\(\int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx\) [20]

Optimal result

Integrand size = 34, antiderivative size = 458 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=-\frac {2 e^{\frac {5}{3} (a-i d)+\frac {5}{3} i (d+b x)}}{b (i f-g) \left (f-e^{2 i (d+b x)} (f+i g)-i g\right )}-\frac {5 i e^{\frac {5}{3} (a-i d)} \arctan \left (\frac {1-\frac {2 e^{\frac {1}{3} i (d+b x)} \sqrt [6]{f+i g}}{\sqrt [6]{f-i g}}}{\sqrt {3}}\right )}{\sqrt {3} b \sqrt [6]{f-i g} (f+i g)^{11/6}}+\frac {5 i e^{\frac {5}{3} (a-i d)} \arctan \left (\frac {1+\frac {2 e^{\frac {1}{3} i (d+b x)} \sqrt [6]{f+i g}}{\sqrt [6]{f-i g}}}{\sqrt {3}}\right )}{\sqrt {3} b \sqrt [6]{f-i g} (f+i g)^{11/6}}-\frac {5 i e^{\frac {5}{3} (a-i d)} \text {arctanh}\left (\frac {e^{\frac {1}{3} i (d+b x)} \sqrt [6]{f-i g}}{\left (e^{\frac {2}{3} i (d+b x)}+\frac {\sqrt [3]{f-i g}}{\sqrt [3]{f+i g}}\right ) \sqrt [6]{f+i g}}\right )}{3 b \sqrt [6]{f-i g} (f+i g)^{11/6}}-\frac {10 i e^{\frac {5}{3} (a-i d)} \text {arctanh}\left (\frac {e^{\frac {1}{3} i (d+b x)} \sqrt [6]{f+i g}}{\sqrt [6]{f-i g}}\right )}{3 b \sqrt [6]{f-i g} (f+i g)^{11/6}} \] Output:

-2*exp(5/3*a-5/3*I*d+5/3*I*(b*x+d))/b/(I*f-g)/(f-exp(2*I*(b*x+d))*(f+I*g)- 
I*g)-5/3*I*exp(5/3*a-5/3*I*d)*arctan(1/3*(1-2*exp(1/3*I*(b*x+d))*(f+I*g)^( 
1/6)/(f-I*g)^(1/6))*3^(1/2))*3^(1/2)/b/(f-I*g)^(1/6)/(f+I*g)^(11/6)+5/3*I* 
exp(5/3*a-5/3*I*d)*arctan(1/3*(1+2*exp(1/3*I*(b*x+d))*(f+I*g)^(1/6)/(f-I*g 
)^(1/6))*3^(1/2))*3^(1/2)/b/(f-I*g)^(1/6)/(f+I*g)^(11/6)-5/3*I*exp(5/3*a-5 
/3*I*d)*arctanh(exp(1/3*I*(b*x+d))*(f-I*g)^(1/6)/(exp(2/3*I*(b*x+d))+(f-I* 
g)^(1/3)/(f+I*g)^(1/3))/(f+I*g)^(1/6))/b/(f-I*g)^(1/6)/(f+I*g)^(11/6)-10/3 
*I*exp(5/3*a-5/3*I*d)*arctanh(exp(1/3*I*(b*x+d))*(f+I*g)^(1/6)/(f-I*g)^(1/ 
6))/b/(f-I*g)^(1/6)/(f+I*g)^(11/6)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.61 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.34 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\frac {e^{5 a/3} \left (-\frac {36 i e^{\frac {5 i b x}{3}}}{(f+i g) \left (\left (-1+e^{2 i (d+b x)}\right ) f+i \left (1+e^{2 i (d+b x)}\right ) g\right )}+\frac {5 e^{-\frac {5 i d}{3}} \text {RootSum}\left [i f+g-i f \text {$\#$1}^6+g \text {$\#$1}^6\&,\frac {2 d+2 b x+3 i \log \left (\left (e^{\frac {1}{3} i (d+b x)}-\text {$\#$1}\right )^2\right )}{\text {$\#$1}}\&\right ]}{(f+i g)^2}\right )}{18 b} \] Input:

Integrate[E^((5*(a + I*b*x))/3)/(g*Cos[d + b*x] + f*Sin[d + b*x])^2,x]
 

Output:

(E^((5*a)/3)*(((-36*I)*E^(((5*I)/3)*b*x))/((f + I*g)*((-1 + E^((2*I)*(d + 
b*x)))*f + I*(1 + E^((2*I)*(d + b*x)))*g)) + (5*RootSum[I*f + g - I*f*#1^6 
 + g*#1^6 & , (2*d + 2*b*x + (3*I)*Log[(E^((I/3)*(d + b*x)) - #1)^2])/#1 & 
 ])/(E^(((5*I)/3)*d)*(f + I*g)^2)))/(18*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.

\(\displaystyle \int \frac {e^{\frac {5}{3} (a+i b x)}}{(f \sin (b x+d)+g \cos (b x+d))^2} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{(f \sin (b x+d)+g \cos (b x+d))^2}dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \frac {e^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{(f \sin (b x+d)+g \cos (b x+d))^2}dx\)

Input:

Int[E^((5*(a + I*b*x))/3)/(g*Cos[d + b*x] + f*Sin[d + b*x])^2,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 [F]

\[\int \frac {{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{\left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}}d x\]

Input:

int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x)
 

Output:

int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x)
 

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. 3275 vs. \(2 (305) = 610\).

Time = 0.11 (sec) , antiderivative size = 3275, normalized size of antiderivative = 7.15 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\text {Too large to display} \] Input:

integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm= 
"fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\text {Timed out} \] Input:

integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm= 
"maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: sign: argument cannot be imagi 
nary; found %i
 

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. 21854 vs. \(2 (305) = 610\).

Time = 40.04 (sec) , antiderivative size = 21854, normalized size of antiderivative = 47.72 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\text {Too large to display} \] Input:

integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm= 
"giac")
 

Output:

-1/6*I*(10*(5*sqrt(3)*cos(-1/12*pi*sgn(-f^2*cos(2*d) + g^2*cos(2*d) + 2*f* 
g*sin(2*d))*sgn(2*f*g*cos(2*d) + f^2*sin(2*d) - g^2*sin(2*d)) + 1/12*pi*sg 
n(2*f*g*cos(2*d) + f^2*sin(2*d) - g^2*sin(2*d)) + 1/6*arctan(f^2*tan(2*d)/ 
(2*f*g*tan(2*d) - f^2 + g^2) - g^2*tan(2*d)/(2*f*g*tan(2*d) - f^2 + g^2) + 
 2*f*g/(2*f*g*tan(2*d) - f^2 + g^2)))^4*cos(2*d)*sin(-1/12*pi*sgn(-f^2*cos 
(2*d) + g^2*cos(2*d) + 2*f*g*sin(2*d))*sgn(2*f*g*cos(2*d) + f^2*sin(2*d) - 
 g^2*sin(2*d)) + 1/12*pi*sgn(2*f*g*cos(2*d) + f^2*sin(2*d) - g^2*sin(2*d)) 
 + 1/6*arctan(f^2*tan(2*d)/(2*f*g*tan(2*d) - f^2 + g^2) - g^2*tan(2*d)/(2* 
f*g*tan(2*d) - f^2 + g^2) + 2*f*g/(2*f*g*tan(2*d) - f^2 + g^2))) - 10*sqrt 
(3)*cos(-1/12*pi*sgn(-f^2*cos(2*d) + g^2*cos(2*d) + 2*f*g*sin(2*d))*sgn(2* 
f*g*cos(2*d) + f^2*sin(2*d) - g^2*sin(2*d)) + 1/12*pi*sgn(2*f*g*cos(2*d) + 
 f^2*sin(2*d) - g^2*sin(2*d)) + 1/6*arctan(f^2*tan(2*d)/(2*f*g*tan(2*d) - 
f^2 + g^2) - g^2*tan(2*d)/(2*f*g*tan(2*d) - f^2 + g^2) + 2*f*g/(2*f*g*tan( 
2*d) - f^2 + g^2)))^2*cos(2*d)*sin(-1/12*pi*sgn(-f^2*cos(2*d) + g^2*cos(2* 
d) + 2*f*g*sin(2*d))*sgn(2*f*g*cos(2*d) + f^2*sin(2*d) - g^2*sin(2*d)) + 1 
/12*pi*sgn(2*f*g*cos(2*d) + f^2*sin(2*d) - g^2*sin(2*d)) + 1/6*arctan(f^2* 
tan(2*d)/(2*f*g*tan(2*d) - f^2 + g^2) - g^2*tan(2*d)/(2*f*g*tan(2*d) - f^2 
 + g^2) + 2*f*g/(2*f*g*tan(2*d) - f^2 + g^2)))^3 + sqrt(3)*cos(2*d)*sin(-1 
/12*pi*sgn(-f^2*cos(2*d) + g^2*cos(2*d) + 2*f*g*sin(2*d))*sgn(2*f*g*cos(2* 
d) + f^2*sin(2*d) - g^2*sin(2*d)) + 1/12*pi*sgn(2*f*g*cos(2*d) + f^2*si...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\int \frac {{\mathrm {e}}^{\frac {5\,a}{3}+\frac {b\,x\,5{}\mathrm {i}}{3}}}{{\left (g\,\cos \left (d+b\,x\right )+f\,\sin \left (d+b\,x\right )\right )}^2} \,d x \] Input:

int(exp((5*a)/3 + (b*x*5i)/3)/(g*cos(d + b*x) + f*sin(d + b*x))^2,x)
 

Output:

int(exp((5*a)/3 + (b*x*5i)/3)/(g*cos(d + b*x) + f*sin(d + b*x))^2, x)
 

Reduce [F]

\[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\int \frac {{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{\left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}}d x \] Input:

int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x)
 

Output:

int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x)