Integrand size = 29, antiderivative size = 91 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^2(d+b x) \, dx=-\frac {i e^{2 (a-i d)-2 i (d+b x)}}{32 b}-\frac {i e^{2 (a-i d)+2 i (d+b x)}}{16 b}+\frac {i e^{2 (a-i d)+6 i (d+b x)}}{96 b} \] Output:
-1/32*I*exp(2*a-2*I*d-2*I*(b*x+d))/b-1/16*I*exp(2*a-2*I*d+2*I*(b*x+d))/b+1 /96*I*exp(2*a-2*I*d+6*I*(b*x+d))/b
Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^2(d+b x) \, dx=\frac {i e^{2 (a-i (2 d+b x))} \left (-3-6 e^{4 i (d+b x)}+e^{8 i (d+b x)}\right )}{96 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Cos[d + b*x]^2*Sin[d + b*x]^2,x]
Output:
((I/96)*E^(2*(a - I*(2*d + b*x)))*(-3 - 6*E^((4*I)*(d + b*x)) + E^((8*I)*( d + b*x))))/b
Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {4972, 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.
\(\displaystyle \int e^{2 (a+i b x)} \sin ^2(b x+d) \cos ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \left (\frac {1}{8} e^{2 (a+i b x)}-\frac {1}{8} e^{2 (a+i b x)} \cos (4 b x+4 d)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{2 (a+i b x)} \sin (4 b x+4 d)}{24 b}-\frac {i e^{2 (a+i b x)} \cos (4 b x+4 d)}{48 b}-\frac {i e^{2 (a+i b x)}}{16 b}\) |
Input:
Int[E^(2*(a + I*b*x))*Cos[d + b*x]^2*Sin[d + b*x]^2,x]
Output:
((-1/16*I)*E^(2*(a + I*b*x)))/b - ((I/48)*E^(2*(a + I*b*x))*Cos[4*d + 4*b* x])/b - (E^(2*(a + I*b*x))*Sin[4*d + 4*b*x])/(24*b)
Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_ .) + (e_.)*(x_)]^(m_.), x_Symbol] :> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Time = 1.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74
method | result | size |
default | \(-\frac {i {\mathrm e}^{2 i b x +2 a}}{16 b}-\frac {i {\mathrm e}^{2 i b x +2 a} \cos \left (4 b x +4 d \right )}{48 b}-\frac {{\mathrm e}^{2 i b x +2 a} \sin \left (4 b x +4 d \right )}{24 b}\) | \(67\) |
parallelrisch | \(-\frac {8 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3} \left (i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-1\right )}{3 b \left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{4}}\) | \(67\) |
norman | \(\frac {-\frac {8 i {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{3 b}+\frac {8 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{3 b}-\frac {8 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}}{3 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{4}}\) | \(97\) |
orering | \(-\frac {i {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{2}}{6 b}-\frac {2 i b \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{2}-2 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{3} b +2 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right ) b}{4 b^{2}}-\frac {i \left (-16 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} b^{2} \sin \left (b x +d \right )^{2}-8 i b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{3}+8 i b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )+2 \,{\mathrm e}^{2 i b x +2 a} b^{2} \sin \left (b x +d \right )^{4}+2 b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{4}\right )}{24 b^{3}}\) | \(268\) |
Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
-1/16*I/b*exp(2*a+2*I*b*x)-1/48*I/b*exp(2*a+2*I*b*x)*cos(4*b*x+4*d)-1/24/b *exp(2*a+2*I*b*x)*sin(4*b*x+4*d)
Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^2(d+b x) \, dx=\frac {{\left (i \, e^{\left (8 i \, b x + 2 \, a + 6 i \, d\right )} - 6 i \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - 3 i \, e^{\left (2 \, a - 2 i \, d\right )}\right )} e^{\left (-2 i \, b x - 2 i \, d\right )}}{96 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^2,x, algorithm="fricas" )
Output:
1/96*(I*e^(8*I*b*x + 2*a + 6*I*d) - 6*I*e^(4*I*b*x + 2*a + 2*I*d) - 3*I*e^ (2*a - 2*I*d))*e^(-2*I*b*x - 2*I*d)/b
Time = 0.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.47 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^2(d+b x) \, dx=\begin {cases} \frac {\left (512 i b^{2} e^{2 a} e^{8 i d} e^{6 i b x} - 3072 i b^{2} e^{2 a} e^{4 i d} e^{2 i b x} - 1536 i b^{2} e^{2 a} e^{- 2 i b x}\right ) e^{- 4 i d}}{49152 b^{3}} & \text {for}\: b^{3} e^{4 i d} \neq 0 \\\frac {x \left (- e^{2 a} e^{8 i d} + 2 e^{2 a} e^{4 i d} - e^{2 a}\right ) e^{- 4 i d}}{16} & \text {otherwise} \end {cases} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)**2*sin(b*x+d)**2,x)
Output:
Piecewise(((512*I*b**2*exp(2*a)*exp(8*I*d)*exp(6*I*b*x) - 3072*I*b**2*exp( 2*a)*exp(4*I*d)*exp(2*I*b*x) - 1536*I*b**2*exp(2*a)*exp(-2*I*b*x))*exp(-4* I*d)/(49152*b**3), Ne(b**3*exp(4*I*d), 0)), (x*(-exp(2*a)*exp(8*I*d) + 2*e xp(2*a)*exp(4*I*d) - exp(2*a))*exp(-4*I*d)/16, True))
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^2(d+b x) \, dx=-\frac {6 i \, \cos \left (2 \, b x\right ) e^{\left (2 \, a\right )} - i \, \cos \left (6 \, b x + 4 \, d\right ) e^{\left (2 \, a\right )} + 3 i \, \cos \left (2 \, b x + 4 \, d\right ) e^{\left (2 \, a\right )} - 6 \, e^{\left (2 \, a\right )} \sin \left (2 \, b x\right ) + e^{\left (2 \, a\right )} \sin \left (6 \, b x + 4 \, d\right ) + 3 \, e^{\left (2 \, a\right )} \sin \left (2 \, b x + 4 \, d\right )}{96 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^2,x, algorithm="maxima" )
Output:
-1/96*(6*I*cos(2*b*x)*e^(2*a) - I*cos(6*b*x + 4*d)*e^(2*a) + 3*I*cos(2*b*x + 4*d)*e^(2*a) - 6*e^(2*a)*sin(2*b*x) + e^(2*a)*sin(6*b*x + 4*d) + 3*e^(2 *a)*sin(2*b*x + 4*d))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (49) = 98\).
Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.25 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^2(d+b x) \, dx=-\frac {-i \, {\left (e^{\left (6 i \, b x + 4 i \, d\right )} + e^{\left (-6 i \, b x - 4 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 3 i \, {\left (e^{\left (2 i \, b x + 4 i \, d\right )} + e^{\left (-2 i \, b x - 4 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 6 i \, {\left (e^{\left (2 i \, b x\right )} + e^{\left (-2 i \, b x\right )}\right )} e^{\left (2 \, a\right )} - 12 \, e^{\left (2 \, a\right )} \sin \left (2 \, b x\right ) + 6 \, e^{\left (2 \, a\right )} \sin \left (2 \, b x + 4 \, d\right ) - 2 \, e^{\left (2 \, a\right )} \sin \left (-6 \, b x - 4 \, d\right )}{192 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^2,x, algorithm="giac")
Output:
-1/192*(-I*(e^(6*I*b*x + 4*I*d) + e^(-6*I*b*x - 4*I*d))*e^(2*a) + 3*I*(e^( 2*I*b*x + 4*I*d) + e^(-2*I*b*x - 4*I*d))*e^(2*a) + 6*I*(e^(2*I*b*x) + e^(- 2*I*b*x))*e^(2*a) - 12*e^(2*a)*sin(2*b*x) + 6*e^(2*a)*sin(2*b*x + 4*d) - 2 *e^(2*a)*sin(-6*b*x - 4*d))/b
Time = 16.77 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^2(d+b x) \, dx=-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (2\,b\,x\right )+\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,b}-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (2\,b\,x\right )-\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (4\,d\right )-\sin \left (4\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,b}+\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (6\,b\,x\right )+\sin \left (6\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (4\,d\right )+\sin \left (4\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{96\,b} \] Input:
int(cos(d + b*x)^2*exp(2*a + b*x*2i)*sin(d + b*x)^2,x)
Output:
(exp(2*a)*(cos(6*b*x) + sin(6*b*x)*1i)*(cos(4*d) + sin(4*d)*1i)*1i)/(96*b) - (exp(2*a)*(cos(2*b*x) - sin(2*b*x)*1i)*(cos(4*d) - sin(4*d)*1i)*1i)/(32 *b) - (exp(2*a)*(cos(2*b*x) + sin(2*b*x)*1i)*1i)/(16*b)
Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.46 \[ \int e^{2 (a+i b x)} \cos ^2(d+b x) \sin ^2(d+b x) \, dx=\frac {e^{2 b i x +2 a} \sin \left (b x +d \right )^{3} \left (2 \cos \left (b x +d \right )-\sin \left (b x +d \right ) i \right )}{6 b} \] Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^2,x)
Output:
(e**(2*a + 2*b*i*x)*sin(b*x + d)**3*(2*cos(b*x + d) - sin(b*x + d)*i))/(6* b)