Integrand size = 29, antiderivative size = 169 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^3(d+b x) \, dx=-\frac {e^{2 (a-i d)-i (d+b x)}}{32 b}+\frac {e^{2 (a-i d)+i (d+b x)}}{16 b}+\frac {e^{2 (a-i d)-3 i (d+b x)}}{96 b}-\frac {e^{2 (a-i d)+3 i (d+b x)}}{48 b}-\frac {e^{2 (a-i d)+5 i (d+b x)}}{160 b}+\frac {e^{2 (a-i d)+7 i (d+b x)}}{224 b} \] Output:
-1/32*exp(2*a-2*I*d-I*(b*x+d))/b+1/16*exp(2*a-2*I*d+I*(b*x+d))/b+1/96*exp( 2*a-2*I*d-3*I*(b*x+d))/b-1/48*exp(2*a-2*I*d+3*I*(b*x+d))/b-1/160*exp(2*a-2 *I*d+5*I*(b*x+d))/b+1/224*exp(2*a-2*I*d+7*I*(b*x+d))/b
Time = 0.37 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.84 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^3(d+b x) \, dx=\frac {e^{2 a-3 i b x} \left (5 \left (7+3 e^{10 i b x}\right ) \cos (5 d)-70 e^{4 i b x} \left (\left (-3+e^{2 i b x}\right ) \cos (d)+i \left (3+e^{2 i b x}\right ) \sin (d)\right )-21 e^{2 i b x} \left (\left (5+e^{6 i b x}\right ) \cos (3 d)+i \left (-5+e^{6 i b x}\right ) \sin (3 d)\right )+5 i \left (-7+3 e^{10 i b x}\right ) \sin (5 d)\right )}{3360 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Cos[d + b*x]^2*Sin[d + b*x]^3,x]
Output:
(E^(2*a - (3*I)*b*x)*(5*(7 + 3*E^((10*I)*b*x))*Cos[5*d] - 70*E^((4*I)*b*x) *((-3 + E^((2*I)*b*x))*Cos[d] + I*(3 + E^((2*I)*b*x))*Sin[d]) - 21*E^((2*I )*b*x)*((5 + E^((6*I)*b*x))*Cos[3*d] + I*(-5 + E^((6*I)*b*x))*Sin[3*d]) + (5*I)*(-7 + 3*E^((10*I)*b*x))*Sin[5*d]))/(3360*b)
Time = 0.35 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {4972, 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.
| \(\displaystyle \int e^{2 (a+i b x)} \sin ^3(b x+d) \cos ^2(b x+d) \, dx\) |
| \(\Big \downarrow \) 4972 |
| \(\displaystyle \int \left (\frac {1}{8} e^{2 (a+i b x)} \sin (b x+d)+\frac {1}{16} e^{2 (a+i b x)} \sin (3 b x+3 d)-\frac {1}{16} e^{2 (a+i b x)} \sin (5 b x+5 d)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {i e^{2 (a+i b x)} \sin (b x+d)}{12 b}+\frac {i e^{2 (a+i b x)} \sin (3 b x+3 d)}{40 b}-\frac {i e^{2 (a+i b x)} \sin (5 b x+5 d)}{168 b}+\frac {e^{2 (a+i b x)} \cos (b x+d)}{24 b}-\frac {3 e^{2 (a+i b x)} \cos (3 b x+3 d)}{80 b}+\frac {5 e^{2 (a+i b x)} \cos (5 b x+5 d)}{336 b}\) |
Input:
Int[E^(2*(a + I*b*x))*Cos[d + b*x]^2*Sin[d + b*x]^3,x]
Output:
(E^(2*(a + I*b*x))*Cos[d + b*x])/(24*b) - (3*E^(2*(a + I*b*x))*Cos[3*d + 3 *b*x])/(80*b) + (5*E^(2*(a + I*b*x))*Cos[5*d + 5*b*x])/(336*b) - ((I/12)*E ^(2*(a + I*b*x))*Sin[d + b*x])/b + ((I/40)*E^(2*(a + I*b*x))*Sin[3*d + 3*b *x])/b - ((I/168)*E^(2*(a + I*b*x))*Sin[5*d + 5*b*x])/b
Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_ .) + (e_.)*(x_)]^(m_.), x_Symbol] :> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Time = 2.36 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.66
| method | result | size |
| parallelrisch | \(-\frac {4 \,{\mathrm e}^{2 i b x +2 a} \left (84 i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}+105 \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}+8 i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}-91 \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}+4 i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )+3 \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-1\right )}{105 b \left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{5}}\) | \(112\) |
| default | \(-\frac {3 \,{\mathrm e}^{2 i b x +2 a} \cos \left (3 b x +3 d \right )}{80 b}+\frac {i {\mathrm e}^{2 i b x +2 a} \sin \left (3 b x +3 d \right )}{40 b}+\frac {{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )}{24 b}-\frac {i {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )}{12 b}+\frac {5 \,{\mathrm e}^{2 i b x +2 a} \cos \left (5 b x +5 d \right )}{336 b}-\frac {i {\mathrm e}^{2 i b x +2 a} \sin \left (5 b x +5 d \right )}{168 b}\) | \(143\) |
| orering | \(-\frac {12 i {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{3}}{35 b}-\frac {341 \left (2 i b \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{3}-2 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{4} b +3 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2} b \right )}{315 b^{2}}-\frac {8 i \left (-21 b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{3}-8 i b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{4}+12 i b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2}+2 \,{\mathrm e}^{2 i b x +2 a} b^{2} \sin \left (b x +d \right )^{5}+6 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right ) b^{2}\right )}{21 b^{3}}-\frac {5 \left (-110 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{3}+68 b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{4}-111 b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2}+12 i b^{3} {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )^{5}+36 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )+6 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{5} b^{3}\right )}{63 b^{4}}-\frac {4 i \left (825 b^{4} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{3}+48 i b^{4} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{5}-696 i b^{4} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2}+416 i b^{4} {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )^{4} \cos \left (b x +d \right )-92 b^{4} {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )^{5}-324 b^{4} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )\right )}{105 b^{5}}+\frac {5402 i b^{5} {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )^{3} \cos \left (b x +d \right )^{2}-2942 b^{5} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{4}+5163 b^{5} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2}-420 b^{5} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{5}-600 i b^{5} {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )^{5}-2280 i b^{5} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )}{315 b^{6}}\) | \(804\) |
Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-4/105*exp(2*a+2*I*b*x)*(84*I*tan(1/2*b*x+1/2*d)^5+105*tan(1/2*b*x+1/2*d)^ 6+8*I*tan(1/2*b*x+1/2*d)^3-91*tan(1/2*b*x+1/2*d)^4+4*I*tan(1/2*b*x+1/2*d)+ 3*tan(1/2*b*x+1/2*d)^2-1)/b/(1+tan(1/2*b*x+1/2*d)^2)^5
Time = 0.07 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.54 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^3(d+b x) \, dx=\frac {{\left (15 \, e^{\left (10 i \, b x + 2 \, a + 8 i \, d\right )} - 21 \, e^{\left (8 i \, b x + 2 \, a + 6 i \, d\right )} - 70 \, e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} + 210 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - 105 \, e^{\left (2 i \, b x + 2 \, a\right )} + 35 \, e^{\left (2 \, a - 2 i \, d\right )}\right )} e^{\left (-3 i \, b x - 3 i \, d\right )}}{3360 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^3,x, algorithm="fricas" )
Output:
1/3360*(15*e^(10*I*b*x + 2*a + 8*I*d) - 21*e^(8*I*b*x + 2*a + 6*I*d) - 70* e^(6*I*b*x + 2*a + 4*I*d) + 210*e^(4*I*b*x + 2*a + 2*I*d) - 105*e^(2*I*b*x + 2*a) + 35*e^(2*a - 2*I*d))*e^(-3*I*b*x - 3*I*d)/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (128) = 256\).
Time = 0.35 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.52 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^3(d+b x) \, dx=\begin {cases} \frac {\left (377487360 b^{5} e^{2 a} e^{14 i d} e^{7 i b x} - 528482304 b^{5} e^{2 a} e^{12 i d} e^{5 i b x} - 1761607680 b^{5} e^{2 a} e^{10 i d} e^{3 i b x} + 5284823040 b^{5} e^{2 a} e^{8 i d} e^{i b x} - 2642411520 b^{5} e^{2 a} e^{6 i d} e^{- i b x} + 880803840 b^{5} e^{2 a} e^{4 i d} e^{- 3 i b x}\right ) e^{- 9 i d}}{84557168640 b^{6}} & \text {for}\: b^{6} e^{9 i d} \neq 0 \\\frac {x \left (i e^{2 a} e^{10 i d} - i e^{2 a} e^{8 i d} - 2 i e^{2 a} e^{6 i d} + 2 i e^{2 a} e^{4 i d} + i e^{2 a} e^{2 i d} - i e^{2 a}\right ) e^{- 5 i d}}{32} & \text {otherwise} \end {cases} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)**2*sin(b*x+d)**3,x)
Output:
Piecewise(((377487360*b**5*exp(2*a)*exp(14*I*d)*exp(7*I*b*x) - 528482304*b **5*exp(2*a)*exp(12*I*d)*exp(5*I*b*x) - 1761607680*b**5*exp(2*a)*exp(10*I* d)*exp(3*I*b*x) + 5284823040*b**5*exp(2*a)*exp(8*I*d)*exp(I*b*x) - 2642411 520*b**5*exp(2*a)*exp(6*I*d)*exp(-I*b*x) + 880803840*b**5*exp(2*a)*exp(4*I *d)*exp(-3*I*b*x))*exp(-9*I*d)/(84557168640*b**6), Ne(b**6*exp(9*I*d), 0)) , (x*(I*exp(2*a)*exp(10*I*d) - I*exp(2*a)*exp(8*I*d) - 2*I*exp(2*a)*exp(6* I*d) + 2*I*exp(2*a)*exp(4*I*d) + I*exp(2*a)*exp(2*I*d) - I*exp(2*a))*exp(- 5*I*d)/32, True))
Time = 0.04 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^3(d+b x) \, dx=\frac {15 \, \cos \left (7 \, b x + 5 \, d\right ) e^{\left (2 \, a\right )} - 21 \, \cos \left (5 \, b x + 3 \, d\right ) e^{\left (2 \, a\right )} + 35 \, \cos \left (3 \, b x + 5 \, d\right ) e^{\left (2 \, a\right )} - 70 \, \cos \left (3 \, b x + d\right ) e^{\left (2 \, a\right )} - 105 \, \cos \left (b x + 3 \, d\right ) e^{\left (2 \, a\right )} + 210 \, \cos \left (b x - d\right ) e^{\left (2 \, a\right )} + 15 i \, e^{\left (2 \, a\right )} \sin \left (7 \, b x + 5 \, d\right ) - 21 i \, e^{\left (2 \, a\right )} \sin \left (5 \, b x + 3 \, d\right ) - 35 i \, e^{\left (2 \, a\right )} \sin \left (3 \, b x + 5 \, d\right ) - 70 i \, e^{\left (2 \, a\right )} \sin \left (3 \, b x + d\right ) + 105 i \, e^{\left (2 \, a\right )} \sin \left (b x + 3 \, d\right ) + 210 i \, e^{\left (2 \, a\right )} \sin \left (b x - d\right )}{3360 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^3,x, algorithm="maxima" )
Output:
1/3360*(15*cos(7*b*x + 5*d)*e^(2*a) - 21*cos(5*b*x + 3*d)*e^(2*a) + 35*cos (3*b*x + 5*d)*e^(2*a) - 70*cos(3*b*x + d)*e^(2*a) - 105*cos(b*x + 3*d)*e^( 2*a) + 210*cos(b*x - d)*e^(2*a) + 15*I*e^(2*a)*sin(7*b*x + 5*d) - 21*I*e^( 2*a)*sin(5*b*x + 3*d) - 35*I*e^(2*a)*sin(3*b*x + 5*d) - 70*I*e^(2*a)*sin(3 *b*x + d) + 105*I*e^(2*a)*sin(b*x + 3*d) + 210*I*e^(2*a)*sin(b*x - d))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (103) = 206\).
Time = 0.18 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.51 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^3(d+b x) \, dx=\frac {15 \, {\left (e^{\left (7 i \, b x + 5 i \, d\right )} - e^{\left (-7 i \, b x - 5 i \, d\right )}\right )} e^{\left (2 \, a\right )} - 21 \, {\left (e^{\left (5 i \, b x + 3 i \, d\right )} - e^{\left (-5 i \, b x - 3 i \, d\right )}\right )} e^{\left (2 \, a\right )} - 35 \, {\left (e^{\left (3 i \, b x + 5 i \, d\right )} - e^{\left (-3 i \, b x - 5 i \, d\right )}\right )} e^{\left (2 \, a\right )} - 70 \, {\left (e^{\left (3 i \, b x + i \, d\right )} - e^{\left (-3 i \, b x - i \, d\right )}\right )} e^{\left (2 \, a\right )} + 105 \, {\left (e^{\left (i \, b x + 3 i \, d\right )} - e^{\left (-i \, b x - 3 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 210 \, {\left (e^{\left (i \, b x - i \, d\right )} - e^{\left (-i \, b x + i \, d\right )}\right )} e^{\left (2 \, a\right )} + 70 \, \cos \left (3 \, b x + 5 \, d\right ) e^{\left (2 \, a\right )} - 210 \, \cos \left (b x + 3 \, d\right ) e^{\left (2 \, a\right )} + 420 \, \cos \left (-b x + d\right ) e^{\left (2 \, a\right )} - 140 \, \cos \left (-3 \, b x - d\right ) e^{\left (2 \, a\right )} - 42 \, \cos \left (-5 \, b x - 3 \, d\right ) e^{\left (2 \, a\right )} + 30 \, \cos \left (-7 \, b x - 5 \, d\right ) e^{\left (2 \, a\right )}}{6720 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^3,x, algorithm="giac")
Output:
1/6720*(15*(e^(7*I*b*x + 5*I*d) - e^(-7*I*b*x - 5*I*d))*e^(2*a) - 21*(e^(5 *I*b*x + 3*I*d) - e^(-5*I*b*x - 3*I*d))*e^(2*a) - 35*(e^(3*I*b*x + 5*I*d) - e^(-3*I*b*x - 5*I*d))*e^(2*a) - 70*(e^(3*I*b*x + I*d) - e^(-3*I*b*x - I* d))*e^(2*a) + 105*(e^(I*b*x + 3*I*d) - e^(-I*b*x - 3*I*d))*e^(2*a) + 210*( e^(I*b*x - I*d) - e^(-I*b*x + I*d))*e^(2*a) + 70*cos(3*b*x + 5*d)*e^(2*a) - 210*cos(b*x + 3*d)*e^(2*a) + 420*cos(-b*x + d)*e^(2*a) - 140*cos(-3*b*x - d)*e^(2*a) - 42*cos(-5*b*x - 3*d)*e^(2*a) + 30*cos(-7*b*x - 5*d)*e^(2*a) )/b
Time = 0.89 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.18 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^3(d+b x) \, dx=-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (b\,x\right )-\sin \left (b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )}{32\,b}+\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (3\,b\,x\right )-\sin \left (3\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,d\right )-\sin \left (5\,d\right )\,1{}\mathrm {i}\right )}{96\,b}-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (5\,b\,x\right )+\sin \left (5\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )}{160\,b}+\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (7\,b\,x\right )+\sin \left (7\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,d\right )+\sin \left (5\,d\right )\,1{}\mathrm {i}\right )}{224\,b}+\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (b\,x\right )+\sin \left (b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )}{16\,b}-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (3\,b\,x\right )+\sin \left (3\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )}{48\,b} \] Input:
int(cos(d + b*x)^2*exp(2*a + b*x*2i)*sin(d + b*x)^3,x)
Output:
(exp(2*a)*(cos(3*b*x) - sin(3*b*x)*1i)*(cos(5*d) - sin(5*d)*1i))/(96*b) - (exp(2*a)*(cos(b*x) - sin(b*x)*1i)*(cos(3*d) - sin(3*d)*1i))/(32*b) - (exp (2*a)*(cos(5*b*x) + sin(5*b*x)*1i)*(cos(3*d) + sin(3*d)*1i))/(160*b) + (ex p(2*a)*(cos(7*b*x) + sin(7*b*x)*1i)*(cos(5*d) + sin(5*d)*1i))/(224*b) + (e xp(2*a)*(cos(b*x) + sin(b*x)*1i)*(cos(d) - sin(d)*1i))/(16*b) - (exp(2*a)* (cos(3*b*x) + sin(3*b*x)*1i)*(cos(d) + sin(d)*1i))/(48*b)
Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.67 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^3(d+b x) \, dx=\frac {e^{2 b i x +2 a} \left (25 \cos \left (b x +d \right ) \sin \left (b x +d \right )^{4}-3 \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}-4 \cos \left (b x +d \right ) \sin \left (b x +d \right ) i +2 \cos \left (b x +d \right )-10 \sin \left (b x +d \right )^{5} i +2 \sin \left (b x +d \right )^{3} i -4 \sin \left (b x +d \right )^{2}-4 \sin \left (b x +d \right ) i +2\right )}{105 b} \] Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^3,x)
Output:
(e**(2*a + 2*b*i*x)*(25*cos(b*x + d)*sin(b*x + d)**4 - 3*cos(b*x + d)*sin( b*x + d)**2 - 4*cos(b*x + d)*sin(b*x + d)*i + 2*cos(b*x + d) - 10*sin(b*x + d)**5*i + 2*sin(b*x + d)**3*i - 4*sin(b*x + d)**2 - 4*sin(b*x + d)*i + 2 ))/(105*b)