Integrand size = 32, antiderivative size = 79 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x)) \, dx=\frac {3 e^{\frac {5}{3} (a-i d)+\frac {2}{3} i (d+b x)} (f-i g)}{4 b}-\frac {3 e^{\frac {5}{3} (a-i d)+\frac {8}{3} i (d+b x)} (f+i g)}{16 b} \] Output:
3/4*exp(5/3*a-5/3*I*d+2/3*I*(b*x+d))*(f-I*g)/b-3/16*exp(5/3*a-5/3*I*d+8/3* I*(b*x+d))*(f+I*g)/b
Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x)) \, dx=-\frac {3 e^{\frac {5 a}{3}+\frac {2 i b x}{3}} \left (\left (\left (-4+e^{2 i b x}\right ) f+i \left (4+e^{2 i b x}\right ) g\right ) \cos (d)+\left (i \left (4+e^{2 i b x}\right ) f-\left (-4+e^{2 i b x}\right ) g\right ) \sin (d)\right )}{16 b} \] Input:
Integrate[E^((5*(a + I*b*x))/3)*(g*Cos[d + b*x] + f*Sin[d + b*x]),x]
Output:
(-3*E^((5*a)/3 + ((2*I)/3)*b*x)*(((-4 + E^((2*I)*b*x))*f + I*(4 + E^((2*I) *b*x))*g)*Cos[d] + (I*(4 + E^((2*I)*b*x))*f - (-4 + E^((2*I)*b*x))*g)*Sin[ d]))/(16*b)
Time = 0.42 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.58, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, 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.
\(\displaystyle \int e^{\frac {5}{3} (a+i b x)} (f \sin (b x+d)+g \cos (b x+d)) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int e^{\frac {5 a}{3}+\frac {5 i b x}{3}} (f \sin (b x+d)+g \cos (b x+d))dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (f e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin (b x+d)+g e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos (b x+d)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {15 i f e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin (b x+d)}{16 b}+\frac {9 f e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos (b x+d)}{16 b}-\frac {9 g e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin (b x+d)}{16 b}-\frac {15 i g e^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos (b x+d)}{16 b}\) |
Input:
Int[E^((5*(a + I*b*x))/3)*(g*Cos[d + b*x] + f*Sin[d + b*x]),x]
Output:
(9*E^((5*a)/3 + ((5*I)/3)*b*x)*f*Cos[d + b*x])/(16*b) - (((15*I)/16)*E^((5 *a)/3 + ((5*I)/3)*b*x)*g*Cos[d + b*x])/b - (((15*I)/16)*E^((5*a)/3 + ((5*I )/3)*b*x)*f*Sin[d + b*x])/b - (9*E^((5*a)/3 + ((5*I)/3)*b*x)*g*Sin[d + b*x ])/(16*b)
Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(\frac {\left (\left (15 i g -9 f \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}+\left (-30 i f -18 g \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )-15 i g +9 f \right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{16 b \left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )}\) | \(77\) |
parts | \(f \left (\frac {9 \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos \left (b x +d \right )}{16 b}-\frac {15 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin \left (b x +d \right )}{16 b}\right )+g \left (-\frac {15 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \cos \left (b x +d \right )}{16 b}-\frac {9 \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \sin \left (b x +d \right )}{16 b}\right )\) | \(94\) |
orering | \(-\frac {15 i {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )}{8 b}+\frac {\frac {15 i b \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )}{16}+\frac {9 \,{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \left (-g b \sin \left (b x +d \right )+f b \cos \left (b x +d \right )\right )}{16}}{b^{2}}\) | \(103\) |
norman | \(\frac {\frac {3 \left (-5 i g +3 f \right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{16 b}-\frac {3 \left (-5 i g +3 f \right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{16 b}-\frac {3 \left (5 i f +3 g \right ) {\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{8 b}}{1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}\) | \(107\) |
default | \({\mathrm e}^{\frac {5 a}{3}} \cos \left (d \right ) g \left (-\frac {15 i {\mathrm e}^{\frac {5 i b x}{3}} \cos \left (b x \right )}{16 b}-\frac {9 \,{\mathrm e}^{\frac {5 i b x}{3}} \sin \left (b x \right )}{16 b}\right )-{\mathrm e}^{\frac {5 a}{3}} \sin \left (d \right ) g \left (\frac {9 \,{\mathrm e}^{\frac {5 i b x}{3}} \cos \left (b x \right )}{16 b}-\frac {15 i {\mathrm e}^{\frac {5 i b x}{3}} \sin \left (b x \right )}{16 b}\right )+{\mathrm e}^{\frac {5 a}{3}} \cos \left (d \right ) f \left (\frac {9 \,{\mathrm e}^{\frac {5 i b x}{3}} \cos \left (b x \right )}{16 b}-\frac {15 i {\mathrm e}^{\frac {5 i b x}{3}} \sin \left (b x \right )}{16 b}\right )+{\mathrm e}^{\frac {5 a}{3}} \sin \left (d \right ) f \left (-\frac {15 i {\mathrm e}^{\frac {5 i b x}{3}} \cos \left (b x \right )}{16 b}-\frac {9 \,{\mathrm e}^{\frac {5 i b x}{3}} \sin \left (b x \right )}{16 b}\right )\) | \(163\) |
Input:
int(exp(5/3*a+5/3*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d)),x,method=_RETURNVERBO SE)
Output:
1/16*((15*I*g-9*f)*tan(1/2*b*x+1/2*d)^2+(-30*I*f-18*g)*tan(1/2*b*x+1/2*d)- 15*I*g+9*f)*exp(5/3*a+5/3*I*b*x)/b/(1+tan(1/2*b*x+1/2*d)^2)
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.54 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x)) \, dx=-\frac {3 \, {\left ({\left (f + i \, g\right )} e^{\left (\frac {8}{3} i \, b x + \frac {5}{3} \, a + i \, d\right )} - 4 \, {\left (f - i \, g\right )} e^{\left (\frac {2}{3} i \, b x + \frac {5}{3} \, a - i \, d\right )}\right )}}{16 \, b} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d)),x, algorithm="f ricas")
Output:
-3/16*((f + I*g)*e^(8/3*I*b*x + 5/3*a + I*d) - 4*(f - I*g)*e^(2/3*I*b*x + 5/3*a - I*d))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (66) = 132\).
Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.05 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x)) \, dx=\begin {cases} \frac {\left (\left (48 b f e^{\frac {5 a}{3}} - 48 i b g e^{\frac {5 a}{3}}\right ) e^{\frac {2 i b x}{3}} + \left (- 12 b f e^{\frac {5 a}{3}} e^{2 i d} - 12 i b g e^{\frac {5 a}{3}} e^{2 i d}\right ) e^{\frac {8 i b x}{3}}\right ) e^{- i d}}{64 b^{2}} & \text {for}\: b^{2} e^{i d} \neq 0 \\\frac {x \left (- i f e^{\frac {5 a}{3}} e^{2 i d} + i f e^{\frac {5 a}{3}} + g e^{\frac {5 a}{3}} e^{2 i d} + g e^{\frac {5 a}{3}}\right ) e^{- i d}}{2} & \text {otherwise} \end {cases} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d)),x)
Output:
Piecewise((((48*b*f*exp(5*a/3) - 48*I*b*g*exp(5*a/3))*exp(2*I*b*x/3) + (-1 2*b*f*exp(5*a/3)*exp(2*I*d) - 12*I*b*g*exp(5*a/3)*exp(2*I*d))*exp(8*I*b*x/ 3))*exp(-I*d)/(64*b**2), Ne(b**2*exp(I*d), 0)), (x*(-I*f*exp(5*a/3)*exp(2* I*d) + I*f*exp(5*a/3) + g*exp(5*a/3)*exp(2*I*d) + g*exp(5*a/3))*exp(-I*d)/ 2, True))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.87 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x)) \, dx=\frac {3 \, {\left (3 \, b \cos \left (b x + d\right ) - 5 i \, b \sin \left (b x + d\right )\right )} f e^{\left (\frac {5}{3} i \, b x + \frac {5}{3} \, a\right )}}{16 \, b^{2}} + \frac {3 \, {\left (-5 i \, b \cos \left (b x + d\right ) - 3 \, b \sin \left (b x + d\right )\right )} g e^{\left (\frac {5}{3} i \, b x + \frac {5}{3} \, a\right )}}{16 \, b^{2}} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d)),x, algorithm="m axima")
Output:
3/16*(3*b*cos(b*x + d) - 5*I*b*sin(b*x + d))*f*e^(5/3*I*b*x + 5/3*a)/b^2 + 3/16*(-5*I*b*cos(b*x + d) - 3*b*sin(b*x + d))*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. 161 vs. \(2 (45) = 90\).
Time = 0.12 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.04 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x)) \, dx=-\frac {3 \, {\left ({\left ({\left (f + i \, g\right )} e^{\left (\frac {8}{3} i \, b x + i \, d\right )} - {\left (f + i \, g\right )} e^{\left (-\frac {8}{3} i \, b x - i \, d\right )}\right )} e^{\left (\frac {5}{3} \, a\right )} - i \, {\left ({\left (i \, f - g\right )} e^{\left (\frac {8}{3} i \, b x + i \, d\right )} - {\left (-i \, f + g\right )} e^{\left (-\frac {8}{3} i \, b x - i \, d\right )}\right )} e^{\left (\frac {5}{3} \, a\right )} - 4 \, {\left ({\left (f - i \, g\right )} e^{\left (\frac {2}{3} i \, b x - i \, d\right )} - {\left (f - i \, g\right )} e^{\left (-\frac {2}{3} i \, b x + i \, d\right )}\right )} e^{\left (\frac {5}{3} \, a\right )} + 4 i \, {\left ({\left (i \, f + g\right )} e^{\left (\frac {2}{3} i \, b x - i \, d\right )} - {\left (-i \, f - g\right )} e^{\left (-\frac {2}{3} i \, b x + i \, d\right )}\right )} e^{\left (\frac {5}{3} \, a\right )}\right )}}{32 \, b} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d)),x, algorithm="g iac")
Output:
-3/32*(((f + I*g)*e^(8/3*I*b*x + I*d) - (f + I*g)*e^(-8/3*I*b*x - I*d))*e^ (5/3*a) - I*((I*f - g)*e^(8/3*I*b*x + I*d) - (-I*f + g)*e^(-8/3*I*b*x - I* d))*e^(5/3*a) - 4*((f - I*g)*e^(2/3*I*b*x - I*d) - (f - I*g)*e^(-2/3*I*b*x + I*d))*e^(5/3*a) + 4*I*((I*f + g)*e^(2/3*I*b*x - I*d) - (-I*f - g)*e^(-2 /3*I*b*x + I*d))*e^(5/3*a))/b
Time = 14.89 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.76 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x)) \, dx=\frac {3\,\cos \left (d+b\,x\right )\,{\mathrm {e}}^{\frac {5\,a}{3}+\frac {b\,x\,5{}\mathrm {i}}{3}}\,\left (3\,f-g\,5{}\mathrm {i}\right )}{16\,b}-\frac {{\mathrm {e}}^{\frac {5\,a}{3}+\frac {b\,x\,5{}\mathrm {i}}{3}}\,\sin \left (d+b\,x\right )\,\left (5\,f-g\,3{}\mathrm {i}\right )\,3{}\mathrm {i}}{16\,b} \] Input:
int(exp((5*a)/3 + (b*x*5i)/3)*(g*cos(d + b*x) + f*sin(d + b*x)),x)
Output:
(3*cos(d + b*x)*exp((5*a)/3 + (b*x*5i)/3)*(3*f - g*5i))/(16*b) - (exp((5*a )/3 + (b*x*5i)/3)*sin(d + b*x)*(5*f - g*3i)*3i)/(16*b)
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int e^{\frac {5}{3} (a+i b x)} (g \cos (d+b x)+f \sin (d+b x)) \, dx=\frac {3 e^{\frac {5 b i x}{3}+\frac {5 a}{3}} \left (3 \cos \left (b x +d \right ) f -5 \cos \left (b x +d \right ) g i -5 \sin \left (b x +d \right ) f i -3 \sin \left (b x +d \right ) g \right )}{16 b} \] Input:
int(exp(5/3*a+5/3*I*b*x)*(g*cos(b*x+d)+f*sin(b*x+d)),x)
Output:
(3*e**((5*a + 5*b*i*x)/3)*(3*cos(b*x + d)*f - 5*cos(b*x + d)*g*i - 5*sin(b *x + d)*f*i - 3*sin(b*x + d)*g))/(16*b)