Integrand size = 30, antiderivative size = 137 \[ \int \frac {e^{a+i b x}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=-\frac {2 e^{a-i d+i (d+b x)}}{b (i f-g) \left (f-e^{2 i (d+b x)} (f+i g)-i g\right )}-\frac {2 i e^{a-i d} \text {arctanh}\left (\frac {e^{i (d+b x)} \sqrt {f+i g}}{\sqrt {f-i g}}\right )}{b \sqrt {f-i g} (f+i g)^{3/2}} \] Output:
-2*exp(a-I*d+I*(b*x+d))/b/(I*f-g)/(f-exp(2*I*(b*x+d))*(f+I*g)-I*g)-2*I*exp (a-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)^(3/2)
Time = 0.51 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98 \[ \int \frac {e^{a+i b x}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\frac {2 e^a \left (-\frac {i e^{i b x}}{(f+i g) \left (\left (-1+e^{2 i (d+b x)}\right ) f+i \left (1+e^{2 i (d+b x)}\right ) g\right )}-\frac {i e^{-i d} \arctan \left (\frac {e^{i (d+b x)} \sqrt {-i f+g}}{\sqrt {i f+g}}\right )}{(-i f+g)^{3/2} \sqrt {i f+g}}\right )}{b} \] Input:
Integrate[E^(a + I*b*x)/(g*Cos[d + b*x] + f*Sin[d + b*x])^2,x]
Output:
(2*E^a*(((-I)*E^(I*b*x))/((f + I*g)*((-1 + E^((2*I)*(d + b*x)))*f + I*(1 + E^((2*I)*(d + b*x)))*g)) - (I*ArcTan[(E^(I*(d + b*x))*Sqrt[(-I)*f + g])/S qrt[I*f + g]])/(E^(I*d)*((-I)*f + g)^(3/2)*Sqrt[I*f + g])))/b
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.
\(\displaystyle \int \frac {e^{a+i b x}}{(f \sin (b x+d)+g \cos (b x+d))^2} \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \frac {e^{a+i b x}}{(f \sin (b x+d)+g \cos (b x+d))^2}dx\) |
Input:
Int[E^(a + I*b*x)/(g*Cos[d + b*x] + f*Sin[d + b*x])^2,x]
Output:
$Aborted
Time = 0.88 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {2 i {\mathrm e}^{a} {\mathrm e}^{i b x}}{b \left (i g +f \right ) \left (f \,{\mathrm e}^{2 i \left (b x +d \right )}+i g \,{\mathrm e}^{2 i \left (b x +d \right )}-f +i g \right )}+\frac {2 i {\mathrm e}^{a} \arctan \left (\frac {f \,{\mathrm e}^{i \left (b x +2 d \right )}+i g \,{\mathrm e}^{i \left (b x +2 d \right )}}{\sqrt {\left (i g -f \right ) {\mathrm e}^{2 i d} \left (i g +f \right )}}\right )}{\left (i g +f \right ) b \sqrt {\left (i g -f \right ) {\mathrm e}^{2 i d} \left (i g +f \right )}}\) | \(149\) |
Input:
int(exp(a+I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x,method=_RETURNVERBOSE)
Output:
-2*I*exp(a)*exp(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)*exp(a)/b/((I*g-f)*exp(2*I*d)*(f+I*g))^(1/2)*arctan(1/(( I*g-f)*exp(2*I*d)*(f+I*g))^(1/2)*(f*exp(I*(b*x+2*d))+I*g*exp(I*(b*x+2*d))) )
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (96) = 192\).
Time = 0.08 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.84 \[ \int \frac {e^{a+i b x}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\frac {{\left (b f^{2} + b g^{2} - {\left (b f^{2} + 2 i \, b f g - b g^{2}\right )} e^{\left (2 i \, b x + 2 i \, d\right )}\right )} \sqrt {-\frac {e^{\left (2 \, a - 2 i \, d\right )}}{b^{2} f^{4} + 2 i \, b^{2} f^{3} g + 2 i \, b^{2} f g^{3} - b^{2} g^{4}}} \log \left ({\left ({\left (i \, b f^{2} + i \, b g^{2}\right )} \sqrt {-\frac {e^{\left (2 \, a - 2 i \, d\right )}}{b^{2} f^{4} + 2 i \, b^{2} f^{3} g + 2 i \, b^{2} f g^{3} - b^{2} g^{4}}} + e^{\left (i \, b x + a\right )}\right )} e^{\left (-a + i \, d\right )}\right ) - {\left (b f^{2} + b g^{2} - {\left (b f^{2} + 2 i \, b f g - b g^{2}\right )} e^{\left (2 i \, b x + 2 i \, d\right )}\right )} \sqrt {-\frac {e^{\left (2 \, a - 2 i \, d\right )}}{b^{2} f^{4} + 2 i \, b^{2} f^{3} g + 2 i \, b^{2} f g^{3} - b^{2} g^{4}}} \log \left ({\left ({\left (-i \, b f^{2} - i \, b g^{2}\right )} \sqrt {-\frac {e^{\left (2 \, a - 2 i \, d\right )}}{b^{2} f^{4} + 2 i \, b^{2} f^{3} g + 2 i \, b^{2} f g^{3} - b^{2} g^{4}}} + e^{\left (i \, b x + a\right )}\right )} e^{\left (-a + i \, d\right )}\right ) + 2 i \, e^{\left (i \, b x + a\right )}}{b f^{2} + b g^{2} - {\left (b f^{2} + 2 i \, b f g - b g^{2}\right )} e^{\left (2 i \, b x + 2 i \, d\right )}} \] Input:
integrate(exp(a+I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm="fricas" )
Output:
((b*f^2 + b*g^2 - (b*f^2 + 2*I*b*f*g - b*g^2)*e^(2*I*b*x + 2*I*d))*sqrt(-e ^(2*a - 2*I*d)/(b^2*f^4 + 2*I*b^2*f^3*g + 2*I*b^2*f*g^3 - b^2*g^4))*log((( I*b*f^2 + I*b*g^2)*sqrt(-e^(2*a - 2*I*d)/(b^2*f^4 + 2*I*b^2*f^3*g + 2*I*b^ 2*f*g^3 - b^2*g^4)) + e^(I*b*x + a))*e^(-a + I*d)) - (b*f^2 + b*g^2 - (b*f ^2 + 2*I*b*f*g - b*g^2)*e^(2*I*b*x + 2*I*d))*sqrt(-e^(2*a - 2*I*d)/(b^2*f^ 4 + 2*I*b^2*f^3*g + 2*I*b^2*f*g^3 - b^2*g^4))*log(((-I*b*f^2 - I*b*g^2)*sq rt(-e^(2*a - 2*I*d)/(b^2*f^4 + 2*I*b^2*f^3*g + 2*I*b^2*f*g^3 - b^2*g^4)) + e^(I*b*x + a))*e^(-a + I*d)) + 2*I*e^(I*b*x + a))/(b*f^2 + b*g^2 - (b*f^2 + 2*I*b*f*g - b*g^2)*e^(2*I*b*x + 2*I*d))
Time = 3.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.26 \[ \int \frac {e^{a+i b x}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\operatorname {RootSum} {\left (z^{2} \left (b^{2} f^{4} e^{2 i d} + 2 i b^{2} f^{3} g e^{2 i d} + 2 i b^{2} f g^{3} e^{2 i d} - b^{2} g^{4} e^{2 i d}\right ) + e^{2 a}, \left ( i \mapsto i \log {\left (\left (i i b f^{2} + i i b g^{2}\right ) e^{- a} + e^{i b x} \right )} \right )\right )} - \frac {2 i e^{a} e^{i b x}}{- b f^{2} - b g^{2} + \left (b f^{2} e^{2 i d} + 2 i b f g e^{2 i d} - b g^{2} e^{2 i d}\right ) e^{2 i b x}} \] Input:
integrate(exp(a+I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))**2,x)
Output:
RootSum(_z**2*(b**2*f**4*exp(2*I*d) + 2*I*b**2*f**3*g*exp(2*I*d) + 2*I*b** 2*f*g**3*exp(2*I*d) - b**2*g**4*exp(2*I*d)) + exp(2*a), Lambda(_i, _i*log( (_i*I*b*f**2 + _i*I*b*g**2)*exp(-a) + exp(I*b*x)))) - 2*I*exp(a)*exp(I*b*x )/(-b*f**2 - b*g**2 + (b*f**2*exp(2*I*d) + 2*I*b*f*g*exp(2*I*d) - b*g**2*e xp(2*I*d))*exp(2*I*b*x))
Timed out. \[ \int \frac {e^{a+i b x}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\text {Timed out} \] Input:
integrate(exp(a+I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm="maxima" )
Output:
Timed out
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (96) = 192\).
Time = 0.21 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.61 \[ \int \frac {e^{a+i b x}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=-2 \, {\left (\frac {\arctan \left (\frac {f e^{\left (i \, b x + 2 i \, d\right )} + i \, g e^{\left (i \, b x + 2 i \, d\right )}}{\sqrt {-f^{2} e^{\left (2 i \, d\right )} - g^{2} e^{\left (2 i \, d\right )}}}\right )}{{\left (i \, b f e^{\left (2 i \, d\right )} - b g e^{\left (2 i \, d\right )}\right )} \sqrt {-f^{2} e^{\left (2 i \, d\right )} - g^{2} e^{\left (2 i \, d\right )}}} - \frac {2 \, e^{\left (i \, b x\right )}}{-2 \, {\left (-i \, b f e^{\left (2 i \, d\right )} + b g e^{\left (2 i \, d\right )}\right )} {\left (f e^{\left (2 i \, b x + 2 i \, d\right )} + i \, g e^{\left (2 i \, b x + 2 i \, d\right )} - f + i \, g\right )}}\right )} e^{\left (a + 2 i \, d\right )} - \frac {4 i \, e^{\left (i \, b x + a\right )}}{{\left (f^{2} e^{\left (2 i \, b x + 2 i \, d\right )} + 2 i \, f g e^{\left (2 i \, b x + 2 i \, d\right )} - g^{2} e^{\left (2 i \, b x + 2 i \, d\right )} - f^{2} - g^{2}\right )} b} \] Input:
integrate(exp(a+I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x, algorithm="giac")
Output:
-2*(arctan((f*e^(I*b*x + 2*I*d) + I*g*e^(I*b*x + 2*I*d))/sqrt(-f^2*e^(2*I* d) - g^2*e^(2*I*d)))/((I*b*f*e^(2*I*d) - b*g*e^(2*I*d))*sqrt(-f^2*e^(2*I*d ) - g^2*e^(2*I*d))) - 2*e^(I*b*x)/((2*I*b*f*e^(2*I*d) - 2*b*g*e^(2*I*d))*( f*e^(2*I*b*x + 2*I*d) + I*g*e^(2*I*b*x + 2*I*d) - f + I*g)))*e^(a + 2*I*d) - 4*I*e^(I*b*x + a)/((f^2*e^(2*I*b*x + 2*I*d) + 2*I*f*g*e^(2*I*b*x + 2*I* d) - g^2*e^(2*I*b*x + 2*I*d) - f^2 - g^2)*b)
Timed out. \[ \int \frac {e^{a+i b x}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\int \frac {{\mathrm {e}}^{a+b\,x\,1{}\mathrm {i}}}{{\left (g\,\cos \left (d+b\,x\right )+f\,\sin \left (d+b\,x\right )\right )}^2} \,d x \] Input:
int(exp(a + b*x*1i)/(g*cos(d + b*x) + f*sin(d + b*x))^2,x)
Output:
int(exp(a + b*x*1i)/(g*cos(d + b*x) + f*sin(d + b*x))^2, x)
\[ \int \frac {e^{a+i b x}}{(g \cos (d+b x)+f \sin (d+b x))^2} \, dx=\int \frac {{\mathrm e}^{i b x +a}}{\left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}}d x \] Input:
int(exp(a+I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x)
Output:
int(exp(a+I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^2,x)