Integrand size = 17, antiderivative size = 144 \[ \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx=\frac {\sin (2 a-3 c+(2 b-3 d) x)}{16 (2 b-3 d)}+\frac {3 \sin (2 a-c+(2 b-d) x)}{16 (2 b-d)}+\frac {3 \sin (c+d x)}{8 d}+\frac {\sin (3 c+3 d x)}{24 d}+\frac {3 \sin (2 a+c+(2 b+d) x)}{16 (2 b+d)}+\frac {\sin (2 a+3 c+(2 b+3 d) x)}{16 (2 b+3 d)} \] Output:
sin(2*a-3*c+(2*b-3*d)*x)/(32*b-48*d)+3*sin(2*a-c+(2*b-d)*x)/(32*b-16*d)+3/ 8*sin(d*x+c)/d+1/24*sin(3*d*x+3*c)/d+3*sin(2*a+c+(2*b+d)*x)/(32*b+16*d)+si n(2*a+3*c+(2*b+3*d)*x)/(32*b+48*d)
Time = 1.10 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.10 \[ \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx=\frac {1}{48} \left (\frac {18 \cos (d x) \sin (c)}{d}+\frac {2 \cos (3 d x) \sin (3 c)}{d}+\frac {18 \cos (c) \sin (d x)}{d}+\frac {2 \cos (3 c) \sin (3 d x)}{d}+\frac {3 \sin (2 a-3 c+2 b x-3 d x)}{2 b-3 d}+\frac {9 \sin (2 a-c+2 b x-d x)}{2 b-d}+\frac {9 \sin (2 a+c+2 b x+d x)}{2 b+d}+\frac {3 \sin (2 a+3 c+2 b x+3 d x)}{2 b+3 d}\right ) \] Input:
Integrate[Cos[a + b*x]^2*Cos[c + d*x]^3,x]
Output:
((18*Cos[d*x]*Sin[c])/d + (2*Cos[3*d*x]*Sin[3*c])/d + (18*Cos[c]*Sin[d*x]) /d + (2*Cos[3*c]*Sin[3*d*x])/d + (3*Sin[2*a - 3*c + 2*b*x - 3*d*x])/(2*b - 3*d) + (9*Sin[2*a - c + 2*b*x - d*x])/(2*b - d) + (9*Sin[2*a + c + 2*b*x + d*x])/(2*b + d) + (3*Sin[2*a + 3*c + 2*b*x + 3*d*x])/(2*b + 3*d))/48
Time = 0.31 (sec) , antiderivative size = 144, 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.
| \(\displaystyle \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx\) |
| \(\Big \downarrow \) 5081 |
| \(\displaystyle \int \left (\frac {1}{16} \cos (2 a+x (2 b-3 d)-3 c)+\frac {3}{16} \cos (2 a+x (2 b-d)-c)+\frac {3}{16} \cos (2 a+x (2 b+d)+c)+\frac {1}{16} \cos (2 a+x (2 b+3 d)+3 c)+\frac {3}{8} \cos (c+d x)+\frac {1}{8} \cos (3 c+3 d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sin (2 a+x (2 b-3 d)-3 c)}{16 (2 b-3 d)}+\frac {3 \sin (2 a+x (2 b-d)-c)}{16 (2 b-d)}+\frac {3 \sin (2 a+x (2 b+d)+c)}{16 (2 b+d)}+\frac {\sin (2 a+x (2 b+3 d)+3 c)}{16 (2 b+3 d)}+\frac {3 \sin (c+d x)}{8 d}+\frac {\sin (3 c+3 d x)}{24 d}\) |
Input:
Int[Cos[a + b*x]^2*Cos[c + d*x]^3,x]
Output:
Sin[2*a - 3*c + (2*b - 3*d)*x]/(16*(2*b - 3*d)) + (3*Sin[2*a - c + (2*b - d)*x])/(16*(2*b - d)) + (3*Sin[c + d*x])/(8*d) + Sin[3*c + 3*d*x]/(24*d) + (3*Sin[2*a + c + (2*b + d)*x])/(16*(2*b + d)) + Sin[2*a + 3*c + (2*b + 3* d)*x]/(16*(2*b + 3*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 = 12.38 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.92
| method | result | size |
| default | \(\frac {3 \sin \left (d x +c \right )}{8 d}+\frac {\sin \left (3 d x +3 c \right )}{24 d}+\frac {\sin \left (2 a -3 c +\left (2 b -3 d \right ) x \right )}{32 b -48 d}+\frac {3 \sin \left (2 a -c +\left (2 b -d \right ) x \right )}{16 \left (2 b -d \right )}+\frac {3 \sin \left (2 a +c +\left (2 b +d \right ) x \right )}{16 \left (2 b +d \right )}+\frac {\sin \left (2 a +3 c +\left (2 b +3 d \right ) x \right )}{32 b +48 d}\) | \(133\) |
| parallelrisch | \(\frac {\left (24 d \,b^{3}+36 b^{2} d^{2}-6 d^{3} b -9 d^{4}\right ) \sin \left (2 a -3 c +\left (2 b -3 d \right ) x \right )+72 \left (d \left (b +\frac {3 d}{2}\right ) \left (b +\frac {d}{2}\right ) \sin \left (2 a -c +\left (2 b -d \right ) x \right )+\left (b -\frac {d}{2}\right ) \left (\frac {d \left (b +\frac {d}{2}\right ) \sin \left (2 a +3 c +\left (2 b +3 d \right ) x \right )}{3}+\left (\sin \left (2 a +c +\left (2 b +d \right ) x \right ) d +4 \left (\sin \left (d x +c \right )+\frac {\sin \left (3 d x +3 c \right )}{9}\right ) \left (b +\frac {d}{2}\right )\right ) \left (b +\frac {3 d}{2}\right )\right )\right ) \left (b -\frac {3 d}{2}\right )}{768 b^{4} d -1920 b^{2} d^{3}+432 d^{5}}\) | \(185\) |
| risch | \(\frac {3 \sin \left (d x +c \right ) b^{2}}{2 d \left (2 b -d \right ) \left (2 b +d \right )}-\frac {3 d \sin \left (d x +c \right )}{8 \left (2 b -d \right ) \left (2 b +d \right )}+\frac {\sin \left (2 b x -3 d x +2 a -3 c \right ) b}{8 \left (2 b -3 d \right ) \left (2 b +3 d \right )}+\frac {3 d \sin \left (2 b x -3 d x +2 a -3 c \right )}{16 \left (2 b -3 d \right ) \left (2 b +3 d \right )}+\frac {3 \sin \left (2 b x -d x +2 a -c \right ) b}{8 \left (2 b -d \right ) \left (2 b +d \right )}+\frac {3 d \sin \left (2 b x -d x +2 a -c \right )}{16 \left (2 b -d \right ) \left (2 b +d \right )}+\frac {3 \sin \left (2 b x +d x +2 a +c \right ) b}{8 \left (2 b -d \right ) \left (2 b +d \right )}-\frac {3 d \sin \left (2 b x +d x +2 a +c \right )}{16 \left (2 b -d \right ) \left (2 b +d \right )}+\frac {\sin \left (2 b x +3 d x +2 a +3 c \right ) b}{8 \left (2 b -3 d \right ) \left (2 b +3 d \right )}-\frac {3 d \sin \left (2 b x +3 d x +2 a +3 c \right )}{16 \left (2 b -3 d \right ) \left (2 b +3 d \right )}+\frac {\sin \left (3 d x +3 c \right ) b^{2}}{6 d \left (2 b -3 d \right ) \left (2 b +3 d \right )}-\frac {3 d \sin \left (3 d x +3 c \right )}{8 \left (2 b -3 d \right ) \left (2 b +3 d \right )}\) | \(404\) |
| orering | \(\text {Expression too large to display}\) | \(3070\) |
Input:
int(cos(b*x+a)^2*cos(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
3/8*sin(d*x+c)/d+1/24*sin(3*d*x+3*c)/d+1/16*sin(2*a-3*c+(2*b-3*d)*x)/(2*b- 3*d)+3/16*sin(2*a-c+(2*b-d)*x)/(2*b-d)+3/16/(2*b+d)*sin(2*a+c+(2*b+d)*x)+1 /16*sin(2*a+3*c+(2*b+3*d)*x)/(2*b+3*d)
Time = 0.08 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.13 \[ \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx=-\frac {6 \, {\left (6 \, b d^{3} \cos \left (b x + a\right ) \cos \left (d x + c\right ) - {\left (4 \, b^{3} d - b d^{3}\right )} \cos \left (b x + a\right ) \cos \left (d x + c\right )^{3}\right )} \sin \left (b x + a\right ) - {\left (18 \, d^{4} \cos \left (b x + a\right )^{2} + 16 \, b^{4} - 40 \, b^{2} d^{2} + {\left (8 \, b^{4} - 2 \, b^{2} d^{2} - 9 \, {\left (4 \, b^{2} d^{2} - d^{4}\right )} \cos \left (b x + a\right )^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (16 \, b^{4} d - 40 \, b^{2} d^{3} + 9 \, d^{5}\right )}} \] Input:
integrate(cos(b*x+a)^2*cos(d*x+c)^3,x, algorithm="fricas")
Output:
-1/3*(6*(6*b*d^3*cos(b*x + a)*cos(d*x + c) - (4*b^3*d - b*d^3)*cos(b*x + a )*cos(d*x + c)^3)*sin(b*x + a) - (18*d^4*cos(b*x + a)^2 + 16*b^4 - 40*b^2* d^2 + (8*b^4 - 2*b^2*d^2 - 9*(4*b^2*d^2 - d^4)*cos(b*x + a)^2)*cos(d*x + c )^2)*sin(d*x + c))/(16*b^4*d - 40*b^2*d^3 + 9*d^5)
Leaf count of result is larger than twice the leaf count of optimal. 2004 vs. \(2 (116) = 232\).
Time = 5.60 (sec) , antiderivative size = 2004, normalized size of antiderivative = 13.92 \[ \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)**2*cos(d*x+c)**3,x)
Output:
Piecewise((x*cos(a)**2*cos(c)**3, Eq(b, 0) & Eq(d, 0)), (3*x*sin(a - 3*d*x /2)**2*sin(c + d*x)**2*cos(c + d*x)/16 - x*sin(a - 3*d*x/2)**2*cos(c + d*x )**3/16 + x*sin(a - 3*d*x/2)*sin(c + d*x)**3*cos(a - 3*d*x/2)/8 - 3*x*sin( a - 3*d*x/2)*sin(c + d*x)*cos(a - 3*d*x/2)*cos(c + d*x)**2/8 - 3*x*sin(c + d*x)**2*cos(a - 3*d*x/2)**2*cos(c + d*x)/16 + x*cos(a - 3*d*x/2)**2*cos(c + d*x)**3/16 + 11*sin(a - 3*d*x/2)**2*sin(c + d*x)**3/(48*d) + sin(a - 3* d*x/2)**2*sin(c + d*x)*cos(c + d*x)**2/d + 3*sin(a - 3*d*x/2)*sin(c + d*x) **2*cos(a - 3*d*x/2)*cos(c + d*x)/(4*d) - 5*sin(a - 3*d*x/2)*cos(a - 3*d*x /2)*cos(c + d*x)**3/(8*d) + 7*sin(c + d*x)**3*cos(a - 3*d*x/2)**2/(16*d), Eq(b, -3*d/2)), (-3*x*sin(a - d*x/2)**2*sin(c + d*x)**2*cos(c + d*x)/16 - 3*x*sin(a - d*x/2)**2*cos(c + d*x)**3/16 - 3*x*sin(a - d*x/2)*sin(c + d*x) **3*cos(a - d*x/2)/8 - 3*x*sin(a - d*x/2)*sin(c + d*x)*cos(a - d*x/2)*cos( c + d*x)**2/8 + 3*x*sin(c + d*x)**2*cos(a - d*x/2)**2*cos(c + d*x)/16 + 3* x*cos(a - d*x/2)**2*cos(c + d*x)**3/16 + 49*sin(a - d*x/2)**2*sin(c + d*x) **3/(48*d) + sin(a - d*x/2)**2*sin(c + d*x)*cos(c + d*x)**2/d - 7*sin(a - d*x/2)*sin(c + d*x)**2*cos(a - d*x/2)*cos(c + d*x)/(4*d) - 13*sin(a - d*x/ 2)*cos(a - d*x/2)*cos(c + d*x)**3/(8*d) - 17*sin(c + d*x)**3*cos(a - d*x/2 )**2/(48*d), Eq(b, -d/2)), (-3*x*sin(a + d*x/2)**2*sin(c + d*x)**2*cos(c + d*x)/16 - 3*x*sin(a + d*x/2)**2*cos(c + d*x)**3/16 + 3*x*sin(a + d*x/2)*s in(c + d*x)**3*cos(a + d*x/2)/8 + 3*x*sin(a + d*x/2)*sin(c + d*x)*cos(a...
Leaf count of result is larger than twice the leaf count of optimal. 1362 vs. \(2 (132) = 264\).
Time = 0.11 (sec) , antiderivative size = 1362, normalized size of antiderivative = 9.46 \[ \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)^2*cos(d*x+c)^3,x, algorithm="maxima")
Output:
-1/96*(3*(8*b^3*d*sin(3*c) - 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) + 3*d^ 4*sin(3*c))*cos((2*b + 3*d)*x + 2*a + 6*c) - 3*(8*b^3*d*sin(3*c) - 12*b^2* d^2*sin(3*c) - 2*b*d^3*sin(3*c) + 3*d^4*sin(3*c))*cos((2*b + 3*d)*x + 2*a) + 9*(8*b^3*d*sin(3*c) - 4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) + 9*d^4*si n(3*c))*cos((2*b + d)*x + 2*a + 4*c) - 9*(8*b^3*d*sin(3*c) - 4*b^2*d^2*sin (3*c) - 18*b*d^3*sin(3*c) + 9*d^4*sin(3*c))*cos((2*b + d)*x + 2*a - 2*c) - 9*(8*b^3*d*sin(3*c) + 4*b^2*d^2*sin(3*c) - 18*b*d^3*sin(3*c) - 9*d^4*sin( 3*c))*cos(-(2*b - d)*x - 2*a + 4*c) + 9*(8*b^3*d*sin(3*c) + 4*b^2*d^2*sin( 3*c) - 18*b*d^3*sin(3*c) - 9*d^4*sin(3*c))*cos(-(2*b - d)*x - 2*a - 2*c) - 3*(8*b^3*d*sin(3*c) + 12*b^2*d^2*sin(3*c) - 2*b*d^3*sin(3*c) - 3*d^4*sin( 3*c))*cos(-(2*b - 3*d)*x - 2*a + 6*c) + 3*(8*b^3*d*sin(3*c) + 12*b^2*d^2*s in(3*c) - 2*b*d^3*sin(3*c) - 3*d^4*sin(3*c))*cos(-(2*b - 3*d)*x - 2*a) - 2 *(16*b^4*sin(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*cos(3*d*x) + 2*( 16*b^4*sin(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*cos(3*d*x + 6*c) + 18*(16*b^4*sin(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*cos(d*x + 4*c ) - 18*(16*b^4*sin(3*c) - 40*b^2*d^2*sin(3*c) + 9*d^4*sin(3*c))*cos(d*x - 2*c) - 3*(8*b^3*d*cos(3*c) - 12*b^2*d^2*cos(3*c) - 2*b*d^3*cos(3*c) + 3*d^ 4*cos(3*c))*sin((2*b + 3*d)*x + 2*a + 6*c) - 3*(8*b^3*d*cos(3*c) - 12*b^2* d^2*cos(3*c) - 2*b*d^3*cos(3*c) + 3*d^4*cos(3*c))*sin((2*b + 3*d)*x + 2*a) - 9*(8*b^3*d*cos(3*c) - 4*b^2*d^2*cos(3*c) - 18*b*d^3*cos(3*c) + 9*d^4...
Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90 \[ \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx=\frac {\sin \left (2 \, b x + 3 \, d x + 2 \, a + 3 \, c\right )}{16 \, {\left (2 \, b + 3 \, d\right )}} + \frac {3 \, \sin \left (2 \, b x + d x + 2 \, a + c\right )}{16 \, {\left (2 \, b + d\right )}} + \frac {3 \, \sin \left (2 \, b x - d x + 2 \, a - c\right )}{16 \, {\left (2 \, b - d\right )}} + \frac {\sin \left (2 \, b x - 3 \, d x + 2 \, a - 3 \, c\right )}{16 \, {\left (2 \, b - 3 \, d\right )}} + \frac {\sin \left (3 \, d x + 3 \, c\right )}{24 \, d} + \frac {3 \, \sin \left (d x + c\right )}{8 \, d} \] Input:
integrate(cos(b*x+a)^2*cos(d*x+c)^3,x, algorithm="giac")
Output:
1/16*sin(2*b*x + 3*d*x + 2*a + 3*c)/(2*b + 3*d) + 3/16*sin(2*b*x + d*x + 2 *a + c)/(2*b + d) + 3/16*sin(2*b*x - d*x + 2*a - c)/(2*b - d) + 1/16*sin(2 *b*x - 3*d*x + 2*a - 3*c)/(2*b - 3*d) + 1/24*sin(3*d*x + 3*c)/d + 3/8*sin( d*x + c)/d
Time = 21.04 (sec) , antiderivative size = 495, normalized size of antiderivative = 3.44 \[ \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx=-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left (\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (24\,b^2-6\,d^2\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}-\frac {3\,d\,\left (2\,b+d\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}+\frac {3\,d\,{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (2\,b-d\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}\right )+{\mathrm {e}}^{a\,2{}\mathrm {i}+c\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {3\,d\,\left (2\,b-d\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (24\,b^2-6\,d^2\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}-\frac {3\,d\,{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (2\,b+d\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}\right )-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,3{}\mathrm {i}+b\,x\,2{}\mathrm {i}-d\,x\,3{}\mathrm {i}}\,\left (-\frac {3\,d\,\left (2\,b+3\,d\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (8\,b^2-18\,d^2\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}+\frac {3\,d\,{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (2\,b-3\,d\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}\right )+{\mathrm {e}}^{a\,2{}\mathrm {i}+c\,3{}\mathrm {i}+b\,x\,2{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\left (\frac {3\,d\,\left (2\,b-3\,d\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (8\,b^2-18\,d^2\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}-\frac {3\,d\,{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (2\,b+3\,d\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}\right ) \] Input:
int(cos(a + b*x)^2*cos(c + d*x)^3,x)
Output:
exp(a*2i + c*1i + b*x*2i + d*x*1i)*((3*d*(2*b - d))/(b^2*d*128i - d^3*32i) + (exp(- a*2i - b*x*2i)*(24*b^2 - 6*d^2))/(b^2*d*128i - d^3*32i) - (3*d*e xp(- a*4i - b*x*4i)*(2*b + d))/(b^2*d*128i - d^3*32i)) - exp(a*2i - c*1i + b*x*2i - d*x*1i)*((exp(- a*2i - b*x*2i)*(24*b^2 - 6*d^2))/(b^2*d*128i - d ^3*32i) - (3*d*(2*b + d))/(b^2*d*128i - d^3*32i) + (3*d*exp(- a*4i - b*x*4 i)*(2*b - d))/(b^2*d*128i - d^3*32i)) - exp(a*2i - c*3i + b*x*2i - d*x*3i) *((exp(- a*2i - b*x*2i)*(8*b^2 - 18*d^2))/(b^2*d*384i - d^3*864i) - (3*d*( 2*b + 3*d))/(b^2*d*384i - d^3*864i) + (3*d*exp(- a*4i - b*x*4i)*(2*b - 3*d ))/(b^2*d*384i - d^3*864i)) + exp(a*2i + c*3i + b*x*2i + d*x*3i)*((3*d*(2* b - 3*d))/(b^2*d*384i - d^3*864i) + (exp(- a*2i - b*x*2i)*(8*b^2 - 18*d^2) )/(b^2*d*384i - d^3*864i) - (3*d*exp(- a*4i - b*x*4i)*(2*b + 3*d))/(b^2*d* 384i - d^3*864i))
Time = 0.26 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.10 \[ \int \cos ^2(a+b x) \cos ^3(c+d x) \, dx=\frac {-24 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right ) \sin \left (d x +c \right )^{2} b^{3} d +6 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right ) \sin \left (d x +c \right )^{2} b \,d^{3}+24 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right ) b^{3} d -42 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right ) b \,d^{3}-36 \sin \left (b x +a \right )^{2} \sin \left (d x +c \right )^{3} b^{2} d^{2}+9 \sin \left (b x +a \right )^{2} \sin \left (d x +c \right )^{3} d^{4}+36 \sin \left (b x +a \right )^{2} \sin \left (d x +c \right ) b^{2} d^{2}-27 \sin \left (b x +a \right )^{2} \sin \left (d x +c \right ) d^{4}-8 \sin \left (d x +c \right )^{3} b^{4}+38 \sin \left (d x +c \right )^{3} b^{2} d^{2}-9 \sin \left (d x +c \right )^{3} d^{4}+24 \sin \left (d x +c \right ) b^{4}-78 \sin \left (d x +c \right ) b^{2} d^{2}+27 \sin \left (d x +c \right ) d^{4}}{3 d \left (16 b^{4}-40 b^{2} d^{2}+9 d^{4}\right )} \] Input:
int(cos(b*x+a)^2*cos(d*x+c)^3,x)
Output:
( - 24*cos(a + b*x)*cos(c + d*x)*sin(a + b*x)*sin(c + d*x)**2*b**3*d + 6*c os(a + b*x)*cos(c + d*x)*sin(a + b*x)*sin(c + d*x)**2*b*d**3 + 24*cos(a + b*x)*cos(c + d*x)*sin(a + b*x)*b**3*d - 42*cos(a + b*x)*cos(c + d*x)*sin(a + b*x)*b*d**3 - 36*sin(a + b*x)**2*sin(c + d*x)**3*b**2*d**2 + 9*sin(a + b*x)**2*sin(c + d*x)**3*d**4 + 36*sin(a + b*x)**2*sin(c + d*x)*b**2*d**2 - 27*sin(a + b*x)**2*sin(c + d*x)*d**4 - 8*sin(c + d*x)**3*b**4 + 38*sin(c + d*x)**3*b**2*d**2 - 9*sin(c + d*x)**3*d**4 + 24*sin(c + d*x)*b**4 - 78*s in(c + d*x)*b**2*d**2 + 27*sin(c + d*x)*d**4)/(3*d*(16*b**4 - 40*b**2*d**2 + 9*d**4))