Integrand size = 27, antiderivative size = 103 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^3(d+b x) \, dx=\frac {e^{2 (a-i d)-2 i (d+b x)}}{32 b}-\frac {e^{2 (a-i d)+4 i (d+b x)}}{32 b}+\frac {e^{2 (a-i d)+6 i (d+b x)}}{96 b}+\frac {1}{8} i e^{2 a-2 i d} x \] Output:
1/32*exp(2*a-2*I*d-2*I*(b*x+d))/b-1/32*exp(2*a-2*I*d+4*I*(b*x+d))/b+1/96*e xp(2*a-2*I*d+6*I*(b*x+d))/b+1/8*I*exp(2*a-2*I*d)*x
Time = 0.47 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^3(d+b x) \, dx=\frac {e^{2 a} \left (-3 \left (e^{4 i b x}-4 i b x\right ) \cos (2 d)-3 i \left (e^{4 i b x}+4 i b x\right ) \sin (2 d)+e^{-2 i b x} \left (\left (3+e^{8 i b x}\right ) \cos (4 d)+i \left (-3+e^{8 i b x}\right ) \sin (4 d)\right )\right )}{96 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Cos[d + b*x]*Sin[d + b*x]^3,x]
Output:
(E^(2*a)*(-3*(E^((4*I)*b*x) - (4*I)*b*x)*Cos[2*d] - (3*I)*(E^((4*I)*b*x) + (4*I)*b*x)*Sin[2*d] + ((3 + E^((8*I)*b*x))*Cos[4*d] + I*(-3 + E^((8*I)*b* x))*Sin[4*d])/E^((2*I)*b*x)))/(96*b)
Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99, 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.
\(\displaystyle \int e^{2 (a+i b x)} \sin ^3(b x+d) \cos (b x+d) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \left (\frac {1}{4} e^{2 (a+i b x)} \sin (2 b x+2 d)-\frac {1}{8} e^{2 (a+i b x)} \sin (4 b x+4 d)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{2 (a+i d)+4 i b x}}{32 b}-\frac {i e^{2 (a+i b x)} \sin (4 b x+4 d)}{48 b}+\frac {e^{2 (a+i b x)} \cos (4 b x+4 d)}{24 b}+\frac {1}{8} i x e^{2 a-2 i d}\) |
Input:
Int[E^(2*(a + I*b*x))*Cos[d + b*x]*Sin[d + b*x]^3,x]
Output:
-1/32*E^(2*(a + I*d) + (4*I)*b*x)/b + (I/8)*E^(2*a - (2*I)*d)*x + (E^(2*(a + I*b*x))*Cos[4*d + 4*b*x])/(24*b) - ((I/48)*E^(2*(a + I*b*x))*Sin[4*d + 4*b*x])/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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 401 vs. \(2 (80 ) = 160\).
Time = 1.66 (sec) , antiderivative size = 402, normalized size of antiderivative = 3.90
method | result | size |
orering | \(\frac {\left (12 b x +i\right ) {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{3}}{12 b}+\frac {i \left (-b x +2 i\right ) \left (2 i b \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{3}-{\mathrm e}^{2 i b x +2 a} b \sin \left (b x +d \right )^{4}+3 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{2} b \right )}{12 b^{2}}-\frac {\left (-8 b x +i\right ) \left (-14 b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{3}-4 i b^{2} {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )^{4}+12 i b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{2}+6 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right ) b^{2}\right )}{48 b^{3}}+\frac {i x \left (-68 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{3}+22 \sin \left (b x +d \right )^{4} {\mathrm e}^{2 i b x +2 a} b^{3}-84 b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{2}+36 i \cos \left (b x +d \right )^{3} \sin \left (b x +d \right ) {\mathrm e}^{2 i b x +2 a} b^{3}+6 \cos \left (b x +d \right )^{4} {\mathrm e}^{2 i b x +2 a} b^{3}\right )}{48 b^{3}}\) | \(402\) |
norman | \(\frac {\frac {x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{2}+\frac {x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{2}-\frac {x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}}{2}-\frac {x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{7}}{2}+\frac {i x \,{\mathrm e}^{2 i b x +2 a}}{8}+\frac {10 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{3 b}-\frac {{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{b}-\frac {{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}}{b}-\frac {i {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{4 b}-\frac {i x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}}{2}-\frac {13 i {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}}{12 b}+\frac {i {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{7}}{4 b}-\frac {i x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{2}+\frac {i x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{8}}{8}-\frac {5 i x \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{4}+\frac {13 i {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{12 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{4}}\) | \(410\) |
Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
1/12*(12*b*x+I)/b*exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^3+1/12*I*(-b*x+2* I)/b^2*(2*I*b*exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^3-exp(2*a+2*I*b*x)*b* sin(b*x+d)^4+3*exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^2*b)-1/48*(-8*b*x+ I)/b^3*(-14*b^2*exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^3-4*I*b^2*exp(2*a+2 *I*b*x)*sin(b*x+d)^4+12*I*b^2*exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^2+6 *exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d)*b^2)+1/48*I/b^3*x*(-68*I*b^3*exp (2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^3+22*sin(b*x+d)^4*exp(2*a+2*I*b*x)*b^3 -84*b^3*exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^2+36*I*cos(b*x+d)^3*sin(b *x+d)*exp(2*a+2*I*b*x)*b^3+6*cos(b*x+d)^4*exp(2*a+2*I*b*x)*b^3)
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.62 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^3(d+b x) \, dx=\frac {{\left (12 i \, b x e^{\left (2 i \, b x + 2 \, a\right )} + e^{\left (8 i \, b x + 2 \, a + 6 i \, d\right )} - 3 \, e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} + 3 \, e^{\left (2 \, a - 2 i \, d\right )}\right )} e^{\left (-2 i \, b x - 2 i \, d\right )}}{96 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^3,x, algorithm="fricas")
Output:
1/96*(12*I*b*x*e^(2*I*b*x + 2*a) + e^(8*I*b*x + 2*a + 6*I*d) - 3*e^(6*I*b* x + 2*a + 4*I*d) + 3*e^(2*a - 2*I*d))*e^(-2*I*b*x - 2*I*d)/b
Time = 0.23 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.77 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^3(d+b x) \, dx=\frac {i x e^{2 a} e^{- 2 i d}}{8} + \begin {cases} \frac {\left (1024 b^{2} e^{2 a} e^{8 i d} e^{6 i b x} - 3072 b^{2} e^{2 a} e^{6 i d} e^{4 i b x} + 3072 b^{2} e^{2 a} e^{- 2 i b x}\right ) e^{- 4 i d}}{98304 b^{3}} & \text {for}\: b^{3} e^{4 i d} \neq 0 \\x \left (\frac {\left (i e^{2 a} e^{8 i d} - 2 i e^{2 a} e^{6 i d} + 2 i e^{2 a} e^{2 i d} - i e^{2 a}\right ) e^{- 4 i d}}{16} - \frac {i e^{2 a} e^{- 2 i d}}{8}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)**3,x)
Output:
I*x*exp(2*a)*exp(-2*I*d)/8 + Piecewise(((1024*b**2*exp(2*a)*exp(8*I*d)*exp (6*I*b*x) - 3072*b**2*exp(2*a)*exp(6*I*d)*exp(4*I*b*x) + 3072*b**2*exp(2*a )*exp(-2*I*b*x))*exp(-4*I*d)/(98304*b**3), Ne(b**3*exp(4*I*d), 0)), (x*((I *exp(2*a)*exp(8*I*d) - 2*I*exp(2*a)*exp(6*I*d) + 2*I*exp(2*a)*exp(2*I*d) - I*exp(2*a))*exp(-4*I*d)/16 - I*exp(2*a)*exp(-2*I*d)/8), True))
Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.18 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^3(d+b x) \, dx=-\frac {12 \, {\left (-i \, b \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - b e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} x - \cos \left (6 \, b x + 4 \, d\right ) e^{\left (2 \, a\right )} + 3 \, \cos \left (4 \, b x + 2 \, d\right ) e^{\left (2 \, a\right )} - 3 \, \cos \left (2 \, b x + 4 \, d\right ) e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (6 \, b x + 4 \, d\right ) + 3 i \, e^{\left (2 \, a\right )} \sin \left (4 \, b x + 2 \, d\right ) + 3 i \, e^{\left (2 \, a\right )} \sin \left (2 \, b x + 4 \, d\right )}{96 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^3,x, algorithm="maxima")
Output:
-1/96*(12*(-I*b*cos(2*d)*e^(2*a) - b*e^(2*a)*sin(2*d))*x - cos(6*b*x + 4*d )*e^(2*a) + 3*cos(4*b*x + 2*d)*e^(2*a) - 3*cos(2*b*x + 4*d)*e^(2*a) - I*e^ (2*a)*sin(6*b*x + 4*d) + 3*I*e^(2*a)*sin(4*b*x + 2*d) + 3*I*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. 162 vs. \(2 (63) = 126\).
Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.57 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^3(d+b x) \, dx=-\frac {-24 i \, {\left (b x + d\right )} \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - 24 \, {\left (b x + d\right )} e^{\left (2 \, a\right )} \sin \left (2 \, d\right ) - {\left (e^{\left (6 i \, b x + 4 i \, d\right )} - e^{\left (-6 i \, b x - 4 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 3 \, {\left (e^{\left (4 i \, b x + 2 i \, d\right )} - e^{\left (-4 i \, b x - 2 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 3 \, {\left (e^{\left (2 i \, b x + 4 i \, d\right )} - e^{\left (-2 i \, b x - 4 i \, d\right )}\right )} e^{\left (2 \, a\right )} - 6 \, \cos \left (2 \, b x + 4 \, d\right ) e^{\left (2 \, a\right )} + 6 \, \cos \left (-4 \, b x - 2 \, d\right ) e^{\left (2 \, a\right )} - 2 \, \cos \left (-6 \, b x - 4 \, d\right ) e^{\left (2 \, a\right )}}{192 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^3,x, algorithm="giac")
Output:
-1/192*(-24*I*(b*x + d)*cos(2*d)*e^(2*a) - 24*(b*x + d)*e^(2*a)*sin(2*d) - (e^(6*I*b*x + 4*I*d) - e^(-6*I*b*x - 4*I*d))*e^(2*a) + 3*(e^(4*I*b*x + 2* I*d) - e^(-4*I*b*x - 2*I*d))*e^(2*a) + 3*(e^(2*I*b*x + 4*I*d) - e^(-2*I*b* x - 4*I*d))*e^(2*a) - 6*cos(2*b*x + 4*d)*e^(2*a) + 6*cos(-4*b*x - 2*d)*e^( 2*a) - 2*cos(-6*b*x - 4*d)*e^(2*a))/b
Time = 0.61 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.22 \[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^3(d+b x) \, dx=\frac {x\,{\mathrm {e}}^{2\,a}\,\left (\cos \left (2\,d\right )-\sin \left (2\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (2\,b\,x\right )-\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (4\,d\right )-\sin \left (4\,d\right )\,1{}\mathrm {i}\right )}{32\,b}-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (4\,b\,x\right )+\sin \left (4\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,d\right )+\sin \left (2\,d\right )\,1{}\mathrm {i}\right )}{32\,b}+\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (6\,b\,x\right )+\sin \left (6\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (4\,d\right )+\sin \left (4\,d\right )\,1{}\mathrm {i}\right )}{96\,b} \] Input:
int(cos(d + b*x)*exp(2*a + b*x*2i)*sin(d + b*x)^3,x)
Output:
(x*exp(2*a)*(cos(2*d) - sin(2*d)*1i)*1i)/8 + (exp(2*a)*(cos(2*b*x) - sin(2 *b*x)*1i)*(cos(4*d) - sin(4*d)*1i))/(32*b) - (exp(2*a)*(cos(4*b*x) + sin(4 *b*x)*1i)*(cos(2*d) + sin(2*d)*1i))/(32*b) + (exp(2*a)*(cos(6*b*x) + sin(6 *b*x)*1i)*(cos(4*d) + sin(4*d)*1i))/(96*b)
\[ \int e^{2 (a+i b x)} \cos (d+b x) \sin ^3(d+b x) \, dx=e^{2 a} \left (\int e^{2 b i x} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{3}d x \right ) \] Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^3,x)
Output:
e**(2*a)*int(e**(2*b*i*x)*cos(b*x + d)*sin(b*x + d)**3,x)