\(\int e^{2 (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx\) [10]

Optimal result

Integrand size = 32, antiderivative size = 102 \[ \int e^{2 (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\frac {i e^{2 (a-i d)+4 i (d+b x)} (f+i g)^2}{16 b}-\frac {i e^{2 (a-i d)+2 i (d+b x)} \left (f^2+g^2\right )}{4 b}-\frac {1}{4} e^{2 a-2 i d} (f-i g)^2 x \] Output:

1/16*I*exp(2*a-2*I*d+4*I*(b*x+d))*(f+I*g)^2/b-1/4*I*exp(2*a-2*I*d+2*I*(b*x 
+d))*(f^2+g^2)/b-1/4*exp(2*a-2*I*d)*(f-I*g)^2*x
 

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88 \[ \int e^{2 (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\frac {e^{2 a-2 i d} \left (-4 d (f-i g)^2+i e^{4 i (d+b x)} (f+i g)^2-4 i e^{2 i (d+b x)} \left (f^2+g^2\right )-4 b (f-i g)^2 x\right )}{16 b} \] Input:

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

Output:

(E^(2*a - (2*I)*d)*(-4*d*(f - I*g)^2 + I*E^((4*I)*(d + b*x))*(f + I*g)^2 - 
 (4*I)*E^((2*I)*(d + b*x))*(f^2 + g^2) - 4*b*(f - I*g)^2*x))/(16*b)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.90, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {7292, 7293, 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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (f^2 e^{2 a+2 i b x} \sin ^2(b x+d)+f g e^{2 a+2 i b x} \sin (2 b x+2 d)+g^2 e^{2 a+2 i b x} \cos ^2(b x+d)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i f^2 e^{2 (a+i d)+4 i b x}}{16 b}-\frac {f g e^{2 (a+i d)+4 i b x}}{8 b}-\frac {i g^2 e^{2 (a+i d)+4 i b x}}{16 b}-\frac {i f^2 e^{2 a+2 i b x}}{4 b}-\frac {i g^2 e^{2 a+2 i b x}}{4 b}-\frac {1}{4} f^2 x e^{2 a-2 i d}+\frac {1}{2} i f g x e^{2 a-2 i d}+\frac {1}{4} g^2 x e^{2 a-2 i d}\)

Input:

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

Output:

((-1/4*I)*E^(2*a + (2*I)*b*x)*f^2)/b + ((I/16)*E^(2*(a + I*d) + (4*I)*b*x) 
*f^2)/b - (E^(2*(a + I*d) + (4*I)*b*x)*f*g)/(8*b) - ((I/4)*E^(2*a + (2*I)* 
b*x)*g^2)/b - ((I/16)*E^(2*(a + I*d) + (4*I)*b*x)*g^2)/b - (E^(2*a - (2*I) 
*d)*f^2*x)/4 + (I/2)*E^(2*a - (2*I)*d)*f*g*x + (E^(2*a - (2*I)*d)*g^2*x)/4
 

Defintions of rubi rules used

rule 2009

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

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(\frac {\left (\left (\left (i b x +\frac {1}{2}\right ) f^{2}-g \left (-2 b x +i\right ) f -g^{2} \left (i b x -\frac {3}{2}\right )\right ) \sin \left (2 b x +2 d \right )+\left (\left (-b x +i\right ) f^{2}+2 i x b f g +g^{2} \left (b x +i\right )\right ) \cos \left (2 b x +2 d \right )-i f^{2}-i g^{2}\right ) {\mathrm e}^{2 i b x +2 a}}{4 b}\)	\(113\)
norman	\(\frac {\left (\frac {1}{4} g^{2}+\frac {1}{2} i f g -\frac {1}{4} f^{2}\right ) x \,{\mathrm e}^{2 i b x +2 a}+\left (-\frac {3}{2} g^{2}-3 i f g +\frac {3}{2} f^{2}\right ) x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}+\left (\frac {1}{4} g^{2}+\frac {1}{2} i f g -\frac {1}{4} f^{2}\right ) x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}+i \left (2 i f g -f^{2}+g^{2}\right ) x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}+i \left (-2 i f g +f^{2}-g^{2}\right ) x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )-\frac {2 \left (i f^{2}+i g^{2}\right ) {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{b}+\frac {\left (-2 i f g +f^{2}+3 g^{2}\right ) {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{2 b}-\frac {\left (-2 i f g +f^{2}+3 g^{2}\right ) {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{2 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\)	\(317\)
orering	\(-\frac {\left (-4 b x +3 i\right ) {\mathrm e}^{2 i b x +2 a} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}}{4 b}-\frac {i \left (-6 b x +i\right ) \left (2 i b \,{\mathrm e}^{2 i b x +2 a} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}+2 \,{\mathrm e}^{2 i b x +2 a} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right ) \left (-g b \sin \left (b x +d \right )+f b \cos \left (b x +d \right )\right )\right )}{8 b^{2}}-\frac {x \left (-4 b^{2} {\mathrm e}^{2 i b x +2 a} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}+8 i b \,{\mathrm e}^{2 i b x +2 a} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right ) \left (-g b \sin \left (b x +d \right )+f b \cos \left (b x +d \right )\right )+2 \,{\mathrm e}^{2 i b x +2 a} \left (-g b \sin \left (b x +d \right )+f b \cos \left (b x +d \right )\right )^{2}+2 \,{\mathrm e}^{2 i b x +2 a} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right ) \left (-g \,b^{2} \cos \left (b x +d \right )-f \,b^{2} \sin \left (b x +d \right )\right )\right )}{8 b^{2}}\)	\(319\)
parts	\(\frac {i g^{2} x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}+\frac {g^{2} x \,{\mathrm e}^{2 i b x +2 a}}{4}-\frac {3 g^{2} x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{2}+\frac {g^{2} x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{4}-\frac {2 i g^{2} {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{b}+\frac {3 g^{2} {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{2 b}-\frac {3 g^{2} {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{2 b}-i g^{2} x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}+\frac {i f^{2} x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )-\frac {f^{2} x \,{\mathrm e}^{2 i b x +2 a}}{4}+\frac {3 f^{2} x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{2}-\frac {f^{2} x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{4}-\frac {2 i f^{2} {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{b}+\frac {f^{2} {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{2 b}-\frac {f^{2} {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{2 b}-i f^{2} x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}+\frac {\frac {i f g \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{b}+2 f g x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )-2 f g x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}+\frac {i f g x \,{\mathrm e}^{2 i b x +2 a}}{2}-\frac {i f g \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{b}-3 i f g x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}+\frac {i f g x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{2}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\)	\(649\)

Input:

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

Output:

1/4*(((I*b*x+1/2)*f^2-g*(-2*b*x+I)*f-g^2*(I*b*x-3/2))*sin(2*b*x+2*d)+((-b* 
x+I)*f^2+2*I*x*b*f*g+g^2*(b*x+I))*cos(2*b*x+2*d)-I*f^2-I*g^2)*exp(2*a+2*I* 
b*x)/b
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int e^{2 (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=-\frac {4 \, {\left (b f^{2} - 2 i \, b f g - b g^{2}\right )} x e^{\left (2 \, a - 2 i \, d\right )} - {\left (i \, f^{2} - 2 \, f g - i \, g^{2}\right )} e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + 4 \, {\left (i \, f^{2} + i \, g^{2}\right )} e^{\left (2 i \, b x + 2 \, a\right )}}{16 \, b} \] Input:

integrate(exp(2*a+2*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm="fri 
cas")
 

Output:

-1/16*(4*(b*f^2 - 2*I*b*f*g - b*g^2)*x*e^(2*a - 2*I*d) - (I*f^2 - 2*f*g - 
I*g^2)*e^(4*I*b*x + 2*a + 2*I*d) + 4*(I*f^2 + I*g^2)*e^(2*I*b*x + 2*a))/b
 

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.03 \[ \int e^{2 (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\frac {x \left (- f^{2} e^{2 a} + 2 i f g e^{2 a} + g^{2} e^{2 a}\right ) e^{- 2 i d}}{4} + \begin {cases} \frac {\left (- 16 i b f^{2} e^{2 a} - 16 i b g^{2} e^{2 a}\right ) e^{2 i b x} + \left (4 i b f^{2} e^{2 a} e^{2 i d} - 8 b f g e^{2 a} e^{2 i d} - 4 i b g^{2} e^{2 a} e^{2 i d}\right ) e^{4 i b x}}{64 b^{2}} & \text {for}\: b^{2} \neq 0 \\x \left (- \frac {\left (- f^{2} e^{2 a} + 2 i f g e^{2 a} + g^{2} e^{2 a}\right ) e^{- 2 i d}}{4} + \frac {\left (- f^{2} e^{2 a} e^{4 i d} + 2 f^{2} e^{2 a} e^{2 i d} - f^{2} e^{2 a} - 2 i f g e^{2 a} e^{4 i d} + 2 i f g e^{2 a} + g^{2} e^{2 a} e^{4 i d} + 2 g^{2} e^{2 a} e^{2 i d} + g^{2} e^{2 a}\right ) e^{- 2 i d}}{4}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

x*(-f**2*exp(2*a) + 2*I*f*g*exp(2*a) + g**2*exp(2*a))*exp(-2*I*d)/4 + Piec 
ewise((((-16*I*b*f**2*exp(2*a) - 16*I*b*g**2*exp(2*a))*exp(2*I*b*x) + (4*I 
*b*f**2*exp(2*a)*exp(2*I*d) - 8*b*f*g*exp(2*a)*exp(2*I*d) - 4*I*b*g**2*exp 
(2*a)*exp(2*I*d))*exp(4*I*b*x))/(64*b**2), Ne(b**2, 0)), (x*(-(-f**2*exp(2 
*a) + 2*I*f*g*exp(2*a) + g**2*exp(2*a))*exp(-2*I*d)/4 + (-f**2*exp(2*a)*ex 
p(4*I*d) + 2*f**2*exp(2*a)*exp(2*I*d) - f**2*exp(2*a) - 2*I*f*g*exp(2*a)*e 
xp(4*I*d) + 2*I*f*g*exp(2*a) + g**2*exp(2*a)*exp(4*I*d) + 2*g**2*exp(2*a)* 
exp(2*I*d) + g**2*exp(2*a))*exp(-2*I*d)/4), True))
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (64) = 128\).

Time = 0.05 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.99 \[ \int e^{2 (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=-\frac {{\left (-4 i \, b x e^{\left (2 \, a\right )} + e^{\left (4 i \, b x + 2 \, a + 4 i \, d\right )}\right )} f g e^{\left (-2 i \, d\right )}}{8 \, b} - \frac {{\left (4 \, {\left (b \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - i \, b e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} x + 4 i \, \cos \left (2 \, b x\right ) e^{\left (2 \, a\right )} - i \, \cos \left (4 \, b x + 2 \, d\right ) e^{\left (2 \, a\right )} - 4 \, e^{\left (2 \, a\right )} \sin \left (2 \, b x\right ) + e^{\left (2 \, a\right )} \sin \left (4 \, b x + 2 \, d\right )\right )} f^{2}}{16 \, b} + \frac {{\left (4 \, {\left (b \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - i \, b e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} x - 4 i \, \cos \left (2 \, b x\right ) e^{\left (2 \, a\right )} - i \, \cos \left (4 \, b x + 2 \, d\right ) e^{\left (2 \, a\right )} + 4 \, e^{\left (2 \, a\right )} \sin \left (2 \, b x\right ) + e^{\left (2 \, a\right )} \sin \left (4 \, b x + 2 \, d\right )\right )} g^{2}}{16 \, b} \] Input:

integrate(exp(2*a+2*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm="max 
ima")
 

Output:

-1/8*(-4*I*b*x*e^(2*a) + e^(4*I*b*x + 2*a + 4*I*d))*f*g*e^(-2*I*d)/b - 1/1 
6*(4*(b*cos(2*d)*e^(2*a) - I*b*e^(2*a)*sin(2*d))*x + 4*I*cos(2*b*x)*e^(2*a 
) - I*cos(4*b*x + 2*d)*e^(2*a) - 4*e^(2*a)*sin(2*b*x) + e^(2*a)*sin(4*b*x 
+ 2*d))*f^2/b + 1/16*(4*(b*cos(2*d)*e^(2*a) - I*b*e^(2*a)*sin(2*d))*x - 4* 
I*cos(2*b*x)*e^(2*a) - I*cos(4*b*x + 2*d)*e^(2*a) + 4*e^(2*a)*sin(2*b*x) + 
 e^(2*a)*sin(4*b*x + 2*d))*g^2/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. 224 vs. \(2 (64) = 128\).

Time = 0.18 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.20 \[ \int e^{2 (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=-\frac {8 \, {\left (f^{2} - 2 i \, f g - g^{2}\right )} {\left (b x + d\right )} \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - 8 \, {\left (i \, f^{2} + 2 \, f g - i \, g^{2}\right )} {\left (b x + d\right )} e^{\left (2 \, a\right )} \sin \left (2 \, d\right ) - 8 \, {\left (f^{2} + g^{2}\right )} e^{\left (2 \, a\right )} \sin \left (2 \, b x\right ) - i \, {\left ({\left (f^{2} + 2 i \, f g - g^{2}\right )} e^{\left (4 i \, b x + 2 i \, d\right )} + {\left (f^{2} + 2 i \, f g - g^{2}\right )} e^{\left (-4 i \, b x - 2 i \, d\right )}\right )} e^{\left (2 \, a\right )} + {\left ({\left (-i \, f^{2} + 2 \, f g + i \, g^{2}\right )} e^{\left (4 i \, b x + 2 i \, d\right )} + {\left (i \, f^{2} - 2 \, f g - i \, g^{2}\right )} e^{\left (-4 i \, b x - 2 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 4 i \, {\left ({\left (f^{2} + g^{2}\right )} e^{\left (2 i \, b x\right )} + {\left (f^{2} + g^{2}\right )} e^{\left (-2 i \, b x\right )}\right )} e^{\left (2 \, a\right )}}{32 \, b} \] Input:

integrate(exp(2*a+2*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm="gia 
c")
 

Output:

-1/32*(8*(f^2 - 2*I*f*g - g^2)*(b*x + d)*cos(2*d)*e^(2*a) - 8*(I*f^2 + 2*f 
*g - I*g^2)*(b*x + d)*e^(2*a)*sin(2*d) - 8*(f^2 + g^2)*e^(2*a)*sin(2*b*x) 
- I*((f^2 + 2*I*f*g - g^2)*e^(4*I*b*x + 2*I*d) + (f^2 + 2*I*f*g - g^2)*e^( 
-4*I*b*x - 2*I*d))*e^(2*a) + ((-I*f^2 + 2*f*g + I*g^2)*e^(4*I*b*x + 2*I*d) 
 + (I*f^2 - 2*f*g - I*g^2)*e^(-4*I*b*x - 2*I*d))*e^(2*a) + 4*I*((f^2 + g^2 
)*e^(2*I*b*x) + (f^2 + g^2)*e^(-2*I*b*x))*e^(2*a))/b
 

Mupad [B] (verification not implemented)

Time = 15.68 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.64 \[ \int e^{2 (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=-\frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\,\left (\frac {f^2\,1{}\mathrm {i}}{2}+\frac {g^2\,1{}\mathrm {i}}{2}\right )}{2\,b}+\frac {x\,{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\,\cos \left (2\,d+2\,b\,x\right )\,{\left (g+f\,1{}\mathrm {i}\right )}^2}{4}+\frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\,\cos \left (2\,d+2\,b\,x\right )\,\left (\frac {f^2\,1{}\mathrm {i}}{2}+\frac {g^2\,1{}\mathrm {i}}{2}\right )}{2\,b}+\frac {x\,{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\,\sin \left (2\,d+2\,b\,x\right )\,{\left (f-g\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{4}+\frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\,\sin \left (2\,d+2\,b\,x\right )\,\left (\frac {f^2}{4}-\frac {f\,g\,1{}\mathrm {i}}{2}+\frac {3\,g^2}{4}\right )}{2\,b} \] Input:

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

Output:

(x*exp(2*a + b*x*2i)*cos(2*d + 2*b*x)*(f*1i + g)^2)/4 - (exp(2*a + b*x*2i) 
*((f^2*1i)/2 + (g^2*1i)/2))/(2*b) + (exp(2*a + b*x*2i)*cos(2*d + 2*b*x)*(( 
f^2*1i)/2 + (g^2*1i)/2))/(2*b) + (x*exp(2*a + b*x*2i)*sin(2*d + 2*b*x)*(f 
- g*1i)^2*1i)/4 + (exp(2*a + b*x*2i)*sin(2*d + 2*b*x)*(f^2/4 - (f*g*1i)/2 
+ (3*g^2)/4))/(2*b)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.40 \[ \int e^{2 (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\frac {e^{2 b i x +2 a} \left (-2 \cos \left (b x +d \right )^{2} b \,f^{2} x +4 \cos \left (b x +d \right )^{2} b f g i x +2 \cos \left (b x +d \right )^{2} b \,g^{2} x -\cos \left (b x +d \right )^{2} f^{2} i -2 \cos \left (b x +d \right )^{2} f g -3 \cos \left (b x +d \right )^{2} g^{2} i +4 \cos \left (b x +d \right ) \sin \left (b x +d \right ) b \,f^{2} i x +8 \cos \left (b x +d \right ) \sin \left (b x +d \right ) b f g x -4 \cos \left (b x +d \right ) \sin \left (b x +d \right ) b \,g^{2} i x +2 \sin \left (b x +d \right )^{2} b \,f^{2} x -4 \sin \left (b x +d \right )^{2} b f g i x -2 \sin \left (b x +d \right )^{2} b \,g^{2} x -3 \sin \left (b x +d \right )^{2} f^{2} i +2 \sin \left (b x +d \right )^{2} f g -\sin \left (b x +d \right )^{2} g^{2} i \right )}{8 b} \] Input:

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

Output:

(e**(2*a + 2*b*i*x)*( - 2*cos(b*x + d)**2*b*f**2*x + 4*cos(b*x + d)**2*b*f 
*g*i*x + 2*cos(b*x + d)**2*b*g**2*x - cos(b*x + d)**2*f**2*i - 2*cos(b*x + 
 d)**2*f*g - 3*cos(b*x + d)**2*g**2*i + 4*cos(b*x + d)*sin(b*x + d)*b*f**2 
*i*x + 8*cos(b*x + d)*sin(b*x + d)*b*f*g*x - 4*cos(b*x + d)*sin(b*x + d)*b 
*g**2*i*x + 2*sin(b*x + d)**2*b*f**2*x - 4*sin(b*x + d)**2*b*f*g*i*x - 2*s 
in(b*x + d)**2*b*g**2*x - 3*sin(b*x + d)**2*f**2*i + 2*sin(b*x + d)**2*f*g 
 - sin(b*x + d)**2*g**2*i))/(8*b)