Integrand size = 17, antiderivative size = 138 \[ \int \cos ^2(c+d x) \sin ^3(a+b x) \, dx=-\frac {3 \cos (a+b x)}{8 b}+\frac {\cos (3 a+3 b x)}{24 b}-\frac {3 \cos (a-2 c+(b-2 d) x)}{16 (b-2 d)}+\frac {\cos (3 a-2 c+(3 b-2 d) x)}{16 (3 b-2 d)}-\frac {3 \cos (a+2 c+(b+2 d) x)}{16 (b+2 d)}+\frac {\cos (3 a+2 c+(3 b+2 d) x)}{16 (3 b+2 d)} \] Output:
-3/8*cos(b*x+a)/b+1/24*cos(3*b*x+3*a)/b-3*cos(a-2*c+(b-2*d)*x)/(16*b-32*d) +cos(3*a-2*c+(3*b-2*d)*x)/(48*b-32*d)-3*cos(a+2*c+(b+2*d)*x)/(16*b+32*d)+c os(3*a+2*c+(3*b+2*d)*x)/(48*b+32*d)
Time = 1.28 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11 \[ \int \cos ^2(c+d x) \sin ^3(a+b x) \, dx=\frac {1}{48} \left (-\frac {18 \cos (a) \cos (b x)}{b}+\frac {2 \cos (3 a) \cos (3 b x)}{b}-\frac {9 \cos (a-2 c+b x-2 d x)}{b-2 d}+\frac {3 \cos (3 a-2 c+3 b x-2 d x)}{3 b-2 d}-\frac {9 \cos (a+2 c+b x+2 d x)}{b+2 d}+\frac {3 \cos (3 a+2 c+3 b x+2 d x)}{3 b+2 d}+\frac {18 \sin (a) \sin (b x)}{b}-\frac {2 \sin (3 a) \sin (3 b x)}{b}\right ) \] Input:
Integrate[Cos[c + d*x]^2*Sin[a + b*x]^3,x]
Output:
((-18*Cos[a]*Cos[b*x])/b + (2*Cos[3*a]*Cos[3*b*x])/b - (9*Cos[a - 2*c + b* x - 2*d*x])/(b - 2*d) + (3*Cos[3*a - 2*c + 3*b*x - 2*d*x])/(3*b - 2*d) - ( 9*Cos[a + 2*c + b*x + 2*d*x])/(b + 2*d) + (3*Cos[3*a + 2*c + 3*b*x + 2*d*x ])/(3*b + 2*d) + (18*Sin[a]*Sin[b*x])/b - (2*Sin[3*a]*Sin[3*b*x])/b)/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 = {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.
| \(\displaystyle \int \sin ^3(a+b x) \cos ^2(c+d x) \, dx\) |
| \(\Big \downarrow \) 5085 |
| \(\displaystyle \int \left (\frac {3}{16} \sin (a+x (b-2 d)-2 c)-\frac {1}{16} \sin (3 a+x (3 b-2 d)-2 c)+\frac {3}{16} \sin (a+x (b+2 d)+2 c)-\frac {1}{16} \sin (3 a+x (3 b+2 d)+2 c)+\frac {3}{8} \sin (a+b x)-\frac {1}{8} \sin (3 a+3 b x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \cos (a+x (b-2 d)-2 c)}{16 (b-2 d)}+\frac {\cos (3 a+x (3 b-2 d)-2 c)}{16 (3 b-2 d)}-\frac {3 \cos (a+x (b+2 d)+2 c)}{16 (b+2 d)}+\frac {\cos (3 a+x (3 b+2 d)+2 c)}{16 (3 b+2 d)}-\frac {3 \cos (a+b x)}{8 b}+\frac {\cos (3 a+3 b x)}{24 b}\) |
Input:
Int[Cos[c + d*x]^2*Sin[a + b*x]^3,x]
Output:
(-3*Cos[a + b*x])/(8*b) + Cos[3*a + 3*b*x]/(24*b) - (3*Cos[a - 2*c + (b - 2*d)*x])/(16*(b - 2*d)) + Cos[3*a - 2*c + (3*b - 2*d)*x]/(16*(3*b - 2*d)) - (3*Cos[a + 2*c + (b + 2*d)*x])/(16*(b + 2*d)) + Cos[3*a + 2*c + (3*b + 2 *d)*x]/(16*(3*b + 2*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 = 12.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.92
| method | result | size |
| default | \(-\frac {3 \cos \left (b x +a \right )}{8 b}+\frac {\cos \left (3 b x +3 a \right )}{24 b}-\frac {3 \cos \left (a -2 c +\left (b -2 d \right ) x \right )}{16 \left (b -2 d \right )}-\frac {3 \cos \left (a +2 c +\left (b +2 d \right ) x \right )}{16 \left (b +2 d \right )}+\frac {\cos \left (3 a -2 c +\left (3 b -2 d \right ) x \right )}{48 b -32 d}+\frac {\cos \left (3 a +2 c +\left (3 b +2 d \right ) x \right )}{48 b +32 d}\) | \(127\) |
| parallelrisch | \(\frac {9 \left (b +\frac {2 d}{3}\right ) \left (b +2 d \right ) b \left (b -2 d \right ) \cos \left (3 a -2 c +\left (3 b -2 d \right ) x \right )+9 \left (b +2 d \right ) \left (b -\frac {2 d}{3}\right ) b \left (b -2 d \right ) \cos \left (3 a +2 c +\left (3 b +2 d \right ) x \right )-81 \left (b +\frac {2 d}{3}\right ) \left (b +2 d \right ) \left (b -\frac {2 d}{3}\right ) b \cos \left (a -2 c +\left (b -2 d \right ) x \right )-81 \left (b +\frac {2 d}{3}\right ) \left (b -\frac {2 d}{3}\right ) b \left (b -2 d \right ) \cos \left (a +2 c +\left (b +2 d \right ) x \right )+\left (18 b^{4}-80 b^{2} d^{2}+32 d^{4}\right ) \cos \left (3 b x +3 a \right )+\left (-162 b^{4}+720 b^{2} d^{2}-288 d^{4}\right ) \cos \left (b x +a \right )-288 b^{4}+640 b^{2} d^{2}-256 d^{4}}{432 b^{5}-1920 b^{3} d^{2}+768 d^{4} b}\) | \(230\) |
| risch | \(-\frac {3 \cos \left (b x +a \right )}{8 b}-\frac {27 \cos \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 \cos \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 \cos \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 \cos \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 \cos \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 \cos \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 \cos \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 \cos \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 \cos \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 {\cos \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 \cos \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 {\cos \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 \cos \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 {\cos \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 \cos \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 {\cos \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 {\cos \left (3 b x +3 a \right )}{24 b}\) | \(859\) |
| orering | \(\text {Expression too large to display}\) | \(3070\) |
Input:
int(cos(d*x+c)^2*sin(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
-3/8*cos(b*x+a)/b+1/24*cos(3*b*x+3*a)/b-3/16/(b-2*d)*cos(a-2*c+(b-2*d)*x)- 3/16/(b+2*d)*cos(a+2*c+(b+2*d)*x)+1/16/(3*b-2*d)*cos(3*a-2*c+(3*b-2*d)*x)+ 1/16/(3*b+2*d)*cos(3*a+2*c+(3*b+2*d)*x)
Time = 0.08 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.30 \[ \int \cos ^2(c+d x) \sin ^3(a+b x) \, dx=-\frac {2 \, {\left (b^{2} d^{2} - 4 \, d^{4}\right )} \cos \left (b x + a\right )^{3} + 6 \, {\left (7 \, b^{3} d - 4 \, b d^{3} - {\left (b^{3} d - 4 \, b d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \cos \left (d x + c\right ) \sin \left (b x + a\right ) \sin \left (d x + c\right ) - 9 \, {\left ({\left (b^{4} - 4 \, b^{2} d^{2}\right )} \cos \left (b x + a\right )^{3} - {\left (3 \, b^{4} - 4 \, b^{2} d^{2}\right )} \cos \left (b x + a\right )\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (7 \, b^{2} d^{2} - 4 \, d^{4}\right )} \cos \left (b x + a\right )}{3 \, {\left (9 \, b^{5} - 40 \, b^{3} d^{2} + 16 \, b d^{4}\right )}} \] Input:
integrate(cos(d*x+c)^2*sin(b*x+a)^3,x, algorithm="fricas")
Output:
-1/3*(2*(b^2*d^2 - 4*d^4)*cos(b*x + a)^3 + 6*(7*b^3*d - 4*b*d^3 - (b^3*d - 4*b*d^3)*cos(b*x + a)^2)*cos(d*x + c)*sin(b*x + a)*sin(d*x + c) - 9*((b^4 - 4*b^2*d^2)*cos(b*x + a)^3 - (3*b^4 - 4*b^2*d^2)*cos(b*x + a))*cos(d*x + c)^2 - 6*(7*b^2*d^2 - 4*d^4)*cos(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 = 5.99 (sec) , antiderivative size = 2020, normalized size of antiderivative = 14.64 \[ \int \cos ^2(c+d x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**2*sin(b*x+a)**3,x)
Output:
Piecewise((x*sin(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))*sin(a)**3, Eq (b, 0)), (-3*x*sin(a - 2*d*x)**3*sin(c + d*x)**2/16 + 3*x*sin(a - 2*d*x)** 3*cos(c + d*x)**2/16 + 3*x*sin(a - 2*d*x)**2*sin(c + d*x)*cos(a - 2*d*x)*c os(c + d*x)/8 - 3*x*sin(a - 2*d*x)*sin(c + d*x)**2*cos(a - 2*d*x)**2/16 + 3*x*sin(a - 2*d*x)*cos(a - 2*d*x)**2*cos(c + d*x)**2/16 + 3*x*sin(c + d*x) *cos(a - 2*d*x)**3*cos(c + d*x)/8 - 3*sin(a - 2*d*x)**3*sin(c + d*x)*cos(c + d*x)/(16*d) + sin(a - 2*d*x)**2*cos(a - 2*d*x)*cos(c + d*x)**2/(2*d) - sin(a - 2*d*x)*sin(c + d*x)*cos(a - 2*d*x)**2*cos(c + d*x)/(8*d) + sin(c + d*x)**2*cos(a - 2*d*x)**3/(96*d) + 31*cos(a - 2*d*x)**3*cos(c + d*x)**2/( 96*d), Eq(b, -2*d)), (-x*sin(a - 2*d*x/3)**3*sin(c + d*x)**2/16 + x*sin(a - 2*d*x/3)**3*cos(c + d*x)**2/16 + 3*x*sin(a - 2*d*x/3)**2*sin(c + d*x)*co s(a - 2*d*x/3)*cos(c + d*x)/8 + 3*x*sin(a - 2*d*x/3)*sin(c + d*x)**2*cos(a - 2*d*x/3)**2/16 - 3*x*sin(a - 2*d*x/3)*cos(a - 2*d*x/3)**2*cos(c + d*x)* *2/16 - x*sin(c + d*x)*cos(a - 2*d*x/3)**3*cos(c + d*x)/8 - sin(a - 2*d*x/ 3)**3*sin(c + d*x)*cos(c + d*x)/(16*d) + 3*sin(a - 2*d*x/3)**2*cos(a - 2*d *x/3)*cos(c + d*x)**2/(2*d) + 15*sin(a - 2*d*x/3)*sin(c + d*x)*cos(a - 2*d *x/3)**2*cos(c + d*x)/(8*d) + 27*sin(c + d*x)**2*cos(a - 2*d*x/3)**3/(32*d ) + 5*cos(a - 2*d*x/3)**3*cos(c + d*x)**2/(32*d), Eq(b, -2*d/3)), (-x*sin( a + 2*d*x/3)**3*sin(c + d*x)**2/16 + x*sin(a + 2*d*x/3)**3*cos(c + d*x)...
Leaf count of result is larger than twice the leaf count of optimal. 1360 vs. \(2 (126) = 252\).
Time = 0.15 (sec) , antiderivative size = 1360, normalized size of antiderivative = 9.86 \[ \int \cos ^2(c+d x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^2*sin(b*x+a)^3,x, algorithm="maxima")
Output:
1/96*(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))*cos((3*b + 2*d)*x + 3*a + 4*c) + 3*(3*b^4*cos(2*c) - 2*b^3*d*co s(2*c) - 12*b^2*d^2*cos(2*c) + 8*b*d^3*cos(2*c))*cos((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*b*d^3*cos (2*c))*cos(-(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))*cos(-(3*b - 2*d)*x - 3*a) - 9*(9*b^4*cos(2*c) - 18*b^3*d*cos(2*c) - 4*b^2*d^2*cos(2*c) + 8*b*d^3*cos(2 *c))*cos((b + 2*d)*x + a + 4*c) - 9*(9*b^4*cos(2*c) - 18*b^3*d*cos(2*c) - 4*b^2*d^2*cos(2*c) + 8*b*d^3*cos(2*c))*cos((b + 2*d)*x + a) - 9*(9*b^4*cos (2*c) + 18*b^3*d*cos(2*c) - 4*b^2*d^2*cos(2*c) - 8*b*d^3*cos(2*c))*cos(-(b - 2*d)*x - a + 4*c) - 9*(9*b^4*cos(2*c) + 18*b^3*d*cos(2*c) - 4*b^2*d^2*c os(2*c) - 8*b*d^3*cos(2*c))*cos(-(b - 2*d)*x - a) + 2*(9*b^4*cos(2*c) - 40 *b^2*d^2*cos(2*c) + 16*d^4*cos(2*c))*cos(3*b*x + 3*a + 2*c) + 2*(9*b^4*cos (2*c) - 40*b^2*d^2*cos(2*c) + 16*d^4*cos(2*c))*cos(3*b*x + 3*a - 2*c) - 18 *(9*b^4*cos(2*c) - 40*b^2*d^2*cos(2*c) + 16*d^4*cos(2*c))*cos(b*x + a + 2* c) - 18*(9*b^4*cos(2*c) - 40*b^2*d^2*cos(2*c) + 16*d^4*cos(2*c))*cos(b*x + a - 2*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))*sin((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))*sin((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*...
Time = 0.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \cos ^2(c+d x) \sin ^3(a+b x) \, dx=\frac {\cos \left (3 \, b x + 2 \, d x + 3 \, a + 2 \, c\right )}{16 \, {\left (3 \, b + 2 \, d\right )}} + \frac {\cos \left (3 \, b x - 2 \, d x + 3 \, a - 2 \, c\right )}{16 \, {\left (3 \, b - 2 \, d\right )}} + \frac {\cos \left (3 \, b x + 3 \, a\right )}{24 \, b} - \frac {3 \, \cos \left (b x + 2 \, d x + a + 2 \, c\right )}{16 \, {\left (b + 2 \, d\right )}} - \frac {3 \, \cos \left (b x - 2 \, d x + a - 2 \, c\right )}{16 \, {\left (b - 2 \, d\right )}} - \frac {3 \, \cos \left (b x + a\right )}{8 \, b} \] Input:
integrate(cos(d*x+c)^2*sin(b*x+a)^3,x, algorithm="giac")
Output:
1/16*cos(3*b*x + 2*d*x + 3*a + 2*c)/(3*b + 2*d) + 1/16*cos(3*b*x - 2*d*x + 3*a - 2*c)/(3*b - 2*d) + 1/24*cos(3*b*x + 3*a)/b - 3/16*cos(b*x + 2*d*x + a + 2*c)/(b + 2*d) - 3/16*cos(b*x - 2*d*x + a - 2*c)/(b - 2*d) - 3/8*cos( b*x + a)/b
Time = 18.38 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.17 \[ \int \cos ^2(c+d x) \sin ^3(a+b x) \, dx=-\frac {81\,b^4\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )+81\,b^4\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )+162\,b^4\,\cos \left (a+b\,x\right )+288\,d^4\,\cos \left (a+b\,x\right )-9\,b^4\,\cos \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )-9\,b^4\,\cos \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )-18\,b^4\,\cos \left (3\,a+3\,b\,x\right )-32\,d^4\,\cos \left (3\,a+3\,b\,x\right )+24\,b\,d^3\,\cos \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )-24\,b\,d^3\,\cos \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )-6\,b^3\,d\,\cos \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )+6\,b^3\,d\,\cos \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )-36\,b^2\,d^2\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )-36\,b^2\,d^2\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )-720\,b^2\,d^2\,\cos \left (a+b\,x\right )+36\,b^2\,d^2\,\cos \left (3\,a-2\,c+3\,b\,x-2\,d\,x\right )+36\,b^2\,d^2\,\cos \left (3\,a+2\,c+3\,b\,x+2\,d\,x\right )+80\,b^2\,d^2\,\cos \left (3\,a+3\,b\,x\right )-72\,b\,d^3\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )+72\,b\,d^3\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )+162\,b^3\,d\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )-162\,b^3\,d\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )}{48\,\left (9\,b^5-40\,b^3\,d^2+16\,b\,d^4\right )} \] Input:
int(cos(c + d*x)^2*sin(a + b*x)^3,x)
Output:
-(81*b^4*cos(a - 2*c + b*x - 2*d*x) + 81*b^4*cos(a + 2*c + b*x + 2*d*x) + 162*b^4*cos(a + b*x) + 288*d^4*cos(a + b*x) - 9*b^4*cos(3*a - 2*c + 3*b*x - 2*d*x) - 9*b^4*cos(3*a + 2*c + 3*b*x + 2*d*x) - 18*b^4*cos(3*a + 3*b*x) - 32*d^4*cos(3*a + 3*b*x) + 24*b*d^3*cos(3*a - 2*c + 3*b*x - 2*d*x) - 24*b *d^3*cos(3*a + 2*c + 3*b*x + 2*d*x) - 6*b^3*d*cos(3*a - 2*c + 3*b*x - 2*d* x) + 6*b^3*d*cos(3*a + 2*c + 3*b*x + 2*d*x) - 36*b^2*d^2*cos(a - 2*c + b*x - 2*d*x) - 36*b^2*d^2*cos(a + 2*c + b*x + 2*d*x) - 720*b^2*d^2*cos(a + b* x) + 36*b^2*d^2*cos(3*a - 2*c + 3*b*x - 2*d*x) + 36*b^2*d^2*cos(3*a + 2*c + 3*b*x + 2*d*x) + 80*b^2*d^2*cos(3*a + 3*b*x) - 72*b*d^3*cos(a - 2*c + b* x - 2*d*x) + 72*b*d^3*cos(a + 2*c + b*x + 2*d*x) + 162*b^3*d*cos(a - 2*c + b*x - 2*d*x) - 162*b^3*d*cos(a + 2*c + b*x + 2*d*x))/(48*(16*b*d^4 + 9*b^ 5 - 40*b^3*d^2))
Time = 0.17 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.12 \[ \int \cos ^2(c+d x) \sin ^3(a+b x) \, dx=\frac {9 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} \sin \left (d x +c \right )^{2} b^{4}-36 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} \sin \left (d x +c \right )^{2} b^{2} d^{2}-9 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b^{4}+38 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} b^{2} d^{2}-8 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} d^{4}+18 \cos \left (b x +a \right ) \sin \left (d x +c \right )^{2} b^{4}-18 \cos \left (b x +a \right ) b^{4}+40 \cos \left (b x +a \right ) b^{2} d^{2}-16 \cos \left (b x +a \right ) d^{4}-6 \cos \left (d x +c \right ) \sin \left (b x +a \right )^{3} \sin \left (d x +c \right ) b^{3} d +24 \cos \left (d x +c \right ) \sin \left (b x +a \right )^{3} \sin \left (d x +c \right ) b \,d^{3}-36 \cos \left (d x +c \right ) \sin \left (b x +a \right ) \sin \left (d x +c \right ) b^{3} d +18 b^{4}-64 b^{2} d^{2}+16 d^{4}}{3 b \left (9 b^{4}-40 b^{2} d^{2}+16 d^{4}\right )} \] Input:
int(cos(d*x+c)^2*sin(b*x+a)^3,x)
Output:
(9*cos(a + b*x)*sin(a + b*x)**2*sin(c + d*x)**2*b**4 - 36*cos(a + b*x)*sin (a + b*x)**2*sin(c + d*x)**2*b**2*d**2 - 9*cos(a + b*x)*sin(a + b*x)**2*b* *4 + 38*cos(a + b*x)*sin(a + b*x)**2*b**2*d**2 - 8*cos(a + b*x)*sin(a + b* x)**2*d**4 + 18*cos(a + b*x)*sin(c + d*x)**2*b**4 - 18*cos(a + b*x)*b**4 + 40*cos(a + b*x)*b**2*d**2 - 16*cos(a + b*x)*d**4 - 6*cos(c + d*x)*sin(a + b*x)**3*sin(c + d*x)*b**3*d + 24*cos(c + d*x)*sin(a + b*x)**3*sin(c + d*x )*b*d**3 - 36*cos(c + d*x)*sin(a + b*x)*sin(c + d*x)*b**3*d + 18*b**4 - 64 *b**2*d**2 + 16*d**4)/(3*b*(9*b**4 - 40*b**2*d**2 + 16*d**4))