2 + x*cos(c + d*x)**2/2 + sin(c + d*x)*cos(c + d*x)/(2*d))*cos(a)**3, Eq (b, 0)), (-3*x*sin(a - 2*d*x)**3*sin(c + d*x)*cos(c + d*x)/8 - 3*x*sin(a - 2*d*x)**2*sin(c + d*x)**2*cos(a - 2*d*x)/16 + 3*x*sin(a - 2*d*x)**2*cos(a - 2*d*x)*cos(c + d*x)**2/16 - 3*x*sin(a - 2*d*x)*sin(c + d*x)*cos(a - 2*d *x)**2*cos(c + d*x)/8 - 3*x*sin(c + d*x)**2*cos(a - 2*d*x)**3/16 + 3*x*cos (a - 2*d*x)**3*cos(c + d*x)**2/16 - sin(a - 2*d*x)**3*sin(c + d*x)**2/(96* d) - 31*sin(a - 2*d*x)**3*cos(c + d*x)**2/(96*d) - sin(a - 2*d*x)**2*sin(c + d*x)*cos(a - 2*d*x)*cos(c + d*x)/(8*d) - sin(a - 2*d*x)*cos(a - 2*d*x)* *2*cos(c + d*x)**2/(2*d) - 3*sin(c + d*x)*cos(a - 2*d*x)**3*cos(c + d*x)/( 16*d), Eq(b, -2*d)), (x*sin(a - 2*d*x/3)**3*sin(c + d*x)*cos(c + d*x)/8 + 3*x*sin(a - 2*d*x/3)**2*sin(c + d*x)**2*cos(a - 2*d*x/3)/16 - 3*x*sin(a - 2*d*x/3)**2*cos(a - 2*d*x/3)*cos(c + d*x)**2/16 - 3*x*sin(a - 2*d*x/3)*sin (c + d*x)*cos(a - 2*d*x/3)**2*cos(c + d*x)/8 - x*sin(c + d*x)**2*cos(a - 2 *d*x/3)**3/16 + x*cos(a - 2*d*x/3)**3*cos(c + d*x)**2/16 - 27*sin(a - 2*d* x/3)**3*sin(c + d*x)**2/(32*d) - 5*sin(a - 2*d*x/3)**3*cos(c + d*x)**2/(32 *d) + 15*sin(a - 2*d*x/3)**2*sin(c + d*x)*cos(a - 2*d*x/3)*cos(c + d*x)/(8 *d) - 3*sin(a - 2*d*x/3)*cos(a - 2*d*x/3)**2*cos(c + d*x)**2/(2*d) - sin(c + d*x)*cos(a - 2*d*x/3)**3*cos(c + d*x)/(16*d), Eq(b, -2*d/3)), (-x*sin(a + 2*d*x/3)**3*sin(c + d*x)*cos(c + d*x)/8 + 3*x*sin(a + 2*d*x/3)**2*si...
Leaf count of result is larger than twice the leaf count of optimal. 1360 vs. \(2 (126) = 252\).
Time = 0.10 (sec) , antiderivative size = 1360, normalized size of antiderivative = 9.86 \[ \int \cos ^3(a+b x) \cos ^2(c+d x) \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)^3*cos(d*x+c)^2,x, algorithm="maxima")
Output:
-1/96*(3*(3*b^4*sin(2*c) - 2*b^3*d*sin(2*c) - 12*b^2*d^2*sin(2*c) + 8*b*d^ 3*sin(2*c))*cos((3*b + 2*d)*x + 3*a + 4*c) - 3*(3*b^4*sin(2*c) - 2*b^3*d*s in(2*c) - 12*b^2*d^2*sin(2*c) + 8*b*d^3*sin(2*c))*cos((3*b + 2*d)*x + 3*a) - 3*(3*b^4*sin(2*c) + 2*b^3*d*sin(2*c) - 12*b^2*d^2*sin(2*c) - 8*b*d^3*si n(2*c))*cos(-(3*b - 2*d)*x - 3*a + 4*c) + 3*(3*b^4*sin(2*c) + 2*b^3*d*sin( 2*c) - 12*b^2*d^2*sin(2*c) - 8*b*d^3*sin(2*c))*cos(-(3*b - 2*d)*x - 3*a) + 9*(9*b^4*sin(2*c) - 18*b^3*d*sin(2*c) - 4*b^2*d^2*sin(2*c) + 8*b*d^3*sin( 2*c))*cos((b + 2*d)*x + a + 4*c) - 9*(9*b^4*sin(2*c) - 18*b^3*d*sin(2*c) - 4*b^2*d^2*sin(2*c) + 8*b*d^3*sin(2*c))*cos((b + 2*d)*x + a) - 9*(9*b^4*si n(2*c) + 18*b^3*d*sin(2*c) - 4*b^2*d^2*sin(2*c) - 8*b*d^3*sin(2*c))*cos(-( b - 2*d)*x - a + 4*c) + 9*(9*b^4*sin(2*c) + 18*b^3*d*sin(2*c) - 4*b^2*d^2* sin(2*c) - 8*b*d^3*sin(2*c))*cos(-(b - 2*d)*x - a) + 2*(9*b^4*sin(2*c) - 4 0*b^2*d^2*sin(2*c) + 16*d^4*sin(2*c))*cos(3*b*x + 3*a + 2*c) - 2*(9*b^4*si n(2*c) - 40*b^2*d^2*sin(2*c) + 16*d^4*sin(2*c))*cos(3*b*x + 3*a - 2*c) + 1 8*(9*b^4*sin(2*c) - 40*b^2*d^2*sin(2*c) + 16*d^4*sin(2*c))*cos(b*x + a + 2 *c) - 18*(9*b^4*sin(2*c) - 40*b^2*d^2*sin(2*c) + 16*d^4*sin(2*c))*cos(b*x + a - 2*c) - 3*(3*b^4*cos(2*c) - 2*b^3*d*cos(2*c) - 12*b^2*d^2*cos(2*c) + 8*b*d^3*cos(2*c))*sin((3*b + 2*d)*x + 3*a + 4*c) - 3*(3*b^4*cos(2*c) - 2*b ^3*d*cos(2*c) - 12*b^2*d^2*cos(2*c) + 8*b*d^3*cos(2*c))*sin((3*b + 2*d)*x + 3*a) + 3*(3*b^4*cos(2*c) + 2*b^3*d*cos(2*c) - 12*b^2*d^2*cos(2*c) - 8...
Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \cos ^3(a+b x) \cos ^2(c+d x) \, dx=\frac {\sin \left (3 \, b x + 2 \, d x + 3 \, a + 2 \, c\right )}{16 \, {\left (3 \, b + 2 \, d\right )}} + \frac {\sin \left (3 \, b x - 2 \, d x + 3 \, a - 2 \, c\right )}{16 \, {\left (3 \, b - 2 \, d\right )}} + \frac {\sin \left (3 \, b x + 3 \, a\right )}{24 \, b} + \frac {3 \, \sin \left (b x + 2 \, d x + a + 2 \, c\right )}{16 \, {\left (b + 2 \, d\right )}} + \frac {3 \, \sin \left (b x - 2 \, d x + a - 2 \, c\right )}{16 \, {\left (b - 2 \, d\right )}} + \frac {3 \, \sin \left (b x + a\right )}{8 \, b} \] Input:
integrate(cos(b*x+a)^3*cos(d*x+c)^2,x, algorithm="giac")
Output:
1/16*sin(3*b*x + 2*d*x + 3*a + 2*c)/(3*b + 2*d) + 1/16*sin(3*b*x - 2*d*x + 3*a - 2*c)/(3*b - 2*d) + 1/24*sin(3*b*x + 3*a)/b + 3/16*sin(b*x + 2*d*x + a + 2*c)/(b + 2*d) + 3/16*sin(b*x - 2*d*x + a - 2*c)/(b - 2*d) + 3/8*sin( b*x + a)/b
Time = 23.11 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.17 \[ \int \cos ^3(a+b x) \cos ^2(c+d x) \, dx=\frac {81\,b^4\,\sin \left (a-2\,c+b\,x-2\,d\,x\right )+81\,b^4\,\sin \left (a+2\,c+b\,x+2\,d\,x\right )+162\,b^4\,\sin \left (a+b\,x\right )+288\,d^4\,\sin \left (a+b\,x\right )+9\,b^4\,\sin \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )+9\,b^4\,\sin \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )+18\,b^4\,\sin \left (3\,a+3\,b\,x\right )+32\,d^4\,\sin \left (3\,a+3\,b\,x\right )-24\,b\,d^3\,\sin \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )+24\,b\,d^3\,\sin \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )+6\,b^3\,d\,\sin \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )-6\,b^3\,d\,\sin \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )-36\,b^2\,d^2\,\sin \left (a-2\,c+b\,x-2\,d\,x\right )-36\,b^2\,d^2\,\sin \left (a+2\,c+b\,x+2\,d\,x\right )-720\,b^2\,d^2\,\sin \left (a+b\,x\right )-36\,b^2\,d^2\,\sin \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )-36\,b^2\,d^2\,\sin \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )-80\,b^2\,d^2\,\sin \left (3\,a+3\,b\,x\right )-72\,b\,d^3\,\sin \left (a-2\,c+b\,x-2\,d\,x\right )+72\,b\,d^3\,\sin \left (a+2\,c+b\,x+2\,d\,x\right )+162\,b^3\,d\,\sin \left (a-2\,c+b\,x-2\,d\,x\right )-162\,b^3\,d\,\sin \left (a+2\,c+b\,x+2\,d\,x\right )}{48\,\left (9\,b^5-40\,b^3\,d^2+16\,b\,d^4\right )} \] Input:
int(cos(a + b*x)^3*cos(c + d*x)^2,x)
Output:
(81*b^4*sin(a - 2*c + b*x - 2*d*x) + 81*b^4*sin(a + 2*c + b*x + 2*d*x) + 1 62*b^4*sin(a + b*x) + 288*d^4*sin(a + b*x) + 9*b^4*sin(3*a - 2*c + 3*b*x - 2*d*x) + 9*b^4*sin(3*a + 2*c + 3*b*x + 2*d*x) + 18*b^4*sin(3*a + 3*b*x) + 32*d^4*sin(3*a + 3*b*x) - 24*b*d^3*sin(3*a - 2*c + 3*b*x - 2*d*x) + 24*b* d^3*sin(3*a + 2*c + 3*b*x + 2*d*x) + 6*b^3*d*sin(3*a - 2*c + 3*b*x - 2*d*x ) - 6*b^3*d*sin(3*a + 2*c + 3*b*x + 2*d*x) - 36*b^2*d^2*sin(a - 2*c + b*x - 2*d*x) - 36*b^2*d^2*sin(a + 2*c + b*x + 2*d*x) - 720*b^2*d^2*sin(a + b*x ) - 36*b^2*d^2*sin(3*a - 2*c + 3*b*x - 2*d*x) - 36*b^2*d^2*sin(3*a + 2*c + 3*b*x + 2*d*x) - 80*b^2*d^2*sin(3*a + 3*b*x) - 72*b*d^3*sin(a - 2*c + b*x - 2*d*x) + 72*b*d^3*sin(a + 2*c + b*x + 2*d*x) + 162*b^3*d*sin(a - 2*c + b*x - 2*d*x) - 162*b^3*d*sin(a + 2*c + b*x + 2*d*x))/(48*(16*b*d^4 + 9*b^5 - 40*b^3*d^2))
Time = 0.22 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.20 \[ \int \cos ^3(a+b x) \cos ^2(c+d x) \, dx=\frac {6 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right )^{2} \sin \left (d x +c \right ) b^{3} d -24 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right )^{2} \sin \left (d x +c \right ) b \,d^{3}-42 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{3} d +24 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) b \,d^{3}+9 \sin \left (b x +a \right )^{3} \sin \left (d x +c \right )^{2} b^{4}-36 \sin \left (b x +a \right )^{3} \sin \left (d x +c \right )^{2} b^{2} d^{2}-9 \sin \left (b x +a \right )^{3} b^{4}+38 \sin \left (b x +a \right )^{3} b^{2} d^{2}-8 \sin \left (b x +a \right )^{3} d^{4}-27 \sin \left (b x +a \right ) \sin \left (d x +c \right )^{2} b^{4}+36 \sin \left (b x +a \right ) \sin \left (d x +c \right )^{2} b^{2} d^{2}+27 \sin \left (b x +a \right ) b^{4}-78 \sin \left (b x +a \right ) b^{2} d^{2}+24 \sin \left (b x +a \right ) d^{4}}{3 b \left (9 b^{4}-40 b^{2} d^{2}+16 d^{4}\right )} \] Input:
int(cos(b*x+a)^3*cos(d*x+c)^2,x)
Output:
(6*cos(a + b*x)*cos(c + d*x)*sin(a + b*x)**2*sin(c + d*x)*b**3*d - 24*cos( a + b*x)*cos(c + d*x)*sin(a + b*x)**2*sin(c + d*x)*b*d**3 - 42*cos(a + b*x )*cos(c + d*x)*sin(c + d*x)*b**3*d + 24*cos(a + b*x)*cos(c + d*x)*sin(c + d*x)*b*d**3 + 9*sin(a + b*x)**3*sin(c + d*x)**2*b**4 - 36*sin(a + b*x)**3* sin(c + d*x)**2*b**2*d**2 - 9*sin(a + b*x)**3*b**4 + 38*sin(a + b*x)**3*b* *2*d**2 - 8*sin(a + b*x)**3*d**4 - 27*sin(a + b*x)*sin(c + d*x)**2*b**4 + 36*sin(a + b*x)*sin(c + d*x)**2*b**2*d**2 + 27*sin(a + b*x)*b**4 - 78*sin( a + b*x)*b**2*d**2 + 24*sin(a + b*x)*d**4)/(3*b*(9*b**4 - 40*b**2*d**2 + 1 6*d**4))