Integrand size = 27, antiderivative size = 101 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=-\frac {3 e^{a-i d-i (d+b x)}}{64 b}-\frac {e^{a-i d+3 i (d+b x)}}{64 b}+\frac {e^{a-i d-5 i (d+b x)}}{320 b}+\frac {e^{a-i d+7 i (d+b x)}}{448 b} \] Output:
-3/64*exp(a-I*d-I*(b*x+d))/b-1/64*exp(a-I*d+3*I*(b*x+d))/b+1/320*exp(a-I*d -5*I*(b*x+d))/b+1/448*exp(a-I*d+7*I*(b*x+d))/b
Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {e^{a-6 i d-5 i b x} \left (7-105 e^{4 i (d+b x)}-35 e^{8 i (d+b x)}+5 e^{12 i (d+b x)}\right )}{2240 b} \] Input:
Integrate[E^(a + I*b*x)*Cos[d + b*x]^3*Sin[d + b*x]^3,x]
Output:
(E^(a - (6*I)*d - (5*I)*b*x)*(7 - 105*E^((4*I)*(d + b*x)) - 35*E^((8*I)*(d + b*x)) + 5*E^((12*I)*(d + b*x))))/(2240*b)
Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, 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.
\(\displaystyle \int e^{a+i b x} \sin ^3(b x+d) \cos ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \left (\frac {3}{32} e^{a+i b x} \sin (2 b x+2 d)-\frac {1}{32} e^{a+i b x} \sin (6 b x+6 d)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i e^{a+i b x} \sin (2 b x+2 d)}{32 b}-\frac {i e^{a+i b x} \sin (6 b x+6 d)}{1120 b}-\frac {e^{a+i b x} \cos (2 b x+2 d)}{16 b}+\frac {3 e^{a+i b x} \cos (6 b x+6 d)}{560 b}\) |
Input:
Int[E^(a + I*b*x)*Cos[d + b*x]^3*Sin[d + b*x]^3,x]
Output:
-1/16*(E^(a + I*b*x)*Cos[2*d + 2*b*x])/b + (3*E^(a + I*b*x)*Cos[6*d + 6*b* x])/(560*b) + ((I/32)*E^(a + I*b*x)*Sin[2*d + 2*b*x])/b - ((I/1120)*E^(a + I*b*x)*Sin[6*d + 6*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 = 3.96 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.77
method | result | size |
parallelrisch | \(-\frac {{\mathrm e}^{i b x +a} \left (70 \cos \left (2 b x +2 d \right )-6 \cos \left (6 b x +6 d \right )-35 i \sin \left (2 b x +2 d \right )-64 i \sin \left (b x +d \right )+i \sin \left (6 b x +6 d \right )+64 \cos \left (b x +d \right )\right )}{1120 b}\) | \(78\) |
default | \(-\frac {{\mathrm e}^{i b x +a} \cos \left (2 b x +2 d \right )}{16 b}+\frac {i {\mathrm e}^{i b x +a} \sin \left (2 b x +2 d \right )}{32 b}+\frac {3 \,{\mathrm e}^{i b x +a} \cos \left (6 b x +6 d \right )}{560 b}-\frac {i {\mathrm e}^{i b x +a} \sin \left (6 b x +6 d \right )}{1120 b}\) | \(92\) |
orering | \(\frac {76 i {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{3}}{105 b}-\frac {34 \left (i b \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{3}-3 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{4} b +3 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )^{2} b \right )}{105 b^{2}}+\frac {4 i \left (-25 b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{3}-6 i b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{4}+6 i b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )^{2}+6 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{5} b^{2}+6 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{5} \sin \left (b x +d \right ) b^{2}\right )}{105 b^{3}}-\frac {-73 i b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{3}+111 b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{4}-111 b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )^{2}+18 i b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{5}+18 i b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{5} \sin \left (b x +d \right )-6 \,{\mathrm e}^{i b x +a} b^{3} \sin \left (b x +d \right )^{6}+6 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{6} b^{3}}{105 b^{4}}\) | \(462\) |
Input:
int(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-1/1120*exp(a+I*b*x)*(70*cos(2*b*x+2*d)-6*cos(6*b*x+6*d)-35*I*sin(2*b*x+2* d)-64*I*sin(b*x+d)+I*sin(6*b*x+6*d)+64*cos(b*x+d))/b
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {{\left (5 \, e^{\left (12 i \, b x + a + 11 i \, d\right )} - 35 \, e^{\left (8 i \, b x + a + 7 i \, d\right )} - 105 \, e^{\left (4 i \, b x + a + 3 i \, d\right )} + 7 \, e^{\left (a - i \, d\right )}\right )} e^{\left (-5 i \, b x - 5 i \, d\right )}}{2240 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x, algorithm="fricas")
Output:
1/2240*(5*e^(12*I*b*x + a + 11*I*d) - 35*e^(8*I*b*x + a + 7*I*d) - 105*e^( 4*I*b*x + a + 3*I*d) + 7*e^(a - I*d))*e^(-5*I*b*x - 5*I*d)/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (73) = 146\).
Time = 0.32 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.65 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\begin {cases} \frac {\left (1310720 b^{3} e^{a} e^{14 i d} e^{7 i b x} - 9175040 b^{3} e^{a} e^{10 i d} e^{3 i b x} - 27525120 b^{3} e^{a} e^{6 i d} e^{- i b x} + 1835008 b^{3} e^{a} e^{2 i d} e^{- 5 i b x}\right ) e^{- 8 i d}}{587202560 b^{4}} & \text {for}\: b^{4} e^{8 i d} \neq 0 \\\frac {x \left (i e^{a} e^{12 i d} - 3 i e^{a} e^{8 i d} + 3 i e^{a} e^{4 i d} - i e^{a}\right ) e^{- 6 i d}}{64} & \text {otherwise} \end {cases} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)**3*sin(b*x+d)**3,x)
Output:
Piecewise(((1310720*b**3*exp(a)*exp(14*I*d)*exp(7*I*b*x) - 9175040*b**3*ex p(a)*exp(10*I*d)*exp(3*I*b*x) - 27525120*b**3*exp(a)*exp(6*I*d)*exp(-I*b*x ) + 1835008*b**3*exp(a)*exp(2*I*d)*exp(-5*I*b*x))*exp(-8*I*d)/(587202560*b **4), Ne(b**4*exp(8*I*d), 0)), (x*(I*exp(a)*exp(12*I*d) - 3*I*exp(a)*exp(8 *I*d) + 3*I*exp(a)*exp(4*I*d) - I*exp(a))*exp(-6*I*d)/64, True))
Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.07 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {5 \, \cos \left (7 \, b x + 6 \, d\right ) e^{a} + 7 \, \cos \left (5 \, b x + 6 \, d\right ) e^{a} - 35 \, \cos \left (3 \, b x + 2 \, d\right ) e^{a} - 105 \, \cos \left (b x + 2 \, d\right ) e^{a} + 5 i \, e^{a} \sin \left (7 \, b x + 6 \, d\right ) - 7 i \, e^{a} \sin \left (5 \, b x + 6 \, d\right ) - 35 i \, e^{a} \sin \left (3 \, b x + 2 \, d\right ) + 105 i \, e^{a} \sin \left (b x + 2 \, d\right )}{2240 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x, algorithm="maxima")
Output:
1/2240*(5*cos(7*b*x + 6*d)*e^a + 7*cos(5*b*x + 6*d)*e^a - 35*cos(3*b*x + 2 *d)*e^a - 105*cos(b*x + 2*d)*e^a + 5*I*e^a*sin(7*b*x + 6*d) - 7*I*e^a*sin( 5*b*x + 6*d) - 35*I*e^a*sin(3*b*x + 2*d) + 105*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 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {5 \, {\left (e^{\left (7 i \, b x + 6 i \, d\right )} - e^{\left (-7 i \, b x - 6 i \, d\right )}\right )} e^{a} - 7 \, {\left (e^{\left (5 i \, b x + 6 i \, d\right )} - e^{\left (-5 i \, b x - 6 i \, d\right )}\right )} e^{a} - 35 \, {\left (e^{\left (3 i \, b x + 2 i \, d\right )} - e^{\left (-3 i \, b x - 2 i \, d\right )}\right )} e^{a} + 105 \, {\left (e^{\left (i \, b x + 2 i \, d\right )} - e^{\left (-i \, b x - 2 i \, d\right )}\right )} e^{a} + 14 \, \cos \left (5 \, b x + 6 \, d\right ) e^{a} - 210 \, \cos \left (b x + 2 \, d\right ) e^{a} - 70 \, \cos \left (-3 \, b x - 2 \, d\right ) e^{a} + 10 \, \cos \left (-7 \, b x - 6 \, d\right ) e^{a}}{4480 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x, algorithm="giac")
Output:
1/4480*(5*(e^(7*I*b*x + 6*I*d) - e^(-7*I*b*x - 6*I*d))*e^a - 7*(e^(5*I*b*x + 6*I*d) - e^(-5*I*b*x - 6*I*d))*e^a - 35*(e^(3*I*b*x + 2*I*d) - e^(-3*I* b*x - 2*I*d))*e^a + 105*(e^(I*b*x + 2*I*d) - e^(-I*b*x - 2*I*d))*e^a + 14* cos(5*b*x + 6*d)*e^a - 210*cos(b*x + 2*d)*e^a - 70*cos(-3*b*x - 2*d)*e^a + 10*cos(-7*b*x - 6*d)*e^a)/b
Time = 15.47 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.30 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=-\frac {3\,{\mathrm {e}}^a\,\left (\cos \left (b\,x\right )-\sin \left (b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,d\right )-\sin \left (2\,d\right )\,1{}\mathrm {i}\right )}{64\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (3\,b\,x\right )+\sin \left (3\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,d\right )+\sin \left (2\,d\right )\,1{}\mathrm {i}\right )}{64\,b}+\frac {{\mathrm {e}}^a\,\left (\cos \left (5\,b\,x\right )-\sin \left (5\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (6\,d\right )-\sin \left (6\,d\right )\,1{}\mathrm {i}\right )}{320\,b}+\frac {{\mathrm {e}}^a\,\left (\cos \left (7\,b\,x\right )+\sin \left (7\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (6\,d\right )+\sin \left (6\,d\right )\,1{}\mathrm {i}\right )}{448\,b} \] Input:
int(cos(d + b*x)^3*exp(a + b*x*1i)*sin(d + b*x)^3,x)
Output:
(exp(a)*(cos(5*b*x) - sin(5*b*x)*1i)*(cos(6*d) - sin(6*d)*1i))/(320*b) - ( exp(a)*(cos(3*b*x) + sin(3*b*x)*1i)*(cos(2*d) + sin(2*d)*1i))/(64*b) - (3* exp(a)*(cos(b*x) - sin(b*x)*1i)*(cos(2*d) - sin(2*d)*1i))/(64*b) + (exp(a) *(cos(7*b*x) + sin(7*b*x)*1i)*(cos(6*d) + sin(6*d)*1i))/(448*b)
Time = 0.17 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.07 \[ \int e^{a+i b x} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {e^{b i x +a} \left (-\cos \left (b x +d \right ) \sin \left (b x +d \right )^{5} i +\cos \left (b x +d \right ) \sin \left (b x +d \right )^{3} i +2 \cos \left (b x +d \right ) \sin \left (b x +d \right ) i +2 \cos \left (b x +d \right )-6 \sin \left (b x +d \right )^{6}+9 \sin \left (b x +d \right )^{4}+\sin \left (b x +d \right )^{2}-2 \sin \left (b x +d \right ) i -2\right )}{35 b} \] Input:
int(exp(a+I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x)
Output:
(e**(a + b*i*x)*( - cos(b*x + d)*sin(b*x + d)**5*i + cos(b*x + d)*sin(b*x + d)**3*i + 2*cos(b*x + d)*sin(b*x + d)*i + 2*cos(b*x + d) - 6*sin(b*x + d )**6 + 9*sin(b*x + d)**4 + sin(b*x + d)**2 - 2*sin(b*x + d)*i - 2))/(35*b)