Integrand size = 27, antiderivative size = 121 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^2(d+b x) \, dx=-\frac {i e^{2 (a-i d)-i (d+b x)}}{8 b}-\frac {i e^{2 (a-i d)+i (d+b x)}}{8 b}-\frac {i e^{2 (a-i d)+3 i (d+b x)}}{24 b}+\frac {i e^{2 (a-i d)+5 i (d+b x)}}{40 b} \] Output:
-1/8*I*exp(2*a-2*I*d-I*(b*x+d))/b-1/8*I*exp(2*a-2*I*d+I*(b*x+d))/b-1/24*I* exp(2*a-2*I*d+3*I*(b*x+d))/b+1/40*I*exp(2*a-2*I*d+5*I*(b*x+d))/b
Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.78 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^2(d+b x) \, dx=\frac {e^{2 a-i b x} \left (3 i \left (-5+e^{6 i b x}\right ) \cos (3 d)+5 e^{2 i b x} \left (-i \left (3+e^{2 i b x}\right ) \cos (d)+\left (-3+e^{2 i b x}\right ) \sin (d)\right )-3 \left (5+e^{6 i b x}\right ) \sin (3 d)\right )}{120 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Cos[d + b*x]*Sin[d + b*x]^2,x]
Output:
(E^(2*a - I*b*x)*((3*I)*(-5 + E^((6*I)*b*x))*Cos[3*d] + 5*E^((2*I)*b*x)*(( -I)*(3 + E^((2*I)*b*x))*Cos[d] + (-3 + E^((2*I)*b*x))*Sin[d]) - 3*(5 + E^( (6*I)*b*x))*Sin[3*d]))/(120*b)
Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, 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.
\(\displaystyle \int e^{2 (a+i b x)} \sin ^2(b x+d) \cos (b x+d) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \left (\frac {1}{4} e^{2 (a+i b x)} \cos (b x+d)-\frac {1}{4} e^{2 (a+i b x)} \cos (3 b x+3 d)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{2 (a+i b x)} \sin (b x+d)}{12 b}-\frac {3 e^{2 (a+i b x)} \sin (3 b x+3 d)}{20 b}-\frac {i e^{2 (a+i b x)} \cos (b x+d)}{6 b}-\frac {i e^{2 (a+i b x)} \cos (3 b x+3 d)}{10 b}\) |
Input:
Int[E^(2*(a + I*b*x))*Cos[d + b*x]*Sin[d + b*x]^2,x]
Output:
((-1/6*I)*E^(2*(a + I*b*x))*Cos[d + b*x])/b - ((I/10)*E^(2*(a + I*b*x))*Co s[3*d + 3*b*x])/b - (E^(2*(a + I*b*x))*Sin[d + b*x])/(12*b) - (3*E^(2*(a + I*b*x))*Sin[3*d + 3*b*x])/(20*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 = 0.69 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69
method | result | size |
parallelrisch | \(\frac {8 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3} \left (i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}-5 i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )-4 \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}+5\right )}{15 b \left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{3}}\) | \(83\) |
default | \(-\frac {i {\mathrm e}^{2 i b x +2 a} \cos \left (3 b x +3 d \right )}{10 b}-\frac {3 \,{\mathrm e}^{2 i b x +2 a} \sin \left (3 b x +3 d \right )}{20 b}-\frac {i {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )}{6 b}-\frac {{\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )}{12 b}\) | \(94\) |
norman | \(\frac {\frac {8 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{3 b}-\frac {8 i {\mathrm e}^{2 i b x +2 a}}{15 b}+\frac {8 i {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{3 b}-\frac {32 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{15 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{3}}\) | \(111\) |
orering | \(-\frac {8 i {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}}{15 b}-\frac {14 \left (2 i b \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}-{\mathrm e}^{2 i b x +2 a} b \sin \left (b x +d \right )^{3}+2 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} b \sin \left (b x +d \right )\right )}{15 b^{2}}-\frac {8 i \left (-11 b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}-4 i b^{2} {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )^{3}+8 i b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )+2 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} b^{2}\right )}{15 b^{3}}+\frac {-50 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}+19 b^{3} {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )^{3}-44 b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )+12 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3}}{15 b^{4}}\) | \(339\) |
Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
8/15*exp(2*a+2*I*b*x)*tan(1/2*b*x+1/2*d)^3*(I*tan(1/2*b*x+1/2*d)^3-5*I*tan (1/2*b*x+1/2*d)-4*tan(1/2*b*x+1/2*d)^2+5)/b/(1+tan(1/2*b*x+1/2*d)^2)^3
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.53 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^2(d+b x) \, dx=\frac {{\left (3 i \, e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} - 5 i \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - 15 i \, e^{\left (2 i \, b x + 2 \, a\right )} - 15 i \, e^{\left (2 \, a - 2 i \, d\right )}\right )} e^{\left (-i \, b x - i \, d\right )}}{120 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x, algorithm="fricas")
Output:
1/120*(3*I*e^(6*I*b*x + 2*a + 4*I*d) - 5*I*e^(4*I*b*x + 2*a + 2*I*d) - 15* I*e^(2*I*b*x + 2*a) - 15*I*e^(2*a - 2*I*d))*e^(-I*b*x - I*d)/b
Time = 0.25 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.43 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^2(d+b x) \, dx=\begin {cases} \frac {\left (1536 i b^{3} e^{2 a} e^{7 i d} e^{5 i b x} - 2560 i b^{3} e^{2 a} e^{5 i d} e^{3 i b x} - 7680 i b^{3} e^{2 a} e^{3 i d} e^{i b x} - 7680 i b^{3} e^{2 a} e^{i d} e^{- i b x}\right ) e^{- 4 i d}}{61440 b^{4}} & \text {for}\: b^{4} e^{4 i d} \neq 0 \\\frac {x \left (- e^{2 a} e^{6 i d} + e^{2 a} e^{4 i d} + e^{2 a} e^{2 i d} - e^{2 a}\right ) e^{- 3 i d}}{8} & \text {otherwise} \end {cases} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)**2,x)
Output:
Piecewise(((1536*I*b**3*exp(2*a)*exp(7*I*d)*exp(5*I*b*x) - 2560*I*b**3*exp (2*a)*exp(5*I*d)*exp(3*I*b*x) - 7680*I*b**3*exp(2*a)*exp(3*I*d)*exp(I*b*x) - 7680*I*b**3*exp(2*a)*exp(I*d)*exp(-I*b*x))*exp(-4*I*d)/(61440*b**4), Ne (b**4*exp(4*I*d), 0)), (x*(-exp(2*a)*exp(6*I*d) + exp(2*a)*exp(4*I*d) + ex p(2*a)*exp(2*I*d) - exp(2*a))*exp(-3*I*d)/8, True))
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.98 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^2(d+b x) \, dx=-\frac {-3 i \, \cos \left (5 \, b x + 3 \, d\right ) e^{\left (2 \, a\right )} + 5 i \, \cos \left (3 \, b x + d\right ) e^{\left (2 \, a\right )} + 15 i \, \cos \left (b x + 3 \, d\right ) e^{\left (2 \, a\right )} + 15 i \, \cos \left (b x - d\right ) e^{\left (2 \, a\right )} + 3 \, e^{\left (2 \, a\right )} \sin \left (5 \, b x + 3 \, d\right ) - 5 \, e^{\left (2 \, a\right )} \sin \left (3 \, b x + d\right ) + 15 \, e^{\left (2 \, a\right )} \sin \left (b x + 3 \, d\right ) - 15 \, e^{\left (2 \, a\right )} \sin \left (b x - d\right )}{120 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x, algorithm="maxima")
Output:
-1/120*(-3*I*cos(5*b*x + 3*d)*e^(2*a) + 5*I*cos(3*b*x + d)*e^(2*a) + 15*I* cos(b*x + 3*d)*e^(2*a) + 15*I*cos(b*x - d)*e^(2*a) + 3*e^(2*a)*sin(5*b*x + 3*d) - 5*e^(2*a)*sin(3*b*x + d) + 15*e^(2*a)*sin(b*x + 3*d) - 15*e^(2*a)* sin(b*x - d))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (69) = 138\).
Time = 0.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.35 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^2(d+b x) \, dx=-\frac {-3 i \, {\left (e^{\left (5 i \, b x + 3 i \, d\right )} + e^{\left (-5 i \, b x - 3 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 5 i \, {\left (e^{\left (3 i \, b x + i \, d\right )} + e^{\left (-3 i \, b x - i \, d\right )}\right )} e^{\left (2 \, a\right )} + 15 i \, {\left (e^{\left (i \, b x + 3 i \, d\right )} + e^{\left (-i \, b x - 3 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 15 i \, {\left (e^{\left (i \, b x - i \, d\right )} + e^{\left (-i \, b x + i \, d\right )}\right )} e^{\left (2 \, a\right )} + 30 \, e^{\left (2 \, a\right )} \sin \left (b x + 3 \, d\right ) + 30 \, e^{\left (2 \, a\right )} \sin \left (-b x + d\right ) + 10 \, e^{\left (2 \, a\right )} \sin \left (-3 \, b x - d\right ) - 6 \, e^{\left (2 \, a\right )} \sin \left (-5 \, b x - 3 \, d\right )}{240 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x, algorithm="giac")
Output:
-1/240*(-3*I*(e^(5*I*b*x + 3*I*d) + e^(-5*I*b*x - 3*I*d))*e^(2*a) + 5*I*(e ^(3*I*b*x + I*d) + e^(-3*I*b*x - I*d))*e^(2*a) + 15*I*(e^(I*b*x + 3*I*d) + e^(-I*b*x - 3*I*d))*e^(2*a) + 15*I*(e^(I*b*x - I*d) + e^(-I*b*x + I*d))*e ^(2*a) + 30*e^(2*a)*sin(b*x + 3*d) + 30*e^(2*a)*sin(-b*x + d) + 10*e^(2*a) *sin(-3*b*x - d) - 6*e^(2*a)*sin(-5*b*x - 3*d))/b
Time = 15.93 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^2(d+b x) \, dx=-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (b\,x\right )-\sin \left (b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,b}+\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (5\,b\,x\right )+\sin \left (5\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{40\,b}-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (b\,x\right )+\sin \left (b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,b}-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (3\,b\,x\right )+\sin \left (3\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{24\,b} \] Input:
int(cos(d + b*x)*exp(2*a + b*x*2i)*sin(d + b*x)^2,x)
Output:
(exp(2*a)*(cos(5*b*x) + sin(5*b*x)*1i)*(cos(3*d) + sin(3*d)*1i)*1i)/(40*b) - (exp(2*a)*(cos(b*x) - sin(b*x)*1i)*(cos(3*d) - sin(3*d)*1i)*1i)/(8*b) - (exp(2*a)*(cos(b*x) + sin(b*x)*1i)*(cos(d) - sin(d)*1i)*1i)/(8*b) - (exp( 2*a)*(cos(3*b*x) + sin(3*b*x)*1i)*(cos(d) + sin(d)*1i)*1i)/(24*b)
Time = 0.19 (sec) , antiderivative size = 856, normalized size of antiderivative = 7.07 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^2(d+b x) \, dx =\text {Too large to display} \] Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^2,x)
Output:
(e**(2*a + 2*b*i*x)*(410*cos(b*x + d)*sin(b*x + d)**2*tan((b*x + d)/2)**6* i + 1230*cos(b*x + d)*sin(b*x + d)**2*tan((b*x + d)/2)**4*i + 1230*cos(b*x + d)*sin(b*x + d)**2*tan((b*x + d)/2)**2*i + 410*cos(b*x + d)*sin(b*x + d )**2*i + 480*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**6*b*i*x + 120*cos (b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**6 + 1440*cos(b*x + d)*sin(b*x + d )*tan((b*x + d)/2)**4*b*i*x + 360*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/ 2)**4 + 1440*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**2*b*i*x + 360*cos (b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**2 + 480*cos(b*x + d)*sin(b*x + d) *b*i*x + 120*cos(b*x + d)*sin(b*x + d) + 580*cos(b*x + d)*tan((b*x + d)/2) **6*i + 1740*cos(b*x + d)*tan((b*x + d)/2)**4*i + 1740*cos(b*x + d)*tan((b *x + d)/2)**2*i + 580*cos(b*x + d)*i + 615*sin(b*x + d)**3*tan((b*x + d)/2 )**6 + 1845*sin(b*x + d)**3*tan((b*x + d)/2)**4 + 1845*sin(b*x + d)**3*tan ((b*x + d)/2)**2 + 615*sin(b*x + d)**3 + 480*sin(b*x + d)**2*tan((b*x + d) /2)**6*b*x - 360*sin(b*x + d)**2*tan((b*x + d)/2)**6*i + 1440*sin(b*x + d) **2*tan((b*x + d)/2)**4*b*x - 1080*sin(b*x + d)**2*tan((b*x + d)/2)**4*i + 1440*sin(b*x + d)**2*tan((b*x + d)/2)**2*b*x - 1080*sin(b*x + d)**2*tan(( b*x + d)/2)**2*i + 480*sin(b*x + d)**2*b*x - 360*sin(b*x + d)**2*i - 120*s in(b*x + d)*tan((b*x + d)/2)**6 - 360*sin(b*x + d)*tan((b*x + d)/2)**4 - 3 60*sin(b*x + d)*tan((b*x + d)/2)**2 - 120*sin(b*x + d) + 580*tan((b*x + d) /2)**6*i + 960*tan((b*x + d)/2)**5*b*i*x + 480*tan((b*x + d)/2)**5 - 19...