Integrand size = 34, antiderivative size = 128 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=-\frac {3 i e^{\frac {5}{3} (a-i d)-\frac {1}{3} i (d+b x)} (f-i g)^2}{4 b}+\frac {3 i e^{\frac {5}{3} (a-i d)+\frac {11}{3} i (d+b x)} (f+i g)^2}{44 b}-\frac {3 i e^{\frac {5}{3} (a-i d)+\frac {5}{3} i (d+b x)} \left (f^2+g^2\right )}{10 b} \] Output:
-3/4*I*exp(5/3*a-5/3*I*d-1/3*I*(b*x+d))*(f-I*g)^2/b+3/44*I*exp(5/3*a-5/3*I *d+11/3*I*(b*x+d))*(f+I*g)^2/b-3/10*I*exp(5/3*a-5/3*I*d+5/3*I*(b*x+d))*(f^ 2+g^2)/b
Time = 0.99 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.89 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\frac {3 i e^{-\frac {1}{3} i (5 i a+6 d+b x)} \left (\left (-55-22 e^{2 i (d+b x)}+5 e^{4 i (d+b x)}\right ) f^2+10 i \left (11+e^{4 i (d+b x)}\right ) f g-\left (-55+22 e^{2 i (d+b x)}+5 e^{4 i (d+b x)}\right ) g^2\right )}{220 b} \] Input:
Integrate[E^((5*(a + I*b*x))/3)*(g*Cos[d + b*x] + f*Sin[d + b*x])^2,x]
Output:
(((3*I)/220)*((-55 - 22*E^((2*I)*(d + b*x)) + 5*E^((4*I)*(d + b*x)))*f^2 + (10*I)*(11 + E^((4*I)*(d + b*x)))*f*g - (-55 + 22*E^((2*I)*(d + b*x)) + 5 *E^((4*I)*(d + b*x)))*g^2))/(b*E^((I/3)*((5*I)*a + 6*d + b*x)))
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(275\) vs. \(2(128)=256\).
Time = 0.63 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.15, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, 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.
| \(\displaystyle \int e^{\frac {5}{3} (a+i b x)} (f \sin (b x+d)+g \cos (b x+d))^2 \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int e^{\frac {5 a}{3}+\frac {5 i b x}{3}} (f \sin (b x+d)+g \cos (b x+d))^2dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (f^2 e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin ^2(b x+d)+f g e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin (2 b x+2 d)+g^2 e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos ^2(b x+d)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {15 i f^2 e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin ^2(b x+d)}{11 b}-\frac {18 f^2 e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin (b x+d) \cos (b x+d)}{11 b}+\frac {15 i f g e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin (2 b x+2 d)}{11 b}-\frac {18 f g e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos (2 b x+2 d)}{11 b}+\frac {15 i g^2 e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos ^2(b x+d)}{11 b}+\frac {18 g^2 e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin (b x+d) \cos (b x+d)}{11 b}-\frac {54 i f^2 e^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{55 b}-\frac {54 i g^2 e^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{55 b}\) |
Input:
Int[E^((5*(a + I*b*x))/3)*(g*Cos[d + b*x] + f*Sin[d + b*x])^2,x]
Output:
(((-54*I)/55)*E^((5*a)/3 + ((5*I)/3)*b*x)*f^2)/b - (((54*I)/55)*E^((5*a)/3 + ((5*I)/3)*b*x)*g^2)/b + (((15*I)/11)*E^((5*a)/3 + ((5*I)/3)*b*x)*g^2*Co s[d + b*x]^2)/b - (18*E^((5*a)/3 + ((5*I)/3)*b*x)*f*g*Cos[2*d + 2*b*x])/(1 1*b) - (18*E^((5*a)/3 + ((5*I)/3)*b*x)*f^2*Cos[d + b*x]*Sin[d + b*x])/(11* b) + (18*E^((5*a)/3 + ((5*I)/3)*b*x)*g^2*Cos[d + b*x]*Sin[d + b*x])/(11*b) + (((15*I)/11)*E^((5*a)/3 + ((5*I)/3)*b*x)*f^2*Sin[d + b*x]^2)/b + (((15* I)/11)*E^((5*a)/3 + ((5*I)/3)*b*x)*f*g*Sin[2*d + 2*b*x])/b
Time = 0.89 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.89
| method | result | size |
| parallelrisch | \(-\frac {3 \left (25 i f^{2} \cos \left (2 b x +2 d \right )-25 i g^{2} \cos \left (2 b x +2 d \right )-50 i f g \sin \left (2 b x +2 d \right )+11 i f^{2}+11 i g^{2}+60 f g \cos \left (2 b x +2 d \right )+30 f^{2} \sin \left (2 b x +2 d \right )-30 g^{2} \sin \left (2 b x +2 d \right )\right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{110 b}\) | \(114\) |
| parts | \(g^{2} \left (-\frac {3 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{10 b}+\frac {15 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos \left (2 b x +2 d \right )}{22 b}+\frac {9 \sin \left (2 b x +2 d \right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{11 b}\right )+f^{2} \left (-\frac {3 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{10 b}-\frac {15 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos \left (2 b x +2 d \right )}{22 b}-\frac {9 \sin \left (2 b x +2 d \right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{11 b}\right )+2 f g \left (-\frac {9 \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos \left (2 b x +2 d \right )}{11 b}+\frac {15 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin \left (2 b x +2 d \right )}{22 b}\right )\) | \(196\) |
| default | \(-\frac {3 i g^{2} {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{10 b}-\frac {3 i f^{2} {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{10 b}-\frac {f^{2} \left (\frac {15 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos \left (2 b x +2 d \right )}{11 b}+\frac {18 \sin \left (2 b x +2 d \right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{11 b}\right )}{2}+\frac {g^{2} \left (\frac {15 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos \left (2 b x +2 d \right )}{11 b}+\frac {18 \sin \left (2 b x +2 d \right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{11 b}\right )}{2}+f g \left (-\frac {18 \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos \left (2 b x +2 d \right )}{11 b}+\frac {15 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin \left (2 b x +2 d \right )}{11 b}\right )\) | \(203\) |
| norman | \(\frac {-\frac {12 \left (-5 i f g +3 f^{2}-3 g^{2}\right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{11 b}+\frac {12 \left (-5 i f g +3 f^{2}-3 g^{2}\right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{11 b}-\frac {3 i \left (-30 i f g +18 f^{2}-7 g^{2}\right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{55 b}+\frac {6 i \left (-90 i f g +32 f^{2}-43 g^{2}\right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{55 b}-\frac {3 i \left (-30 i f g +18 f^{2}-7 g^{2}\right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{55 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\) | \(218\) |
| orering | \(\frac {117 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}}{55 b}-\frac {27 \left (\frac {5 i b \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}}{3}+2 \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right ) \left (-g b \sin \left (b x +d \right )+f b \cos \left (b x +d \right )\right )\right )}{11 b^{2}}-\frac {27 i \left (-\frac {25 b^{2} {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{2}}{9}+\frac {20 i b \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right ) \left (-g b \sin \left (b x +d \right )+f b \cos \left (b x +d \right )\right )}{3}+2 \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (-g b \sin \left (b x +d \right )+f b \cos \left (b x +d \right )\right )^{2}+2 \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right ) \left (-g \,b^{2} \cos \left (b x +d \right )-f \,b^{2} \sin \left (b x +d \right )\right )\right )}{55 b^{3}}\) | \(305\) |
Input:
int(exp(5/3*a+5/3*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d))^2,x,method=_RETURNVER BOSE)
Output:
-3/110*(25*I*f^2*cos(2*b*x+2*d)-25*I*g^2*cos(2*b*x+2*d)-50*I*f*g*sin(2*b*x +2*d)+11*I*f^2+11*I*g^2+60*f*g*cos(2*b*x+2*d)+30*f^2*sin(2*b*x+2*d)-30*g^2 *sin(2*b*x+2*d))*exp(5/3*a+5/3*I*b*x)/b
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.73 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=-\frac {3 \, {\left (5 \, {\left (-i \, f^{2} + 2 \, f g + i \, g^{2}\right )} e^{\left (4 i \, b x + \frac {5}{3} \, a + \frac {7}{3} i \, d\right )} + 22 \, {\left (i \, f^{2} + i \, g^{2}\right )} e^{\left (2 i \, b x + \frac {5}{3} \, a + \frac {1}{3} i \, d\right )} + 55 \, {\left (i \, f^{2} + 2 \, f g - i \, g^{2}\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} i \, d\right )}\right )} e^{\left (-\frac {1}{3} i \, b x - \frac {1}{3} i \, d\right )}}{220 \, b} \] 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:
-3/220*(5*(-I*f^2 + 2*f*g + I*g^2)*e^(4*I*b*x + 5/3*a + 7/3*I*d) + 22*(I*f ^2 + I*g^2)*e^(2*I*b*x + 5/3*a + 1/3*I*d) + 55*(I*f^2 + 2*f*g - I*g^2)*e^( 5/3*a - 5/3*I*d))*e^(-1/3*I*b*x - 1/3*I*d)/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (109) = 218\).
Time = 0.41 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.75 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\begin {cases} \frac {\left (\left (- 528 i b^{2} f^{2} e^{\frac {5 a}{3}} e^{2 i d} - 528 i b^{2} g^{2} e^{\frac {5 a}{3}} e^{2 i d}\right ) e^{\frac {5 i b x}{3}} + \left (- 1320 i b^{2} f^{2} e^{\frac {5 a}{3}} - 2640 b^{2} f g e^{\frac {5 a}{3}} + 1320 i b^{2} g^{2} e^{\frac {5 a}{3}}\right ) e^{- \frac {i b x}{3}} + \left (120 i b^{2} f^{2} e^{\frac {5 a}{3}} e^{4 i d} - 240 b^{2} f g e^{\frac {5 a}{3}} e^{4 i d} - 120 i b^{2} g^{2} e^{\frac {5 a}{3}} e^{4 i d}\right ) e^{\frac {11 i b x}{3}}\right ) e^{- 2 i d}}{1760 b^{3}} & \text {for}\: b^{3} e^{2 i d} \neq 0 \\\frac {x \left (- f^{2} e^{\frac {5 a}{3}} e^{4 i d} + 2 f^{2} e^{\frac {5 a}{3}} e^{2 i d} - f^{2} e^{\frac {5 a}{3}} - 2 i f g e^{\frac {5 a}{3}} e^{4 i d} + 2 i f g e^{\frac {5 a}{3}} + g^{2} e^{\frac {5 a}{3}} e^{4 i d} + 2 g^{2} e^{\frac {5 a}{3}} e^{2 i d} + g^{2} e^{\frac {5 a}{3}}\right ) e^{- 2 i d}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d))**2,x)
Output:
Piecewise((((-528*I*b**2*f**2*exp(5*a/3)*exp(2*I*d) - 528*I*b**2*g**2*exp( 5*a/3)*exp(2*I*d))*exp(5*I*b*x/3) + (-1320*I*b**2*f**2*exp(5*a/3) - 2640*b **2*f*g*exp(5*a/3) + 1320*I*b**2*g**2*exp(5*a/3))*exp(-I*b*x/3) + (120*I*b **2*f**2*exp(5*a/3)*exp(4*I*d) - 240*b**2*f*g*exp(5*a/3)*exp(4*I*d) - 120* I*b**2*g**2*exp(5*a/3)*exp(4*I*d))*exp(11*I*b*x/3))*exp(-2*I*d)/(1760*b**3 ), Ne(b**3*exp(2*I*d), 0)), (x*(-f**2*exp(5*a/3)*exp(4*I*d) + 2*f**2*exp(5 *a/3)*exp(2*I*d) - f**2*exp(5*a/3) - 2*I*f*g*exp(5*a/3)*exp(4*I*d) + 2*I*f *g*exp(5*a/3) + g**2*exp(5*a/3)*exp(4*I*d) + 2*g**2*exp(5*a/3)*exp(2*I*d) + g**2*exp(5*a/3))*exp(-2*I*d)/4, True))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (70) = 140\).
Time = 0.05 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.75 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\frac {3 \, {\left (-22 i \, \cos \left (\frac {5}{3} \, b x\right ) e^{\left (\frac {5}{3} \, a\right )} + 5 i \, \cos \left (\frac {11}{3} \, b x + 2 \, d\right ) e^{\left (\frac {5}{3} \, a\right )} - 55 i \, \cos \left (\frac {1}{3} \, b x + 2 \, d\right ) e^{\left (\frac {5}{3} \, a\right )} + 22 \, e^{\left (\frac {5}{3} \, a\right )} \sin \left (\frac {5}{3} \, b x\right ) - 5 \, e^{\left (\frac {5}{3} \, a\right )} \sin \left (\frac {11}{3} \, b x + 2 \, d\right ) - 55 \, e^{\left (\frac {5}{3} \, a\right )} \sin \left (\frac {1}{3} \, b x + 2 \, d\right )\right )} f^{2}}{220 \, b} - \frac {3 \, {\left (22 i \, \cos \left (\frac {5}{3} \, b x\right ) e^{\left (\frac {5}{3} \, a\right )} + 5 i \, \cos \left (\frac {11}{3} \, b x + 2 \, d\right ) e^{\left (\frac {5}{3} \, a\right )} - 55 i \, \cos \left (\frac {1}{3} \, b x + 2 \, d\right ) e^{\left (\frac {5}{3} \, a\right )} - 22 \, e^{\left (\frac {5}{3} \, a\right )} \sin \left (\frac {5}{3} \, b x\right ) - 5 \, e^{\left (\frac {5}{3} \, a\right )} \sin \left (\frac {11}{3} \, b x + 2 \, d\right ) - 55 \, e^{\left (\frac {5}{3} \, a\right )} \sin \left (\frac {1}{3} \, b x + 2 \, d\right )\right )} g^{2}}{220 \, b} - \frac {3 \, {\left (6 \, b \cos \left (2 \, b x + 2 \, d\right ) - 5 i \, b \sin \left (2 \, b x + 2 \, d\right )\right )} f g e^{\left (\frac {5}{3} i \, b x + \frac {5}{3} \, a\right )}}{11 \, b^{2}} \] 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:
3/220*(-22*I*cos(5/3*b*x)*e^(5/3*a) + 5*I*cos(11/3*b*x + 2*d)*e^(5/3*a) - 55*I*cos(1/3*b*x + 2*d)*e^(5/3*a) + 22*e^(5/3*a)*sin(5/3*b*x) - 5*e^(5/3*a )*sin(11/3*b*x + 2*d) - 55*e^(5/3*a)*sin(1/3*b*x + 2*d))*f^2/b - 3/220*(22 *I*cos(5/3*b*x)*e^(5/3*a) + 5*I*cos(11/3*b*x + 2*d)*e^(5/3*a) - 55*I*cos(1 /3*b*x + 2*d)*e^(5/3*a) - 22*e^(5/3*a)*sin(5/3*b*x) - 5*e^(5/3*a)*sin(11/3 *b*x + 2*d) - 55*e^(5/3*a)*sin(1/3*b*x + 2*d))*g^2/b - 3/11*(6*b*cos(2*b*x + 2*d) - 5*I*b*sin(2*b*x + 2*d))*f*g*e^(5/3*I*b*x + 5/3*a)/b^2
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (70) = 140\).
Time = 0.20 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.16 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\frac {3 \, {\left (44 \, {\left (f^{2} + g^{2}\right )} e^{\left (\frac {5}{3} \, a\right )} \sin \left (\frac {5}{3} \, b x\right ) + 5 i \, {\left ({\left (f^{2} + 2 i \, f g - g^{2}\right )} e^{\left (\frac {11}{3} i \, b x + 2 i \, d\right )} + {\left (f^{2} + 2 i \, f g - g^{2}\right )} e^{\left (-\frac {11}{3} i \, b x - 2 i \, d\right )}\right )} e^{\left (\frac {5}{3} \, a\right )} - 5 \, {\left ({\left (-i \, f^{2} + 2 \, f g + i \, g^{2}\right )} e^{\left (\frac {11}{3} i \, b x + 2 i \, d\right )} + {\left (i \, f^{2} - 2 \, f g - i \, g^{2}\right )} e^{\left (-\frac {11}{3} i \, b x - 2 i \, d\right )}\right )} e^{\left (\frac {5}{3} \, a\right )} - 55 i \, {\left ({\left (f^{2} - 2 i \, f g - g^{2}\right )} e^{\left (\frac {1}{3} i \, b x + 2 i \, d\right )} + {\left (f^{2} - 2 i \, f g - g^{2}\right )} e^{\left (-\frac {1}{3} i \, b x - 2 i \, d\right )}\right )} e^{\left (\frac {5}{3} \, a\right )} - 55 \, {\left ({\left (-i \, f^{2} - 2 \, f g + i \, g^{2}\right )} e^{\left (\frac {1}{3} i \, b x + 2 i \, d\right )} + {\left (i \, f^{2} + 2 \, f g - i \, g^{2}\right )} e^{\left (-\frac {1}{3} i \, b x - 2 i \, d\right )}\right )} e^{\left (\frac {5}{3} \, a\right )} - 22 i \, {\left ({\left (f^{2} + g^{2}\right )} e^{\left (\frac {5}{3} i \, b x\right )} + {\left (f^{2} + g^{2}\right )} e^{\left (-\frac {5}{3} i \, b x\right )}\right )} e^{\left (\frac {5}{3} \, a\right )}\right )}}{440 \, b} \] 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:
3/440*(44*(f^2 + g^2)*e^(5/3*a)*sin(5/3*b*x) + 5*I*((f^2 + 2*I*f*g - g^2)* e^(11/3*I*b*x + 2*I*d) + (f^2 + 2*I*f*g - g^2)*e^(-11/3*I*b*x - 2*I*d))*e^ (5/3*a) - 5*((-I*f^2 + 2*f*g + I*g^2)*e^(11/3*I*b*x + 2*I*d) + (I*f^2 - 2* f*g - I*g^2)*e^(-11/3*I*b*x - 2*I*d))*e^(5/3*a) - 55*I*((f^2 - 2*I*f*g - g ^2)*e^(1/3*I*b*x + 2*I*d) + (f^2 - 2*I*f*g - g^2)*e^(-1/3*I*b*x - 2*I*d))* e^(5/3*a) - 55*((-I*f^2 - 2*f*g + I*g^2)*e^(1/3*I*b*x + 2*I*d) + (I*f^2 + 2*f*g - I*g^2)*e^(-1/3*I*b*x - 2*I*d))*e^(5/3*a) - 22*I*((f^2 + g^2)*e^(5/ 3*I*b*x) + (f^2 + g^2)*e^(-5/3*I*b*x))*e^(5/3*a))/b
Time = 16.75 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.97 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\frac {{\cos \left (d+b\,x\right )}^2\,{\mathrm {e}}^{\frac {5\,a}{3}+\frac {b\,x\,5{}\mathrm {i}}{3}}\,\left (-18\,f^2+f\,g\,30{}\mathrm {i}+7\,g^2\right )\,3{}\mathrm {i}}{55\,b}-\frac {{\mathrm {e}}^{\frac {5\,a}{3}+\frac {b\,x\,5{}\mathrm {i}}{3}}\,{\sin \left (d+b\,x\right )}^2\,\left (-21\,f^2+f\,g\,90{}\mathrm {i}+54\,g^2\right )\,1{}\mathrm {i}}{55\,b}+\frac {6\,\cos \left (d+b\,x\right )\,{\mathrm {e}}^{\frac {5\,a}{3}+\frac {b\,x\,5{}\mathrm {i}}{3}}\,\sin \left (d+b\,x\right )\,\left (-3\,f^2+f\,g\,5{}\mathrm {i}+3\,g^2\right )}{11\,b} \] Input:
int(exp((5*a)/3 + (b*x*5i)/3)*(g*cos(d + b*x) + f*sin(d + b*x))^2,x)
Output:
(cos(d + b*x)^2*exp((5*a)/3 + (b*x*5i)/3)*(f*g*30i - 18*f^2 + 7*g^2)*3i)/( 55*b) - (exp((5*a)/3 + (b*x*5i)/3)*sin(d + b*x)^2*(f*g*90i - 21*f^2 + 54*g ^2)*1i)/(55*b) + (6*cos(d + b*x)*exp((5*a)/3 + (b*x*5i)/3)*sin(d + b*x)*(f *g*5i - 3*f^2 + 3*g^2))/(11*b)
Time = 0.16 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.16 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x))^2 \, dx=\frac {3 e^{\frac {5 b i x}{3}+\frac {5 a}{3}} \left (-18 \cos \left (b x +d \right )^{2} f^{2} i -30 \cos \left (b x +d \right )^{2} f g +7 \cos \left (b x +d \right )^{2} g^{2} i -30 \cos \left (b x +d \right ) \sin \left (b x +d \right ) f^{2}+50 \cos \left (b x +d \right ) \sin \left (b x +d \right ) f g i +30 \cos \left (b x +d \right ) \sin \left (b x +d \right ) g^{2}+7 \sin \left (b x +d \right )^{2} f^{2} i +30 \sin \left (b x +d \right )^{2} f g -18 \sin \left (b x +d \right )^{2} g^{2} i \right )}{55 b} \] Input:
int(exp(5/3*a+5/3*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d))^2,x)
Output:
(3*e**((5*a + 5*b*i*x)/3)*( - 18*cos(b*x + d)**2*f**2*i - 30*cos(b*x + d)* *2*f*g + 7*cos(b*x + d)**2*g**2*i - 30*cos(b*x + d)*sin(b*x + d)*f**2 + 50 *cos(b*x + d)*sin(b*x + d)*f*g*i + 30*cos(b*x + d)*sin(b*x + d)*g**2 + 7*s in(b*x + d)**2*f**2*i + 30*sin(b*x + d)**2*f*g - 18*sin(b*x + d)**2*g**2*i ))/(55*b)