Integrand size = 32, antiderivative size = 117 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\frac {2 e^{2 a-2 i d} (i f+g)}{b \left (f-e^{2 i (d+b x)} (f+i g)-i g\right ) (f+i g)^2}+\frac {2 i e^{2 a-2 i d} \log \left (f-e^{2 i (d+b x)} (f+i g)-i g\right )}{b (f+i g)^2} \] Output:
2*exp(2*a-2*I*d)*(I*f+g)/b/(f-exp(2*I*(b*x+d))*(f+I*g)-I*g)/(f+I*g)^2+2*I* exp(2*a-2*I*d)*ln(f-exp(2*I*(b*x+d))*(f+I*g)-I*g)/b/(f+I*g)^2
Time = 0.63 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.65 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\frac {e^{2 a-2 i d} \left (\frac {2 (f-i g)}{i \left (-1+e^{2 i (d+b x)}\right ) f-\left (1+e^{2 i (d+b x)}\right ) g}-2 \arctan \left (\frac {\left (-1+e^{2 i (d+b x)}\right ) f^3+\left (3+e^{2 i (d+b x)}\right ) f g^2}{\left (-3+e^{2 i (d+b x)}\right ) f^2 g+\left (1+e^{2 i (d+b x)}\right ) g^3}\right )+i \log \left (\left (-1+e^{2 i (d+b x)}\right )^2 f^2+\left (1+e^{2 i (d+b x)}\right )^2 g^2\right )\right )}{b (f+i g)^2} \] 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)*((2*(f - I*g))/(I*(-1 + E^((2*I)*(d + b*x)))*f - (1 + E ^((2*I)*(d + b*x)))*g) - 2*ArcTan[((-1 + E^((2*I)*(d + b*x)))*f^3 + (3 + E ^((2*I)*(d + b*x)))*f*g^2)/((-3 + E^((2*I)*(d + b*x)))*f^2*g + (1 + E^((2* I)*(d + b*x)))*g^3)] + I*Log[(-1 + E^((2*I)*(d + b*x)))^2*f^2 + (1 + E^((2 *I)*(d + b*x)))^2*g^2]))/(b*(f + I*g)^2)
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.
\(\displaystyle \int \frac {e^{2 (a+i b x)}}{(f \sin (b x+d)+g \cos (b x+d))^2} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 a+2 i b x}}{(f \sin (b x+d)+g \cos (b x+d))^2}dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \frac {e^{2 a+2 i b x}}{(f \sin (b x+d)+g \cos (b x+d))^2}dx\) |
Input:
Int[E^(2*(a + I*b*x))/(g*Cos[d + b*x] + f*Sin[d + b*x])^2,x]
Output:
$Aborted
Time = 0.85 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99
method | result | size |
risch | \(-\frac {2 i {\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{b \left (i g +f \right ) \left (f \,{\mathrm e}^{2 i \left (b x +d \right )}+i g \,{\mathrm e}^{2 i \left (b x +d \right )}-f +i g \right )}+\frac {2 i {\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \ln \left (f \,{\mathrm e}^{2 i \left (b x +d \right )}+i g \,{\mathrm e}^{2 i \left (b x +d \right )}-f +i g \right )}{\left (i g +f \right )^{2} b}\) | \(116\) |
Input:
int(exp(2*a+2*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x,method=_RETURNVERBOSE )
Output:
-2*I*exp(2*a)*exp(2*I*b*x)/b/(f+I*g)/(f*exp(2*I*(b*x+d))+I*g*exp(2*I*(b*x+ d))-f+I*g)+2*I/(f+I*g)^2*exp(2*a)/b*exp(-2*I*d)*ln(f*exp(2*I*(b*x+d))+I*g* exp(2*I*(b*x+d))-f+I*g)
Time = 0.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.21 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\frac {2 \, {\left ({\left (f - i \, g\right )} e^{\left (2 \, a - 2 i \, d\right )} - {\left ({\left (f + i \, g\right )} e^{\left (2 i \, b x + 2 \, a\right )} - {\left (f - i \, g\right )} e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (\frac {{\left (-i \, f + g\right )} e^{\left (2 i \, b x + 2 i \, d\right )} + i \, f + g}{-i \, f + g}\right )\right )}}{-i \, b f^{3} + b f^{2} g - i \, b f g^{2} + b g^{3} + {\left (i \, b f^{3} - 3 \, b f^{2} g - 3 i \, b f g^{2} + b g^{3}\right )} e^{\left (2 i \, b x + 2 i \, d\right )}} \] 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:
2*((f - I*g)*e^(2*a - 2*I*d) - ((f + I*g)*e^(2*I*b*x + 2*a) - (f - I*g)*e^ (2*a - 2*I*d))*log(((-I*f + g)*e^(2*I*b*x + 2*I*d) + I*f + g)/(-I*f + g))) /(-I*b*f^3 + b*f^2*g - I*b*f*g^2 + b*g^3 + (I*b*f^3 - 3*b*f^2*g - 3*I*b*f* g^2 + b*g^3)*e^(2*I*b*x + 2*I*d))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (88) = 176\).
Time = 1.86 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.68 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\frac {- 2 i f e^{2 a} - 2 g e^{2 a}}{- b f^{3} e^{2 i d} - i b f^{2} g e^{2 i d} - b f g^{2} e^{2 i d} - i b g^{3} e^{2 i d} + \left (b f^{3} e^{4 i d} + 3 i b f^{2} g e^{4 i d} - 3 b f g^{2} e^{4 i d} - i b g^{3} e^{4 i d}\right ) e^{2 i b x}} + \frac {2 i e^{2 a} e^{- 2 i d} \log {\left (\frac {- f + i g}{f e^{2 i d} + i g e^{2 i d}} + e^{2 i b x} \right )}}{b \left (f + i g\right )^{2}} \] Input:
integrate(exp(2*a+2*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))**2,x)
Output:
(-2*I*f*exp(2*a) - 2*g*exp(2*a))/(-b*f**3*exp(2*I*d) - I*b*f**2*g*exp(2*I* d) - b*f*g**2*exp(2*I*d) - I*b*g**3*exp(2*I*d) + (b*f**3*exp(4*I*d) + 3*I* b*f**2*g*exp(4*I*d) - 3*b*f*g**2*exp(4*I*d) - I*b*g**3*exp(4*I*d))*exp(2*I *b*x)) + 2*I*exp(2*a)*exp(-2*I*d)*log((-f + I*g)/(f*exp(2*I*d) + I*g*exp(2 *I*d)) + exp(2*I*b*x))/(b*(f + I*g)**2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 863 vs. \(2 (92) = 184\).
Time = 0.16 (sec) , antiderivative size = 863, normalized size of antiderivative = 7.38 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\text {Too large to display} \] Input:
integrate(exp(2*a+2*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm="max ima")
Output:
-(2*((cos(2*d)^2*e^(2*a) + e^(2*a)*sin(2*d)^2)*f + (-I*cos(2*d)^2*e^(2*a) - I*e^(2*a)*sin(2*d)^2)*g - ((cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*f - ( -I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*g)*cos(2*b*x + 4*d) + ((-I*cos(2*d )*e^(2*a) - e^(2*a)*sin(2*d))*f + (cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))* g)*sin(2*b*x + 4*d))*arctan2((2*f*g*cos(2*d) + f^2*sin(2*d) - g^2*sin(2*d) - (f^2*sin(2*d) + g^2*sin(2*d))*cos(2*b*x + 2*d) + (f^2*cos(2*d) + g^2*co s(2*d))*sin(2*b*x + 2*d))/((cos(2*d)^2 + sin(2*d)^2)*f^2 + (cos(2*d)^2 + s in(2*d)^2)*g^2), -(f^2*cos(2*d) - g^2*cos(2*d) - 2*f*g*sin(2*d) - (f^2*cos (2*d) + g^2*cos(2*d))*cos(2*b*x + 2*d) - (f^2*sin(2*d) + g^2*sin(2*d))*sin (2*b*x + 2*d))/((cos(2*d)^2 + sin(2*d)^2)*f^2 + (cos(2*d)^2 + sin(2*d)^2)* g^2)) - 2*I*f*e^(2*a) - 2*g*e^(2*a) - ((I*cos(2*d)^2*e^(2*a) + I*e^(2*a)*s in(2*d)^2)*f + (cos(2*d)^2*e^(2*a) + e^(2*a)*sin(2*d)^2)*g + ((-I*cos(2*d) *e^(2*a) - e^(2*a)*sin(2*d))*f + (cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*g )*cos(2*b*x + 4*d) + ((cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*f + (I*cos(2 *d)*e^(2*a) + e^(2*a)*sin(2*d))*g)*sin(2*b*x + 4*d))*log(((f^2 + g^2)*cos( 2*b*x + 2*d)^2 + 4*f*g*sin(2*b*x + 2*d) + (f^2 + g^2)*sin(2*b*x + 2*d)^2 + f^2 + g^2 - 2*(f^2 - g^2)*cos(2*b*x + 2*d))/((cos(2*d)^2 + sin(2*d)^2)*f^ 2 + (cos(2*d)^2 + sin(2*d)^2)*g^2)))/((b*cos(2*d) + I*b*sin(2*d))*f^3 - (- I*b*cos(2*d) + b*sin(2*d))*f^2*g + (b*cos(2*d) + I*b*sin(2*d))*f*g^2 - (-I *b*cos(2*d) + b*sin(2*d))*g^3 - (b*f^3 + 3*I*b*f^2*g - 3*b*f*g^2 - I*b*...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (92) = 184\).
Time = 0.18 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.37 \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=-\frac {2 \, {\left (-i \, f e^{\left (2 i \, b x + 2 \, a\right )} \log \left (f e^{\left (2 i \, b x + 2 i \, d\right )} + i \, g e^{\left (2 i \, b x + 2 i \, d\right )} - f + i \, g\right ) + g e^{\left (2 i \, b x + 2 \, a\right )} \log \left (f e^{\left (2 i \, b x + 2 i \, d\right )} + i \, g e^{\left (2 i \, b x + 2 i \, d\right )} - f + i \, g\right ) + i \, f e^{\left (2 \, a - 2 i \, d\right )} \log \left (f e^{\left (2 i \, b x + 2 i \, d\right )} + i \, g e^{\left (2 i \, b x + 2 i \, d\right )} - f + i \, g\right ) + g e^{\left (2 \, a - 2 i \, d\right )} \log \left (f e^{\left (2 i \, b x + 2 i \, d\right )} + i \, g e^{\left (2 i \, b x + 2 i \, d\right )} - f + i \, g\right ) + i \, f e^{\left (2 \, a - 2 i \, d\right )} + g e^{\left (2 \, a - 2 i \, d\right )}\right )}}{{\left (f^{3} e^{\left (2 i \, b x + 2 i \, d\right )} + 3 i \, f^{2} g e^{\left (2 i \, b x + 2 i \, d\right )} - 3 \, f g^{2} e^{\left (2 i \, b x + 2 i \, d\right )} - i \, g^{3} e^{\left (2 i \, b x + 2 i \, d\right )} - f^{3} - i \, f^{2} g - f g^{2} - i \, g^{3}\right )} b} \] 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:
-2*(-I*f*e^(2*I*b*x + 2*a)*log(f*e^(2*I*b*x + 2*I*d) + I*g*e^(2*I*b*x + 2* I*d) - f + I*g) + g*e^(2*I*b*x + 2*a)*log(f*e^(2*I*b*x + 2*I*d) + I*g*e^(2 *I*b*x + 2*I*d) - f + I*g) + I*f*e^(2*a - 2*I*d)*log(f*e^(2*I*b*x + 2*I*d) + I*g*e^(2*I*b*x + 2*I*d) - f + I*g) + g*e^(2*a - 2*I*d)*log(f*e^(2*I*b*x + 2*I*d) + I*g*e^(2*I*b*x + 2*I*d) - f + I*g) + I*f*e^(2*a - 2*I*d) + g*e ^(2*a - 2*I*d))/((f^3*e^(2*I*b*x + 2*I*d) + 3*I*f^2*g*e^(2*I*b*x + 2*I*d) - 3*f*g^2*e^(2*I*b*x + 2*I*d) - I*g^3*e^(2*I*b*x + 2*I*d) - f^3 - I*f^2*g - f*g^2 - I*g^3)*b)
Timed out. \[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\int \frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}}{{\left (g\,\cos \left (d+b\,x\right )+f\,\sin \left (d+b\,x\right )\right )}^2} \,d x \] Input:
int(exp(2*a + b*x*2i)/(g*cos(d + b*x) + f*sin(d + b*x))^2,x)
Output:
int(exp(2*a + b*x*2i)/(g*cos(d + b*x) + f*sin(d + b*x))^2, x)
\[ \int \frac {e^{2 (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\int \frac {{\mathrm e}^{2 i b x +2 a}}{\left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}}d x \] Input:
int(exp(2*a+2*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x)
Output:
int(exp(2*a+2*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x)