\(\int e^{a+i b x} \cos ^3(d+b x) \sin ^2(d+b x) \, dx\) [16]

Optimal result

Integrand size = 27, antiderivative size = 150 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^2(d+b x) \, dx=-\frac {i e^{a-i d-2 i (d+b x)}}{64 b}-\frac {i e^{a-i d+2 i (d+b x)}}{32 b}-\frac {i e^{a-i d-4 i (d+b x)}}{128 b}+\frac {i e^{a-i d+4 i (d+b x)}}{128 b}+\frac {i e^{a-i d+6 i (d+b x)}}{192 b}+\frac {1}{16} e^{a-i d} x \] Output:

-1/64*I*exp(a-I*d-2*I*(b*x+d))/b-1/32*I*exp(a-I*d+2*I*(b*x+d))/b-1/128*I*e 
xp(a-I*d-4*I*(b*x+d))/b+1/128*I*exp(a-I*d+4*I*(b*x+d))/b+1/192*I*exp(a-I*d 
+6*I*(b*x+d))/b+1/16*exp(a-I*d)*x
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.96 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^2(d+b x) \, dx=\frac {e^a \left (12 \left (\left (-i e^{2 i b x}+2 b x\right ) \cos (d)+\left (e^{2 i b x}-2 i b x\right ) \sin (d)\right )+e^{-2 i b x} \left (3 i \left (-2+e^{6 i b x}\right ) \cos (3 d)-3 \left (2+e^{6 i b x}\right ) \sin (3 d)\right )+e^{-4 i b x} \left (i \left (-3+2 e^{10 i b x}\right ) \cos (5 d)-\left (3+2 e^{10 i b x}\right ) \sin (5 d)\right )\right )}{384 b} \] Input:

Integrate[E^(a + I*b*x)*Cos[d + b*x]^3*Sin[d + b*x]^2,x]
 

Output:

(E^a*(12*(((-I)*E^((2*I)*b*x) + 2*b*x)*Cos[d] + (E^((2*I)*b*x) - (2*I)*b*x 
)*Sin[d]) + ((3*I)*(-2 + E^((6*I)*b*x))*Cos[3*d] - 3*(2 + E^((6*I)*b*x))*S 
in[3*d])/E^((2*I)*b*x) + (I*(-3 + 2*E^((10*I)*b*x))*Cos[5*d] - (3 + 2*E^(( 
10*I)*b*x))*Sin[5*d])/E^((4*I)*b*x)))/(384*b)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, 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.

\(\displaystyle \int e^{a+i b x} \sin ^2(b x+d) \cos ^3(b x+d) \, dx\)

\(\Big \downarrow \) 4972

\(\displaystyle \int \left (\frac {1}{8} e^{a+i b x} \cos (b x+d)-\frac {1}{16} e^{a+i b x} \cos (3 b x+3 d)-\frac {1}{16} e^{a+i b x} \cos (5 b x+5 d)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i e^{a+2 i b x+i d}}{32 b}-\frac {3 e^{a+i b x} \sin (3 b x+3 d)}{128 b}-\frac {5 e^{a+i b x} \sin (5 b x+5 d)}{384 b}-\frac {i e^{a+i b x} \cos (3 b x+3 d)}{128 b}-\frac {i e^{a+i b x} \cos (5 b x+5 d)}{384 b}+\frac {1}{16} x e^{a-i d}\)

Input:

Int[E^(a + I*b*x)*Cos[d + b*x]^3*Sin[d + b*x]^2,x]
 

Output:

((-1/32*I)*E^(a + I*d + (2*I)*b*x))/b + (E^(a - I*d)*x)/16 - ((I/128)*E^(a 
 + I*b*x)*Cos[3*d + 3*b*x])/b - ((I/384)*E^(a + I*b*x)*Cos[5*d + 5*b*x])/b 
 - (3*E^(a + I*b*x)*Sin[3*d + 3*b*x])/(128*b) - (5*E^(a + I*b*x)*Sin[5*d + 
 5*b*x])/(384*b)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4972

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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (116 ) = 232\).

Time = 3.90 (sec) , antiderivative size = 786, normalized size of antiderivative = 5.24

method	result	size
orering	\(-\frac {\left (-6 b x +i\right ) {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2}}{6 b}+\frac {i \left (8 b x +15 i\right ) \left (i b \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2}-3 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{3} b +2 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right ) b \right )}{48 b^{2}}-\frac {5 \left (-6 b x +i\right ) \left (-18 b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2}-6 i b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{3}+4 i b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )+6 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{4} b^{2}+2 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{5} b^{2}\right )}{96 b^{3}}+\frac {i \left (10 b x +3 i\right ) \left (-52 i b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2}+84 \sin \left (b x +d \right )^{3} \cos \left (b x +d \right )^{2} {\mathrm e}^{i b x +a} b^{3}-50 \sin \left (b x +d \right ) \cos \left (b x +d \right )^{4} {\mathrm e}^{i b x +a} b^{3}+18 i b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{4}+6 i b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{5}-6 \sin \left (b x +d \right )^{5} {\mathrm e}^{i b x +a} b^{3}\right )}{192 b^{4}}-\frac {\left (-6 b x +i\right ) \left (504 \sin \left (b x +d \right )^{2} \cos \left (b x +d \right )^{3} {\mathrm e}^{i b x +a} b^{4}+312 i b^{4} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{3}-184 i b^{4} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )-216 \sin \left (b x +d \right )^{4} \cos \left (b x +d \right ) {\mathrm e}^{i b x +a} b^{4}-24 i \sin \left (b x +d \right )^{5} {\mathrm e}^{i b x +a} b^{4}-56 \cos \left (b x +d \right )^{5} {\mathrm e}^{i b x +a} b^{4}\right )}{384 b^{5}}+\frac {i x \left (1472 \sin \left (b x +d \right ) \cos \left (b x +d \right )^{4} {\mathrm e}^{i b x +a} b^{5}-2688 \sin \left (b x +d \right )^{3} \cos \left (b x +d \right )^{2} {\mathrm e}^{i b x +a} b^{5}-960 i \sin \left (b x +d \right )^{4} {\mathrm e}^{i b x +a} b^{5} \cos \left (b x +d \right )-240 i \cos \left (b x +d \right )^{5} b^{5} {\mathrm e}^{i b x +a}+2176 i b^{5} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{2}+240 \sin \left (b x +d \right )^{5} b^{5} {\mathrm e}^{i b x +a}\right )}{384 b^{5}}\)	\(786\)

Input:

int(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/6*(-6*b*x+I)/b*exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^2+1/48*I*(8*b*x+15* 
I)/b^2*(I*b*exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^2-3*exp(a+I*b*x)*cos(b*x+ 
d)^2*sin(b*x+d)^3*b+2*exp(a+I*b*x)*cos(b*x+d)^4*sin(b*x+d)*b)-5/96*(-6*b*x 
+I)/b^3*(-18*b^2*exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^2-6*I*b^2*exp(a+I*b* 
x)*cos(b*x+d)^2*sin(b*x+d)^3+4*I*b^2*exp(a+I*b*x)*cos(b*x+d)^4*sin(b*x+d)+ 
6*exp(a+I*b*x)*cos(b*x+d)*sin(b*x+d)^4*b^2+2*exp(a+I*b*x)*cos(b*x+d)^5*b^2 
)+1/192*I*(10*b*x+3*I)/b^4*(-52*I*b^3*exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d) 
^2+84*sin(b*x+d)^3*cos(b*x+d)^2*exp(a+I*b*x)*b^3-50*sin(b*x+d)*cos(b*x+d)^ 
4*exp(a+I*b*x)*b^3+18*I*b^3*exp(a+I*b*x)*cos(b*x+d)*sin(b*x+d)^4+6*I*b^3*e 
xp(a+I*b*x)*cos(b*x+d)^5-6*sin(b*x+d)^5*exp(a+I*b*x)*b^3)-1/384*(-6*b*x+I) 
/b^5*(504*sin(b*x+d)^2*cos(b*x+d)^3*exp(a+I*b*x)*b^4+312*I*b^4*exp(a+I*b*x 
)*cos(b*x+d)^2*sin(b*x+d)^3-184*I*b^4*exp(a+I*b*x)*cos(b*x+d)^4*sin(b*x+d) 
-216*sin(b*x+d)^4*cos(b*x+d)*exp(a+I*b*x)*b^4-24*I*sin(b*x+d)^5*exp(a+I*b* 
x)*b^4-56*cos(b*x+d)^5*exp(a+I*b*x)*b^4)+1/384*I/b^5*x*(1472*sin(b*x+d)*co 
s(b*x+d)^4*exp(a+I*b*x)*b^5-2688*sin(b*x+d)^3*cos(b*x+d)^2*exp(a+I*b*x)*b^ 
5-960*I*sin(b*x+d)^4*exp(a+I*b*x)*b^5*cos(b*x+d)-240*I*cos(b*x+d)^5*b^5*ex 
p(a+I*b*x)+2176*I*b^5*exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^2+240*sin(b*x+d 
)^5*b^5*exp(a+I*b*x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.57 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^2(d+b x) \, dx=\frac {{\left (24 \, b x e^{\left (4 i \, b x + a + 3 i \, d\right )} + 2 i \, e^{\left (10 i \, b x + a + 9 i \, d\right )} + 3 i \, e^{\left (8 i \, b x + a + 7 i \, d\right )} - 12 i \, e^{\left (6 i \, b x + a + 5 i \, d\right )} - 6 i \, e^{\left (2 i \, b x + a + i \, d\right )} - 3 i \, e^{\left (a - i \, d\right )}\right )} e^{\left (-4 i \, b x - 4 i \, d\right )}}{384 \, b} \] Input:

integrate(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^2,x, algorithm="fricas")
 

Output:

1/384*(24*b*x*e^(4*I*b*x + a + 3*I*d) + 2*I*e^(10*I*b*x + a + 9*I*d) + 3*I 
*e^(8*I*b*x + a + 7*I*d) - 12*I*e^(6*I*b*x + a + 5*I*d) - 6*I*e^(2*I*b*x + 
 a + I*d) - 3*I*e^(a - I*d))*e^(-4*I*b*x - 4*I*d)/b
 

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.57 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^2(d+b x) \, dx=\frac {x e^{a} e^{- i d}}{16} + \begin {cases} \frac {\left (33554432 i b^{4} e^{a} e^{13 i d} e^{6 i b x} + 50331648 i b^{4} e^{a} e^{11 i d} e^{4 i b x} - 201326592 i b^{4} e^{a} e^{9 i d} e^{2 i b x} - 100663296 i b^{4} e^{a} e^{5 i d} e^{- 2 i b x} - 50331648 i b^{4} e^{a} e^{3 i d} e^{- 4 i b x}\right ) e^{- 8 i d}}{6442450944 b^{5}} & \text {for}\: b^{5} e^{8 i d} \neq 0 \\x \left (\frac {\left (- e^{a} e^{10 i d} - e^{a} e^{8 i d} + 2 e^{a} e^{6 i d} + 2 e^{a} e^{4 i d} - e^{a} e^{2 i d} - e^{a}\right ) e^{- 5 i d}}{32} - \frac {e^{a} e^{- i d}}{16}\right ) & \text {otherwise} \end {cases} \] Input:

integrate(exp(a+I*b*x)*cos(b*x+d)**3*sin(b*x+d)**2,x)
                                                                                    
                                                                                    
 

Output:

x*exp(a)*exp(-I*d)/16 + Piecewise(((33554432*I*b**4*exp(a)*exp(13*I*d)*exp 
(6*I*b*x) + 50331648*I*b**4*exp(a)*exp(11*I*d)*exp(4*I*b*x) - 201326592*I* 
b**4*exp(a)*exp(9*I*d)*exp(2*I*b*x) - 100663296*I*b**4*exp(a)*exp(5*I*d)*e 
xp(-2*I*b*x) - 50331648*I*b**4*exp(a)*exp(3*I*d)*exp(-4*I*b*x))*exp(-8*I*d 
)/(6442450944*b**5), Ne(b**5*exp(8*I*d), 0)), (x*((-exp(a)*exp(10*I*d) - e 
xp(a)*exp(8*I*d) + 2*exp(a)*exp(6*I*d) + 2*exp(a)*exp(4*I*d) - exp(a)*exp( 
2*I*d) - exp(a))*exp(-5*I*d)/32 - exp(a)*exp(-I*d)/16), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^2(d+b x) \, dx=\frac {24 \, {\left (b \cos \left (d\right ) e^{a} - i \, b e^{a} \sin \left (d\right )\right )} x + 2 i \, \cos \left (6 \, b x + 5 \, d\right ) e^{a} - 3 i \, \cos \left (4 \, b x + 5 \, d\right ) e^{a} + 3 i \, \cos \left (4 \, b x + 3 \, d\right ) e^{a} - 6 i \, \cos \left (2 \, b x + 3 \, d\right ) e^{a} - 12 i \, \cos \left (2 \, b x + d\right ) e^{a} - 2 \, e^{a} \sin \left (6 \, b x + 5 \, d\right ) - 3 \, e^{a} \sin \left (4 \, b x + 5 \, d\right ) - 3 \, e^{a} \sin \left (4 \, b x + 3 \, d\right ) - 6 \, e^{a} \sin \left (2 \, b x + 3 \, d\right ) + 12 \, e^{a} \sin \left (2 \, b x + d\right )}{384 \, b} \] Input:

integrate(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^2,x, algorithm="maxima")
 

Output:

1/384*(24*(b*cos(d)*e^a - I*b*e^a*sin(d))*x + 2*I*cos(6*b*x + 5*d)*e^a - 3 
*I*cos(4*b*x + 5*d)*e^a + 3*I*cos(4*b*x + 3*d)*e^a - 6*I*cos(2*b*x + 3*d)* 
e^a - 12*I*cos(2*b*x + d)*e^a - 2*e^a*sin(6*b*x + 5*d) - 3*e^a*sin(4*b*x + 
 5*d) - 3*e^a*sin(4*b*x + 3*d) - 6*e^a*sin(2*b*x + 3*d) + 12*e^a*sin(2*b*x 
 + d))/b
 

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (85) = 170\).

Time = 0.21 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.39 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^2(d+b x) \, dx=\frac {48 \, {\left (b x + d\right )} \cos \left (d\right ) e^{a} - 48 i \, {\left (b x + d\right )} e^{a} \sin \left (d\right ) + 2 i \, {\left (e^{\left (6 i \, b x + 5 i \, d\right )} + e^{\left (-6 i \, b x - 5 i \, d\right )}\right )} e^{a} - 3 i \, {\left (e^{\left (4 i \, b x + 5 i \, d\right )} + e^{\left (-4 i \, b x - 5 i \, d\right )}\right )} e^{a} + 3 i \, {\left (e^{\left (4 i \, b x + 3 i \, d\right )} + e^{\left (-4 i \, b x - 3 i \, d\right )}\right )} e^{a} - 6 i \, {\left (e^{\left (2 i \, b x + 3 i \, d\right )} + e^{\left (-2 i \, b x - 3 i \, d\right )}\right )} e^{a} - 12 i \, {\left (e^{\left (2 i \, b x + i \, d\right )} + e^{\left (-2 i \, b x - i \, d\right )}\right )} e^{a} - 6 \, e^{a} \sin \left (4 \, b x + 5 \, d\right ) - 12 \, e^{a} \sin \left (2 \, b x + 3 \, d\right ) - 24 \, e^{a} \sin \left (-2 \, b x - d\right ) + 6 \, e^{a} \sin \left (-4 \, b x - 3 \, d\right ) + 4 \, e^{a} \sin \left (-6 \, b x - 5 \, d\right )}{768 \, b} \] Input:

integrate(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^2,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

1/768*(48*(b*x + d)*cos(d)*e^a - 48*I*(b*x + d)*e^a*sin(d) + 2*I*(e^(6*I*b 
*x + 5*I*d) + e^(-6*I*b*x - 5*I*d))*e^a - 3*I*(e^(4*I*b*x + 5*I*d) + e^(-4 
*I*b*x - 5*I*d))*e^a + 3*I*(e^(4*I*b*x + 3*I*d) + e^(-4*I*b*x - 3*I*d))*e^ 
a - 6*I*(e^(2*I*b*x + 3*I*d) + e^(-2*I*b*x - 3*I*d))*e^a - 12*I*(e^(2*I*b* 
x + I*d) + e^(-2*I*b*x - I*d))*e^a - 6*e^a*sin(4*b*x + 5*d) - 12*e^a*sin(2 
*b*x + 3*d) - 24*e^a*sin(-2*b*x - d) + 6*e^a*sin(-4*b*x - 3*d) + 4*e^a*sin 
(-6*b*x - 5*d))/b
 

Mupad [B] (verification not implemented)

Time = 15.72 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.20 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^2(d+b x) \, dx=\frac {x\,{\mathrm {e}}^a\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )}{16}-\frac {{\mathrm {e}}^a\,\left (\cos \left (2\,b\,x\right )-\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{64\,b}+\frac {{\mathrm {e}}^a\,\left (\cos \left (4\,b\,x\right )+\sin \left (4\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{128\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (4\,b\,x\right )-\sin \left (4\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,d\right )-\sin \left (5\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{128\,b}+\frac {{\mathrm {e}}^a\,\left (\cos \left (6\,b\,x\right )+\sin \left (6\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,d\right )+\sin \left (5\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{192\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (2\,b\,x\right )+\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,b} \] Input:

int(cos(d + b*x)^3*exp(a + b*x*1i)*sin(d + b*x)^2,x)
 

Output:

(x*exp(a)*(cos(d) - sin(d)*1i))/16 - (exp(a)*(cos(2*b*x) - sin(2*b*x)*1i)* 
(cos(3*d) - sin(3*d)*1i)*1i)/(64*b) + (exp(a)*(cos(4*b*x) + sin(4*b*x)*1i) 
*(cos(3*d) + sin(3*d)*1i)*1i)/(128*b) - (exp(a)*(cos(4*b*x) - sin(4*b*x)*1 
i)*(cos(5*d) - sin(5*d)*1i)*1i)/(128*b) + (exp(a)*(cos(6*b*x) + sin(6*b*x) 
*1i)*(cos(5*d) + sin(5*d)*1i)*1i)/(192*b) - (exp(a)*(cos(2*b*x) + sin(2*b* 
x)*1i)*(cos(d) + sin(d)*1i)*1i)/(32*b)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.65 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^2(d+b x) \, dx=\frac {e^{b i x +a} \left (-2 \cos \left (b x +d \right ) \sin \left (b x +d \right )^{4} i +3 \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} i +3 \cos \left (b x +d \right ) b x -10 \sin \left (b x +d \right )^{5}+17 \sin \left (b x +d \right )^{3}-3 \sin \left (b x +d \right ) b i x -3 \sin \left (b x +d \right )\right )}{48 b} \] Input:

int(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^2,x)
 

Output:

(e**(a + b*i*x)*( - 2*cos(b*x + d)*sin(b*x + d)**4*i + 3*cos(b*x + d)*sin( 
b*x + d)**2*i + 3*cos(b*x + d)*b*x - 10*sin(b*x + d)**5 + 17*sin(b*x + d)* 
*3 - 3*sin(b*x + d)*b*i*x - 3*sin(b*x + d)))/(48*b)