Integrand size = 25, antiderivative size = 101 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin (d+b x) \, dx=-\frac {e^{a-i d-i (d+b x)}}{8 b}-\frac {e^{a-i d-3 i (d+b x)}}{48 b}-\frac {e^{a-i d+3 i (d+b x)}}{24 b}-\frac {e^{a-i d+5 i (d+b x)}}{80 b} \] Output:
-1/8*exp(a-I*d-I*(b*x+d))/b-1/48*exp(a-I*d-3*I*(b*x+d))/b-1/24*exp(a-I*d+3 *I*(b*x+d))/b-1/80*exp(a-I*d+5*I*(b*x+d))/b
Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin (d+b x) \, dx=-\frac {e^{a-i b x} \left (\left (3+e^{4 i b x}\right ) \cos (2 d)+i \left (-3+e^{4 i b x}\right ) \sin (2 d)\right )}{24 b}-\frac {e^{a-3 i b x} \left (\left (5+3 e^{8 i b x}\right ) \cos (4 d)+i \left (-5+3 e^{8 i b x}\right ) \sin (4 d)\right )}{240 b} \] Input:
Integrate[E^(a + I*b*x)*Cos[d + b*x]^3*Sin[d + b*x],x]
Output:
-1/24*(E^(a - I*b*x)*((3 + E^((4*I)*b*x))*Cos[2*d] + I*(-3 + E^((4*I)*b*x) )*Sin[2*d]))/b - (E^(a - (3*I)*b*x)*((5 + 3*E^((8*I)*b*x))*Cos[4*d] + I*(- 5 + 3*E^((8*I)*b*x))*Sin[4*d]))/(240*b)
Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.
\(\displaystyle \int e^{a+i b x} \sin (b x+d) \cos ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \left (\frac {1}{4} e^{a+i b x} \sin (2 b x+2 d)+\frac {1}{8} e^{a+i b x} \sin (4 b x+4 d)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i e^{a+i b x} \sin (2 b x+2 d)}{12 b}+\frac {i e^{a+i b x} \sin (4 b x+4 d)}{120 b}-\frac {e^{a+i b x} \cos (2 b x+2 d)}{6 b}-\frac {e^{a+i b x} \cos (4 b x+4 d)}{30 b}\) |
Input:
Int[E^(a + I*b*x)*Cos[d + b*x]^3*Sin[d + b*x],x]
Output:
-1/6*(E^(a + I*b*x)*Cos[2*d + 2*b*x])/b - (E^(a + I*b*x)*Cos[4*d + 4*b*x]) /(30*b) + ((I/12)*E^(a + I*b*x)*Sin[2*d + 2*b*x])/b + ((I/120)*E^(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]
Time = 1.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.91
method | result | size |
default | \(-\frac {{\mathrm e}^{i b x +a} \cos \left (4 b x +4 d \right )}{30 b}+\frac {i {\mathrm e}^{i b x +a} \sin \left (4 b x +4 d \right )}{120 b}-\frac {{\mathrm e}^{i b x +a} \cos \left (2 b x +2 d \right )}{6 b}+\frac {i {\mathrm e}^{i b x +a} \sin \left (2 b x +2 d \right )}{12 b}\) | \(92\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{i b x +a} \left (10 i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}+15 \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}+8 i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}-5 \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}+6 i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )+9 \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-3\right )}{15 b \left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{4}}\) | \(110\) |
norman | \(\frac {-\frac {2 \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{3 b}-\frac {2 \,{\mathrm e}^{i b x +a}}{5 b}+\frac {2 \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}}{b}+\frac {6 \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{5 b}+\frac {4 i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{5 b}+\frac {4 i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}}{3 b}+\frac {16 i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{15 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{4}}\) | \(176\) |
orering | \(\frac {4 i {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )}{5 b}-\frac {14 \left (i b \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )-3 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{2} b +{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{4} b \right )}{45 b^{2}}+\frac {4 i \left (-11 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right ) b^{2}-6 i b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{2}+2 i b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{4}+6 b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{3}\right )}{45 b^{3}}-\frac {-31 i b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )+57 b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{2}-13 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{4} b^{3}+18 i b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{3}-6 b^{3} {\mathrm e}^{i b x +a} \sin \left (b x +d \right )^{4}}{45 b^{4}}\) | \(353\) |
Input:
int(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d),x,method=_RETURNVERBOSE)
Output:
-1/30/b*exp(a+I*b*x)*cos(4*b*x+4*d)+1/120*I/b*exp(a+I*b*x)*sin(4*b*x+4*d)- 1/6/b*exp(a+I*b*x)*cos(2*b*x+2*d)+1/12*I/b*exp(a+I*b*x)*sin(2*b*x+2*d)
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin (d+b x) \, dx=-\frac {{\left (3 \, e^{\left (8 i \, b x + a + 7 i \, d\right )} + 10 \, e^{\left (6 i \, b x + a + 5 i \, d\right )} + 30 \, e^{\left (2 i \, b x + a + i \, d\right )} + 5 \, e^{\left (a - i \, d\right )}\right )} e^{\left (-3 i \, b x - 3 i \, d\right )}}{240 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d),x, algorithm="fricas")
Output:
-1/240*(3*e^(8*I*b*x + a + 7*I*d) + 10*e^(6*I*b*x + a + 5*I*d) + 30*e^(2*I *b*x + a + I*d) + 5*e^(a - I*d))*e^(-3*I*b*x - 3*I*d)/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (73) = 146\).
Time = 0.30 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.66 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin (d+b x) \, dx=\begin {cases} \frac {\left (- 9216 b^{3} e^{a} e^{10 i d} e^{5 i b x} - 30720 b^{3} e^{a} e^{8 i d} e^{3 i b x} - 92160 b^{3} e^{a} e^{4 i d} e^{- i b x} - 15360 b^{3} e^{a} e^{2 i d} e^{- 3 i b x}\right ) e^{- 6 i d}}{737280 b^{4}} & \text {for}\: b^{4} e^{6 i d} \neq 0 \\\frac {x \left (- i e^{a} e^{8 i d} - 2 i e^{a} e^{6 i d} + 2 i e^{a} e^{2 i d} + i e^{a}\right ) e^{- 4 i d}}{16} & \text {otherwise} \end {cases} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)**3*sin(b*x+d),x)
Output:
Piecewise(((-9216*b**3*exp(a)*exp(10*I*d)*exp(5*I*b*x) - 30720*b**3*exp(a) *exp(8*I*d)*exp(3*I*b*x) - 92160*b**3*exp(a)*exp(4*I*d)*exp(-I*b*x) - 1536 0*b**3*exp(a)*exp(2*I*d)*exp(-3*I*b*x))*exp(-6*I*d)/(737280*b**4), Ne(b**4 *exp(6*I*d), 0)), (x*(-I*exp(a)*exp(8*I*d) - 2*I*exp(a)*exp(6*I*d) + 2*I*e xp(a)*exp(2*I*d) + I*exp(a))*exp(-4*I*d)/16, True))
Time = 0.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.07 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin (d+b x) \, dx=-\frac {3 \, \cos \left (5 \, b x + 4 \, d\right ) e^{a} + 5 \, \cos \left (3 \, b x + 4 \, d\right ) e^{a} + 10 \, \cos \left (3 \, b x + 2 \, d\right ) e^{a} + 30 \, \cos \left (b x + 2 \, d\right ) e^{a} + 3 i \, e^{a} \sin \left (5 \, b x + 4 \, d\right ) - 5 i \, e^{a} \sin \left (3 \, b x + 4 \, d\right ) + 10 i \, e^{a} \sin \left (3 \, b x + 2 \, d\right ) - 30 i \, e^{a} \sin \left (b x + 2 \, d\right )}{240 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d),x, algorithm="maxima")
Output:
-1/240*(3*cos(5*b*x + 4*d)*e^a + 5*cos(3*b*x + 4*d)*e^a + 10*cos(3*b*x + 2 *d)*e^a + 30*cos(b*x + 2*d)*e^a + 3*I*e^a*sin(5*b*x + 4*d) - 5*I*e^a*sin(3 *b*x + 4*d) + 10*I*e^a*sin(3*b*x + 2*d) - 30*I*e^a*sin(b*x + 2*d))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (61) = 122\).
Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.55 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin (d+b x) \, dx=-\frac {3 \, {\left (e^{\left (5 i \, b x + 4 i \, d\right )} - e^{\left (-5 i \, b x - 4 i \, d\right )}\right )} e^{a} - 5 \, {\left (e^{\left (3 i \, b x + 4 i \, d\right )} - e^{\left (-3 i \, b x - 4 i \, d\right )}\right )} e^{a} + 10 \, {\left (e^{\left (3 i \, b x + 2 i \, d\right )} - e^{\left (-3 i \, b x - 2 i \, d\right )}\right )} e^{a} - 30 \, {\left (e^{\left (i \, b x + 2 i \, d\right )} - e^{\left (-i \, b x - 2 i \, d\right )}\right )} e^{a} + 10 \, \cos \left (3 \, b x + 4 \, d\right ) e^{a} + 60 \, \cos \left (b x + 2 \, d\right ) e^{a} + 20 \, \cos \left (-3 \, b x - 2 \, d\right ) e^{a} + 6 \, \cos \left (-5 \, b x - 4 \, d\right ) e^{a}}{480 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d),x, algorithm="giac")
Output:
-1/480*(3*(e^(5*I*b*x + 4*I*d) - e^(-5*I*b*x - 4*I*d))*e^a - 5*(e^(3*I*b*x + 4*I*d) - e^(-3*I*b*x - 4*I*d))*e^a + 10*(e^(3*I*b*x + 2*I*d) - e^(-3*I* b*x - 2*I*d))*e^a - 30*(e^(I*b*x + 2*I*d) - e^(-I*b*x - 2*I*d))*e^a + 10*c os(3*b*x + 4*d)*e^a + 60*cos(b*x + 2*d)*e^a + 20*cos(-3*b*x - 2*d)*e^a + 6 *cos(-5*b*x - 4*d)*e^a)/b
Time = 15.44 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.30 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin (d+b x) \, dx=-\frac {{\mathrm {e}}^a\,\left (\cos \left (b\,x\right )-\sin \left (b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,d\right )-\sin \left (2\,d\right )\,1{}\mathrm {i}\right )}{8\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (3\,b\,x\right )+\sin \left (3\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,d\right )+\sin \left (2\,d\right )\,1{}\mathrm {i}\right )}{24\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (3\,b\,x\right )-\sin \left (3\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (4\,d\right )-\sin \left (4\,d\right )\,1{}\mathrm {i}\right )}{48\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (5\,b\,x\right )+\sin \left (5\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (4\,d\right )+\sin \left (4\,d\right )\,1{}\mathrm {i}\right )}{80\,b} \] Input:
int(cos(d + b*x)^3*exp(a + b*x*1i)*sin(d + b*x),x)
Output:
- (exp(a)*(cos(b*x) - sin(b*x)*1i)*(cos(2*d) - sin(2*d)*1i))/(8*b) - (exp( a)*(cos(3*b*x) + sin(3*b*x)*1i)*(cos(2*d) + sin(2*d)*1i))/(24*b) - (exp(a) *(cos(3*b*x) - sin(3*b*x)*1i)*(cos(4*d) - sin(4*d)*1i))/(48*b) - (exp(a)*( cos(5*b*x) + sin(5*b*x)*1i)*(cos(4*d) + sin(4*d)*1i))/(80*b)
\[ \int e^{a+i b x} \cos ^3(d+b x) \sin (d+b x) \, dx=e^{a} \left (\int e^{b i x} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )d x \right ) \] Input:
int(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d),x)
Output:
e**a*int(e**(b*i*x)*cos(b*x + d)**3*sin(b*x + d),x)