Integrand size = 17, antiderivative size = 144 \[ \int \cos ^3(c+d x) \sin ^2(a+b x) \, dx=-\frac {\sin (2 a-3 c+(2 b-3 d) x)}{16 (2 b-3 d)}-\frac {3 \sin (2 a-c+(2 b-d) x)}{16 (2 b-d)}+\frac {3 \sin (c+d x)}{8 d}+\frac {\sin (3 c+3 d x)}{24 d}-\frac {3 \sin (2 a+c+(2 b+d) x)}{16 (2 b+d)}-\frac {\sin (2 a+3 c+(2 b+3 d) x)}{16 (2 b+3 d)} \] Output:
-1/16*sin(2*a-3*c+(2*b-3*d)*x)/(2*b-3*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 )-sin(2*a+3*c+(2*b+3*d)*x)/(32*b+48*d)
Time = 1.15 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.10 \[ \int \cos ^3(c+d x) \sin ^2(a+b x) \, dx=\frac {1}{48} \left (\frac {18 \cos (d x) \sin (c)}{d}+\frac {2 \cos (3 d x) \sin (3 c)}{d}+\frac {18 \cos (c) \sin (d x)}{d}+\frac {2 \cos (3 c) \sin (3 d x)}{d}-\frac {3 \sin (2 a-3 c+2 b x-3 d x)}{2 b-3 d}-\frac {9 \sin (2 a-c+2 b x-d x)}{2 b-d}-\frac {9 \sin (2 a+c+2 b x+d x)}{2 b+d}-\frac {3 \sin (2 a+3 c+2 b x+3 d x)}{2 b+3 d}\right ) \] Input:
Integrate[Cos[c + d*x]^3*Sin[a + b*x]^2,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 = 0.33 (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 = {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.
| \(\displaystyle \int \sin ^2(a+b x) \cos ^3(c+d x) \, dx\) |
| \(\Big \downarrow \) 5085 |
| \(\displaystyle \int \left (-\frac {1}{16} \cos (2 a+x (2 b-3 d)-3 c)-\frac {3}{16} \cos (2 a+x (2 b-d)-c)-\frac {3}{16} \cos (2 a+x (2 b+d)+c)-\frac {1}{16} \cos (2 a+x (2 b+3 d)+3 c)+\frac {3}{8} \cos (c+d x)+\frac {1}{8} \cos (3 c+3 d x)\right )dx\) |
| \(\Big \downarrow \) 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[c + d*x]^3*Sin[a + b*x]^2,x]
Output:
-1/16*Sin[2*a - 3*c + (2*b - 3*d)*x]/(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[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 = 13.14 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.92
| method | result | size |
| default | \(\frac {3 \sin \left (d x +c \right )}{8 d}+\frac {\sin \left (3 d x +3 c \right )}{24 d}-\frac {\sin \left (2 a -3 c +\left (2 b -3 d \right ) x \right )}{16 \left (2 b -3 d \right )}-\frac {3 \sin \left (2 a -c +\left (2 b -d \right ) x \right )}{16 \left (2 b -d \right )}-\frac {3 \sin \left (2 a +c +\left (2 b +d \right ) x \right )}{16 \left (2 b +d \right )}-\frac {\sin \left (2 a +3 c +\left (2 b +3 d \right ) x \right )}{16 \left (2 b +3 d \right )}\) | \(133\) |
| parallelrisch | \(\frac {\left (-24 d \,b^{3}-36 b^{2} d^{2}+6 b \,d^{3}+9 d^{4}\right ) \sin \left (2 a -3 c +\left (2 b -3 d \right ) x \right )-72 \left (\left (b +\frac {3 d}{2}\right ) d \left (b +\frac {d}{2}\right ) \sin \left (2 a -c +\left (2 b -d \right ) x \right )+\left (b -\frac {d}{2}\right ) \left (\frac {d \left (b +\frac {d}{2}\right ) \sin \left (2 a +3 c +\left (2 b +3 d \right ) x \right )}{3}+\left (\sin \left (2 a +c +\left (2 b +d \right ) x \right ) d -4 \left (\sin \left (d x +c \right )+\frac {\sin \left (3 d x +3 c \right )}{9}\right ) \left (b +\frac {d}{2}\right )\right ) \left (b +\frac {3 d}{2}\right )\right )\right ) \left (b -\frac {3 d}{2}\right )}{768 b^{4} d -1920 d^{3} b^{2}+432 d^{5}}\) | \(185\) |
| 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(d*x+c)^3*sin(b*x+a)^2,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 = 174, normalized size of antiderivative = 1.21 \[ \int \cos ^3(c+d x) \sin ^2(a+b x) \, dx=\frac {6 \, {\left (6 \, b d^{3} \cos \left (b x + a\right ) \cos \left (d x + c\right ) - {\left (4 \, b^{3} d - b d^{3}\right )} \cos \left (b x + a\right ) \cos \left (d x + c\right )^{3}\right )} \sin \left (b x + a\right ) - {\left (18 \, d^{4} \cos \left (b x + a\right )^{2} - 16 \, b^{4} + 40 \, b^{2} d^{2} - 18 \, d^{4} - {\left (8 \, b^{4} - 38 \, b^{2} d^{2} + 9 \, d^{4} + 9 \, {\left (4 \, b^{2} d^{2} - d^{4}\right )} \cos \left (b x + a\right )^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (16 \, b^{4} d - 40 \, b^{2} d^{3} + 9 \, d^{5}\right )}} \] Input:
integrate(cos(d*x+c)^3*sin(b*x+a)^2,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 - 18*d^4 - (8*b^4 - 38*b^2*d^2 + 9*d^4 + 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 = 5.68 (sec) , antiderivative size = 2004, normalized size of antiderivative = 13.92 \[ \int \cos ^3(c+d x) \sin ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**3*sin(b*x+a)**2,x)
Output:
Piecewise((x*sin(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 + 5*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 + 5*sin(a - 3*d*x/2)*sin(c + d*x) **2*cos(a - 3*d*x/2)*cos(c + d*x)/(4*d) - sin(a - 3*d*x/2)*cos(a - 3*d*x/2 )*cos(c + d*x)**3/(24*d) + 9*sin(c + d*x)**3*cos(a - 3*d*x/2)**2/(16*d), E q(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 + 31*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 - sin(a - d*x/ 2)*sin(c + d*x)**2*cos(a - d*x/2)*cos(c + d*x)/(4*d) - 3*sin(a - d*x/2)*co s(a - d*x/2)*cos(c + d*x)**3/(8*d) + 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)*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)...
Leaf count of result is larger than twice the leaf count of optimal. 1362 vs. \(2 (132) = 264\).
Time = 0.19 (sec) , antiderivative size = 1362, normalized size of antiderivative = 9.46 \[ \int \cos ^3(c+d x) \sin ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^3*sin(b*x+a)^2,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*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) - 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*si n(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*(1 6*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 = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90 \[ \int \cos ^3(c+d x) \sin ^2(a+b x) \, dx=-\frac {\sin \left (2 \, b x + 3 \, d x + 2 \, a + 3 \, c\right )}{16 \, {\left (2 \, b + 3 \, d\right )}} - \frac {3 \, \sin \left (2 \, b x + d x + 2 \, a + c\right )}{16 \, {\left (2 \, b + d\right )}} - \frac {3 \, \sin \left (2 \, b x - d x + 2 \, a - c\right )}{16 \, {\left (2 \, b - d\right )}} - \frac {\sin \left (2 \, b x - 3 \, d x + 2 \, a - 3 \, c\right )}{16 \, {\left (2 \, b - 3 \, d\right )}} + \frac {\sin \left (3 \, d x + 3 \, c\right )}{24 \, d} + \frac {3 \, \sin \left (d x + c\right )}{8 \, d} \] Input:
integrate(cos(d*x+c)^3*sin(b*x+a)^2,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 = 18.99 (sec) , antiderivative size = 495, normalized size of antiderivative = 3.44 \[ \int \cos ^3(c+d x) \sin ^2(a+b x) \, dx=-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left (\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (24\,b^2-6\,d^2\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}+\frac {3\,d\,\left (2\,b+d\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}-\frac {3\,d\,{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (2\,b-d\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}\right )+{\mathrm {e}}^{a\,2{}\mathrm {i}+c\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (-\frac {3\,d\,\left (2\,b-d\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (24\,b^2-6\,d^2\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}+\frac {3\,d\,{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (2\,b+d\right )}{b^2\,d\,128{}\mathrm {i}-d^3\,32{}\mathrm {i}}\right )-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,3{}\mathrm {i}+b\,x\,2{}\mathrm {i}-d\,x\,3{}\mathrm {i}}\,\left (\frac {3\,d\,\left (2\,b+3\,d\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (8\,b^2-18\,d^2\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}-\frac {3\,d\,{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (2\,b-3\,d\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}\right )+{\mathrm {e}}^{a\,2{}\mathrm {i}+c\,3{}\mathrm {i}+b\,x\,2{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\left (-\frac {3\,d\,\left (2\,b-3\,d\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}+\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,\left (8\,b^2-18\,d^2\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}+\frac {3\,d\,{\mathrm {e}}^{-a\,4{}\mathrm {i}-b\,x\,4{}\mathrm {i}}\,\left (2\,b+3\,d\right )}{b^2\,d\,384{}\mathrm {i}-d^3\,864{}\mathrm {i}}\right ) \] Input:
int(cos(c + d*x)^3*sin(a + b*x)^2,x)
Output:
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*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) *((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)) + 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))
Time = 0.16 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.94 \[ \int \cos ^3(c+d x) \sin ^2(a+b x) \, dx=\frac {24 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right ) \sin \left (d x +c \right )^{2} b^{3} d -6 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right ) \sin \left (d x +c \right )^{2} b \,d^{3}-24 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right ) b^{3} d +42 \cos \left (b x +a \right ) \cos \left (d x +c \right ) \sin \left (b x +a \right ) b \,d^{3}+36 \sin \left (b x +a \right )^{2} \sin \left (d x +c \right )^{3} b^{2} d^{2}-9 \sin \left (b x +a \right )^{2} \sin \left (d x +c \right )^{3} d^{4}-36 \sin \left (b x +a \right )^{2} \sin \left (d x +c \right ) b^{2} d^{2}+27 \sin \left (b x +a \right )^{2} \sin \left (d x +c \right ) d^{4}-8 \sin \left (d x +c \right )^{3} b^{4}+2 \sin \left (d x +c \right )^{3} b^{2} d^{2}+24 \sin \left (d x +c \right ) b^{4}-42 \sin \left (d x +c \right ) b^{2} d^{2}}{3 d \left (16 b^{4}-40 b^{2} d^{2}+9 d^{4}\right )} \] Input:
int(cos(d*x+c)^3*sin(b*x+a)^2,x)
Output:
(24*cos(a + b*x)*cos(c + d*x)*sin(a + b*x)*sin(c + d*x)**2*b**3*d - 6*cos( 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 + 2*sin(c + d* x)**3*b**2*d**2 + 24*sin(c + d*x)*b**4 - 42*sin(c + d*x)*b**2*d**2)/(3*d*( 16*b**4 - 40*b**2*d**2 + 9*d**4))