\(\int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx\) [14]

Optimal result

Integrand size = 32, antiderivative size = 202 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=-\frac {3 e^{2 (a-i d)+i (d+b x)}}{b \left (f-e^{2 i (d+b x)} (f+i g)-i g\right ) (f+i g)^2}+\frac {2 e^{2 (a-i d)+3 i (d+b x)}}{b \left (f-e^{2 i (d+b x)} (f+i g)-i g\right )^2 (f+i g)}+\frac {3 e^{2 a-2 i d} \text {arctanh}\left (\frac {e^{i (d+b x)} \sqrt {f+i g}}{\sqrt {f-i g}}\right )}{b \sqrt {f-i g} (f+i g)^{5/2}} \] Output:

-3*exp(2*a-2*I*d+I*(b*x+d))/b/(f-exp(2*I*(b*x+d))*(f+I*g)-I*g)/(f+I*g)^2+2 
*exp(2*a-2*I*d+3*I*(b*x+d))/b/(f-exp(2*I*(b*x+d))*(f+I*g)-I*g)^2/(f+I*g)+3 
*exp(2*a-2*I*d)*arctanh(exp(I*(b*x+d))*(f+I*g)^(1/2)/(f-I*g)^(1/2))/b/(f-I 
*g)^(1/2)/(f+I*g)^(5/2)
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\frac {e^{2 a-2 i d} \left (\frac {e^{i (d+b x)} \left (\left (-3+5 e^{2 i (d+b x)}\right ) f+i \left (3+5 e^{2 i (d+b x)}\right ) g\right )}{(f+i g)^2 \left (\left (-1+e^{2 i (d+b x)}\right ) f+i \left (1+e^{2 i (d+b x)}\right ) g\right )^2}-\frac {3 i \arctan \left (\frac {e^{i (d+b x)} \sqrt {-i f+g}}{\sqrt {i f+g}}\right )}{(-i f+g)^{5/2} \sqrt {i f+g}}\right )}{b} \] Input:

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

Output:

(E^(2*a - (2*I)*d)*((E^(I*(d + b*x))*((-3 + 5*E^((2*I)*(d + b*x)))*f + I*( 
3 + 5*E^((2*I)*(d + b*x)))*g))/((f + I*g)^2*((-1 + E^((2*I)*(d + b*x)))*f 
+ I*(1 + E^((2*I)*(d + b*x)))*g)^2) - ((3*I)*ArcTan[(E^(I*(d + b*x))*Sqrt[ 
(-I)*f + g])/Sqrt[I*f + g]])/(((-I)*f + g)^(5/2)*Sqrt[I*f + g])))/b
 

Rubi [F]

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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7299

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 7292

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

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (168 ) = 336\).

Time = 4.12 (sec) , antiderivative size = 428, normalized size of antiderivative = 2.12

method	result	size
risch	\(\frac {3 i g \,{\mathrm e}^{2 a} {\mathrm e}^{5 i b x} {\mathrm e}^{3 i d}+3 f \,{\mathrm e}^{2 a} {\mathrm e}^{5 i b x} {\mathrm e}^{3 i d}+i g \,{\mathrm e}^{2 a} {\mathrm e}^{3 i b x} {\mathrm e}^{i d}-f \,{\mathrm e}^{2 a} {\mathrm e}^{3 i b x} {\mathrm e}^{i d}}{\left (-i g +f \right ) \left (f \,{\mathrm e}^{2 i \left (b x +d \right )}+i g \,{\mathrm e}^{2 i \left (b x +d \right )}-f +i g \right )^{2} b \left (i g +f \right )}+\frac {3 i {\mathrm e}^{2 a} {\mathrm e}^{-i d} \arctan \left (\frac {f \,{\mathrm e}^{i \left (b x +2 d \right )}}{\sqrt {-{\mathrm e}^{2 i d} f^{2}-g^{2} {\mathrm e}^{2 i d}}}+\frac {i g \,{\mathrm e}^{i \left (b x +2 d \right )}}{\sqrt {-{\mathrm e}^{2 i d} f^{2}-g^{2} {\mathrm e}^{2 i d}}}\right ) g}{b \left (i g +f \right )^{2} \left (-i g +f \right ) \sqrt {-{\mathrm e}^{2 i d} f^{2}-g^{2} {\mathrm e}^{2 i d}}}-\frac {3 \,{\mathrm e}^{2 a} {\mathrm e}^{-i d} {\mathrm e}^{i b x}}{b \left (i g +f \right )^{2} \left (-i g +f \right )}-\frac {3 \,{\mathrm e}^{2 a} {\mathrm e}^{-i d} \arctan \left (\frac {f \,{\mathrm e}^{i \left (b x +2 d \right )}}{\sqrt {-{\mathrm e}^{2 i d} f^{2}-g^{2} {\mathrm e}^{2 i d}}}+\frac {i g \,{\mathrm e}^{i \left (b x +2 d \right )}}{\sqrt {-{\mathrm e}^{2 i d} f^{2}-g^{2} {\mathrm e}^{2 i d}}}\right ) f}{b \left (i g +f \right )^{2} \left (-i g +f \right ) \sqrt {-{\mathrm e}^{2 i d} f^{2}-g^{2} {\mathrm e}^{2 i d}}}\)	\(428\)

Input:

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

Output:

(3*I*g*exp(2*a)*exp(5*I*b*x)*exp(3*I*d)+3*f*exp(2*a)*exp(5*I*b*x)*exp(3*I* 
d)+I*g*exp(2*a)*exp(3*I*b*x)*exp(I*d)-f*exp(2*a)*exp(3*I*b*x)*exp(I*d))/(f 
-I*g)/(f*exp(2*I*(b*x+d))+I*g*exp(2*I*(b*x+d))-f+I*g)^2/b/(f+I*g)+3*I/b/(f 
+I*g)^2/(f-I*g)/(-exp(2*I*d)*f^2-g^2*exp(2*I*d))^(1/2)*exp(2*a)*exp(-I*d)* 
arctan(f/(-exp(2*I*d)*f^2-g^2*exp(2*I*d))^(1/2)*exp(I*(b*x+2*d))+I*g/(-exp 
(2*I*d)*f^2-g^2*exp(2*I*d))^(1/2)*exp(I*(b*x+2*d)))*g-3/b/(f+I*g)^2/(f-I*g 
)*exp(2*a)*exp(-I*d)*exp(I*b*x)-3/b/(f+I*g)^2/(f-I*g)/(-exp(2*I*d)*f^2-g^2 
*exp(2*I*d))^(1/2)*exp(2*a)*exp(-I*d)*arctan(f/(-exp(2*I*d)*f^2-g^2*exp(2* 
I*d))^(1/2)*exp(I*(b*x+2*d))+I*g/(-exp(2*I*d)*f^2-g^2*exp(2*I*d))^(1/2)*ex 
p(I*(b*x+2*d)))*f
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 711 vs. \(2 (149) = 298\).

Time = 0.09 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.52 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx =\text {Too large to display} \] Input:

integrate(exp(2*a+2*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x, algorithm="fri 
cas")
 

Output:

1/2*(3*(b*f^4 + 2*b*f^2*g^2 + b*g^4 + (b*f^4 + 4*I*b*f^3*g - 6*b*f^2*g^2 - 
 4*I*b*f*g^3 + b*g^4)*e^(4*I*b*x + 4*I*d) - 2*(b*f^4 + 2*I*b*f^3*g + 2*I*b 
*f*g^3 - b*g^4)*e^(2*I*b*x + 2*I*d))*e^(2*a - 2*I*d)*log(((b*f^3 + I*b*f^2 
*g + b*f*g^2 + I*b*g^3)*e^(2*a - 2*I*d)/sqrt(b^2*f^6 + 4*I*b^2*f^5*g - 5*b 
^2*f^4*g^2 - 5*b^2*f^2*g^4 - 4*I*b^2*f*g^5 + b^2*g^6) + e^(I*b*x + 2*a - I 
*d))*e^(-2*a + 2*I*d))/sqrt(b^2*f^6 + 4*I*b^2*f^5*g - 5*b^2*f^4*g^2 - 5*b^ 
2*f^2*g^4 - 4*I*b^2*f*g^5 + b^2*g^6) - 3*(b*f^4 + 2*b*f^2*g^2 + b*g^4 + (b 
*f^4 + 4*I*b*f^3*g - 6*b*f^2*g^2 - 4*I*b*f*g^3 + b*g^4)*e^(4*I*b*x + 4*I*d 
) - 2*(b*f^4 + 2*I*b*f^3*g + 2*I*b*f*g^3 - b*g^4)*e^(2*I*b*x + 2*I*d))*e^( 
2*a - 2*I*d)*log(-((b*f^3 + I*b*f^2*g + b*f*g^2 + I*b*g^3)*e^(2*a - 2*I*d) 
/sqrt(b^2*f^6 + 4*I*b^2*f^5*g - 5*b^2*f^4*g^2 - 5*b^2*f^2*g^4 - 4*I*b^2*f* 
g^5 + b^2*g^6) - e^(I*b*x + 2*a - I*d))*e^(-2*a + 2*I*d))/sqrt(b^2*f^6 + 4 
*I*b^2*f^5*g - 5*b^2*f^4*g^2 - 5*b^2*f^2*g^4 - 4*I*b^2*f*g^5 + b^2*g^6) + 
10*(f + I*g)*e^(3*I*b*x + 2*a + I*d) - 6*(f - I*g)*e^(I*b*x + 2*a - I*d))/ 
(b*f^4 + 2*b*f^2*g^2 + b*g^4 + (b*f^4 + 4*I*b*f^3*g - 6*b*f^2*g^2 - 4*I*b* 
f*g^3 + b*g^4)*e^(4*I*b*x + 4*I*d) - 2*(b*f^4 + 2*I*b*f^3*g + 2*I*b*f*g^3 
- b*g^4)*e^(2*I*b*x + 2*I*d))
 

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 447 vs. \(2 (153) = 306\).

Time = 6.37 (sec) , antiderivative size = 447, normalized size of antiderivative = 2.21 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\frac {\left (- 3 f e^{2 a} + 3 i g e^{2 a}\right ) e^{i b x} + \left (5 f e^{2 a} e^{2 i d} + 5 i g e^{2 a} e^{2 i d}\right ) e^{3 i b x}}{b f^{4} e^{i d} + 2 b f^{2} g^{2} e^{i d} + b g^{4} e^{i d} + \left (- 2 b f^{4} e^{3 i d} - 4 i b f^{3} g e^{3 i d} - 4 i b f g^{3} e^{3 i d} + 2 b g^{4} e^{3 i d}\right ) e^{2 i b x} + \left (b f^{4} e^{5 i d} + 4 i b f^{3} g e^{5 i d} - 6 b f^{2} g^{2} e^{5 i d} - 4 i b f g^{3} e^{5 i d} + b g^{4} e^{5 i d}\right ) e^{4 i b x}} + \operatorname {RootSum} {\left (z^{2} \cdot \left (4 b^{2} f^{6} e^{4 i d} + 16 i b^{2} f^{5} g e^{4 i d} - 20 b^{2} f^{4} g^{2} e^{4 i d} - 20 b^{2} f^{2} g^{4} e^{4 i d} - 16 i b^{2} f g^{5} e^{4 i d} + 4 b^{2} g^{6} e^{4 i d}\right ) - 9 e^{4 a}, \left ( i \mapsto i \log {\left (\frac {\left (2 i b f^{3} e^{i d} + 2 i i b f^{2} g e^{i d} + 2 i b f g^{2} e^{i d} + 2 i i b g^{3} e^{i d}\right ) e^{- 2 a}}{3} + e^{i b x} \right )} \right )\right )} \] Input:

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

Output:

((-3*f*exp(2*a) + 3*I*g*exp(2*a))*exp(I*b*x) + (5*f*exp(2*a)*exp(2*I*d) + 
5*I*g*exp(2*a)*exp(2*I*d))*exp(3*I*b*x))/(b*f**4*exp(I*d) + 2*b*f**2*g**2* 
exp(I*d) + b*g**4*exp(I*d) + (-2*b*f**4*exp(3*I*d) - 4*I*b*f**3*g*exp(3*I* 
d) - 4*I*b*f*g**3*exp(3*I*d) + 2*b*g**4*exp(3*I*d))*exp(2*I*b*x) + (b*f**4 
*exp(5*I*d) + 4*I*b*f**3*g*exp(5*I*d) - 6*b*f**2*g**2*exp(5*I*d) - 4*I*b*f 
*g**3*exp(5*I*d) + b*g**4*exp(5*I*d))*exp(4*I*b*x)) + RootSum(_z**2*(4*b** 
2*f**6*exp(4*I*d) + 16*I*b**2*f**5*g*exp(4*I*d) - 20*b**2*f**4*g**2*exp(4* 
I*d) - 20*b**2*f**2*g**4*exp(4*I*d) - 16*I*b**2*f*g**5*exp(4*I*d) + 4*b**2 
*g**6*exp(4*I*d)) - 9*exp(4*a), Lambda(_i, _i*log((2*_i*b*f**3*exp(I*d) + 
2*_i*I*b*f**2*g*exp(I*d) + 2*_i*b*f*g**2*exp(I*d) + 2*_i*I*b*g**3*exp(I*d) 
)*exp(-2*a)/3 + exp(I*b*x))))
 

Maxima [F(-1)]

Timed out. \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\text {Timed out} \] Input:

integrate(exp(2*a+2*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x, algorithm="max 
ima")
 

Output:

Timed out
 

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. 427 vs. \(2 (149) = 298\).

Time = 0.35 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.11 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=-\frac {{\left (\frac {3 \, \log \left (\frac {f e^{\left (i \, b x + i \, d\right )} + i \, g e^{\left (i \, b x + i \, d\right )} - \sqrt {f^{2} + g^{2}}}{f e^{\left (i \, b x + i \, d\right )} + i \, g e^{\left (i \, b x + i \, d\right )} + \sqrt {f^{2} + g^{2}}}\right )}{{\left (f^{2} + 2 i \, f g - g^{2}\right )} \sqrt {f^{2} + g^{2}}} + \frac {16 \, {\left (5 i \, f e^{\left (3 i \, b x + 3 i \, d\right )} - 5 \, g e^{\left (3 i \, b x + 3 i \, d\right )} - 3 i \, f e^{\left (i \, b x + i \, d\right )} - 3 \, g e^{\left (i \, b x + i \, d\right )}\right )}}{-8 \, {\left (i \, f^{2} - 2 \, f g - i \, g^{2}\right )} {\left (f e^{\left (2 i \, b x + 2 i \, d\right )} + i \, g e^{\left (2 i \, b x + 2 i \, d\right )} - f + i \, g\right )}^{2}}\right )} e^{\left (2 \, a - 2 i \, d\right )}}{2 \, b} + \frac {2 \, {\left (5 \, f e^{\left (3 i \, b x + 2 \, a + i \, d\right )} + 5 i \, g e^{\left (3 i \, b x + 2 \, a + i \, d\right )} - 3 \, f e^{\left (i \, b x + 2 \, a - i \, d\right )} + 3 i \, g e^{\left (i \, b x + 2 \, a - i \, d\right )}\right )}}{{\left (f^{4} e^{\left (4 i \, b x + 4 i \, d\right )} + 4 i \, f^{3} g e^{\left (4 i \, b x + 4 i \, d\right )} - 6 \, f^{2} g^{2} e^{\left (4 i \, b x + 4 i \, d\right )} - 4 i \, f g^{3} e^{\left (4 i \, b x + 4 i \, d\right )} + g^{4} e^{\left (4 i \, b x + 4 i \, d\right )} - 2 \, f^{4} e^{\left (2 i \, b x + 2 i \, d\right )} - 4 i \, f^{3} g e^{\left (2 i \, b x + 2 i \, d\right )} - 4 i \, f g^{3} e^{\left (2 i \, b x + 2 i \, d\right )} + 2 \, g^{4} e^{\left (2 i \, b x + 2 i \, d\right )} + f^{4} + 2 \, f^{2} g^{2} + g^{4}\right )} b} \] Input:

integrate(exp(2*a+2*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x, algorithm="gia 
c")
 

Output:

-1/2*(3*log((f*e^(I*b*x + I*d) + I*g*e^(I*b*x + I*d) - sqrt(f^2 + g^2))/(f 
*e^(I*b*x + I*d) + I*g*e^(I*b*x + I*d) + sqrt(f^2 + g^2)))/((f^2 + 2*I*f*g 
 - g^2)*sqrt(f^2 + g^2)) + 16*(5*I*f*e^(3*I*b*x + 3*I*d) - 5*g*e^(3*I*b*x 
+ 3*I*d) - 3*I*f*e^(I*b*x + I*d) - 3*g*e^(I*b*x + I*d))/((-8*I*f^2 + 16*f* 
g + 8*I*g^2)*(f*e^(2*I*b*x + 2*I*d) + I*g*e^(2*I*b*x + 2*I*d) - f + I*g)^2 
))*e^(2*a - 2*I*d)/b + 2*(5*f*e^(3*I*b*x + 2*a + I*d) + 5*I*g*e^(3*I*b*x + 
 2*a + I*d) - 3*f*e^(I*b*x + 2*a - I*d) + 3*I*g*e^(I*b*x + 2*a - I*d))/((f 
^4*e^(4*I*b*x + 4*I*d) + 4*I*f^3*g*e^(4*I*b*x + 4*I*d) - 6*f^2*g^2*e^(4*I* 
b*x + 4*I*d) - 4*I*f*g^3*e^(4*I*b*x + 4*I*d) + g^4*e^(4*I*b*x + 4*I*d) - 2 
*f^4*e^(2*I*b*x + 2*I*d) - 4*I*f^3*g*e^(2*I*b*x + 2*I*d) - 4*I*f*g^3*e^(2* 
I*b*x + 2*I*d) + 2*g^4*e^(2*I*b*x + 2*I*d) + f^4 + 2*f^2*g^2 + g^4)*b)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\int \frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}}{{\left (g\,\cos \left (d+b\,x\right )+f\,\sin \left (d+b\,x\right )\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\int \frac {{\mathrm e}^{2 i b x +2 a}}{\left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{3}}d x \] Input:

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

Output:

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