Integrand size = 29, antiderivative size = 103 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {e^{2 (a-i d)-4 i (d+b x)}}{256 b}-\frac {3 e^{2 (a-i d)+4 i (d+b x)}}{256 b}+\frac {e^{2 (a-i d)+8 i (d+b x)}}{512 b}+\frac {3}{64} i e^{2 a-2 i d} x \] Output:
1/256*exp(2*a-2*I*d-4*I*(b*x+d))/b-3/256*exp(2*a-2*I*d+4*I*(b*x+d))/b+1/51 2*exp(2*a-2*I*d+8*I*(b*x+d))/b+3/64*I*exp(2*a-2*I*d)*x
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {e^{2 a-6 i d-4 i b x} \left (2+24 i d e^{4 i (d+b x)}-6 e^{8 i (d+b x)}+e^{12 i (d+b x)}+24 i b e^{4 i (d+b x)} x\right )}{512 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Cos[d + b*x]^3*Sin[d + b*x]^3,x]
Output:
(E^(2*a - (6*I)*d - (4*I)*b*x)*(2 + (24*I)*d*E^((4*I)*(d + b*x)) - 6*E^((8 *I)*(d + b*x)) + E^((12*I)*(d + b*x)) + (24*I)*b*E^((4*I)*(d + b*x))*x))/( 512*b)
Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, 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.
\(\displaystyle \int e^{2 (a+i b x)} \sin ^3(b x+d) \cos ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \left (\frac {3}{32} e^{2 (a+i b x)} \sin (2 b x+2 d)-\frac {1}{32} e^{2 (a+i b x)} \sin (6 b x+6 d)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 e^{2 (a+i d)+4 i b x}}{256 b}-\frac {i e^{2 (a+i b x)} \sin (6 b x+6 d)}{512 b}+\frac {3 e^{2 (a+i b x)} \cos (6 b x+6 d)}{512 b}+\frac {3}{64} i x e^{2 a-2 i d}\) |
Input:
Int[E^(2*(a + I*b*x))*Cos[d + b*x]^3*Sin[d + b*x]^3,x]
Output:
(-3*E^(2*(a + I*d) + (4*I)*b*x))/(256*b) + ((3*I)/64)*E^(2*a - (2*I)*d)*x + (3*E^(2*(a + I*b*x))*Cos[6*d + 6*b*x])/(512*b) - ((I/512)*E^(2*(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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (80 ) = 160\).
Time = 7.51 (sec) , antiderivative size = 516, normalized size of antiderivative = 5.01
method | result | size |
orering | \(-\frac {\left (-8 b x +i\right ) {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{3}}{8 b}+\frac {i \left (2 b x +i\right ) \left (2 i b \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{3}-3 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{4} b +3 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )^{2} b \right )}{16 b^{2}}-\frac {\left (-8 b x +i\right ) \left (-28 b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{3}-12 i b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{4}+12 i b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )^{2}+6 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{5} b^{2}+6 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{5} \sin \left (b x +d \right ) b^{2}\right )}{128 b^{3}}+\frac {i x \left (-152 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{3}+138 b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )^{4}-138 b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{4} \sin \left (b x +d \right )^{2}+36 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{5}+36 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{5} \sin \left (b x +d \right )-6 \,{\mathrm e}^{2 i b x +2 a} b^{3} \sin \left (b x +d \right )^{6}+6 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{6} b^{3}\right )}{128 b^{3}}\) | \(516\) |
Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-1/8*(-8*b*x+I)/b*exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3+1/16*I*(2*b*x +I)/b^2*(2*I*b*exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3-3*exp(2*a+2*I*b* x)*cos(b*x+d)^2*sin(b*x+d)^4*b+3*exp(2*a+2*I*b*x)*cos(b*x+d)^4*sin(b*x+d)^ 2*b)-1/128*(-8*b*x+I)/b^3*(-28*b^2*exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d )^3-12*I*b^2*exp(2*a+2*I*b*x)*cos(b*x+d)^2*sin(b*x+d)^4+12*I*b^2*exp(2*a+2 *I*b*x)*cos(b*x+d)^4*sin(b*x+d)^2+6*exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d) ^5*b^2+6*exp(2*a+2*I*b*x)*cos(b*x+d)^5*sin(b*x+d)*b^2)+1/128*I/b^3*x*(-152 *I*b^3*exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3+138*b^3*exp(2*a+2*I*b*x) *cos(b*x+d)^2*sin(b*x+d)^4-138*b^3*exp(2*a+2*I*b*x)*cos(b*x+d)^4*sin(b*x+d )^2+36*I*b^3*exp(2*a+2*I*b*x)*cos(b*x+d)*sin(b*x+d)^5+36*I*b^3*exp(2*a+2*I *b*x)*cos(b*x+d)^5*sin(b*x+d)-6*exp(2*a+2*I*b*x)*b^3*sin(b*x+d)^6+6*exp(2* a+2*I*b*x)*cos(b*x+d)^6*b^3)
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.65 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {{\left (24 i \, b x e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + e^{\left (12 i \, b x + 2 \, a + 10 i \, d\right )} - 6 \, e^{\left (8 i \, b x + 2 \, a + 6 i \, d\right )} + 2 \, e^{\left (2 \, a - 2 i \, d\right )}\right )} e^{\left (-4 i \, b x - 4 i \, d\right )}}{512 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x, algorithm="fricas" )
Output:
1/512*(24*I*b*x*e^(4*I*b*x + 2*a + 2*I*d) + e^(12*I*b*x + 2*a + 10*I*d) - 6*e^(8*I*b*x + 2*a + 6*I*d) + 2*e^(2*a - 2*I*d))*e^(-4*I*b*x - 4*I*d)/b
Time = 0.21 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.80 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {3 i x e^{2 a} e^{- 2 i d}}{64} + \begin {cases} \frac {\left (65536 b^{2} e^{2 a} e^{12 i d} e^{8 i b x} - 393216 b^{2} e^{2 a} e^{8 i d} e^{4 i b x} + 131072 b^{2} e^{2 a} e^{- 4 i b x}\right ) e^{- 6 i d}}{33554432 b^{3}} & \text {for}\: b^{3} e^{6 i d} \neq 0 \\x \left (\frac {\left (i e^{2 a} e^{12 i d} - 3 i e^{2 a} e^{8 i d} + 3 i e^{2 a} e^{4 i d} - i e^{2 a}\right ) e^{- 6 i d}}{64} - \frac {3 i e^{2 a} e^{- 2 i d}}{64}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)**3*sin(b*x+d)**3,x)
Output:
3*I*x*exp(2*a)*exp(-2*I*d)/64 + Piecewise(((65536*b**2*exp(2*a)*exp(12*I*d )*exp(8*I*b*x) - 393216*b**2*exp(2*a)*exp(8*I*d)*exp(4*I*b*x) + 131072*b** 2*exp(2*a)*exp(-4*I*b*x))*exp(-6*I*d)/(33554432*b**3), Ne(b**3*exp(6*I*d), 0)), (x*((I*exp(2*a)*exp(12*I*d) - 3*I*exp(2*a)*exp(8*I*d) + 3*I*exp(2*a) *exp(4*I*d) - I*exp(2*a))*exp(-6*I*d)/64 - 3*I*exp(2*a)*exp(-2*I*d)/64), T rue))
Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.18 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=-\frac {24 \, {\left (-i \, b \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - b e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} x - \cos \left (8 \, b x + 6 \, d\right ) e^{\left (2 \, a\right )} - 2 \, \cos \left (4 \, b x + 6 \, d\right ) e^{\left (2 \, a\right )} + 6 \, \cos \left (4 \, b x + 2 \, d\right ) e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (8 \, b x + 6 \, d\right ) + 2 i \, e^{\left (2 \, a\right )} \sin \left (4 \, b x + 6 \, d\right ) + 6 i \, e^{\left (2 \, a\right )} \sin \left (4 \, b x + 2 \, d\right )}{512 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x, algorithm="maxima" )
Output:
-1/512*(24*(-I*b*cos(2*d)*e^(2*a) - b*e^(2*a)*sin(2*d))*x - cos(8*b*x + 6* d)*e^(2*a) - 2*cos(4*b*x + 6*d)*e^(2*a) + 6*cos(4*b*x + 2*d)*e^(2*a) - I*e ^(2*a)*sin(8*b*x + 6*d) + 2*I*e^(2*a)*sin(4*b*x + 6*d) + 6*I*e^(2*a)*sin(4 *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. 162 vs. \(2 (63) = 126\).
Time = 0.16 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.57 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=-\frac {-48 i \, {\left (b x + d\right )} \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - 48 \, {\left (b x + d\right )} e^{\left (2 \, a\right )} \sin \left (2 \, d\right ) - {\left (e^{\left (8 i \, b x + 6 i \, d\right )} - e^{\left (-8 i \, b x - 6 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 2 \, {\left (e^{\left (4 i \, b x + 6 i \, d\right )} - e^{\left (-4 i \, b x - 6 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 6 \, {\left (e^{\left (4 i \, b x + 2 i \, d\right )} - e^{\left (-4 i \, b x - 2 i \, d\right )}\right )} e^{\left (2 \, a\right )} - 4 \, \cos \left (4 \, b x + 6 \, d\right ) e^{\left (2 \, a\right )} + 12 \, \cos \left (-4 \, b x - 2 \, d\right ) e^{\left (2 \, a\right )} - 2 \, \cos \left (-8 \, b x - 6 \, d\right ) e^{\left (2 \, a\right )}}{1024 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x, algorithm="giac")
Output:
-1/1024*(-48*I*(b*x + d)*cos(2*d)*e^(2*a) - 48*(b*x + d)*e^(2*a)*sin(2*d) - (e^(8*I*b*x + 6*I*d) - e^(-8*I*b*x - 6*I*d))*e^(2*a) + 2*(e^(4*I*b*x + 6 *I*d) - e^(-4*I*b*x - 6*I*d))*e^(2*a) + 6*(e^(4*I*b*x + 2*I*d) - e^(-4*I*b *x - 2*I*d))*e^(2*a) - 4*cos(4*b*x + 6*d)*e^(2*a) + 12*cos(-4*b*x - 2*d)*e ^(2*a) - 2*cos(-8*b*x - 6*d)*e^(2*a))/b
Time = 15.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.22 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=\frac {x\,{\mathrm {e}}^{2\,a}\,\left (\cos \left (2\,d\right )-\sin \left (2\,d\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{64}-\frac {3\,{\mathrm {e}}^{2\,a}\,\left (\cos \left (4\,b\,x\right )+\sin \left (4\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,d\right )+\sin \left (2\,d\right )\,1{}\mathrm {i}\right )}{256\,b}+\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (4\,b\,x\right )-\sin \left (4\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (6\,d\right )-\sin \left (6\,d\right )\,1{}\mathrm {i}\right )}{256\,b}+\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (8\,b\,x\right )+\sin \left (8\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (6\,d\right )+\sin \left (6\,d\right )\,1{}\mathrm {i}\right )}{512\,b} \] Input:
int(cos(d + b*x)^3*exp(2*a + b*x*2i)*sin(d + b*x)^3,x)
Output:
(x*exp(2*a)*(cos(2*d) - sin(2*d)*1i)*3i)/64 - (3*exp(2*a)*(cos(4*b*x) + si n(4*b*x)*1i)*(cos(2*d) + sin(2*d)*1i))/(256*b) + (exp(2*a)*(cos(4*b*x) - s in(4*b*x)*1i)*(cos(6*d) - sin(6*d)*1i))/(256*b) + (exp(2*a)*(cos(8*b*x) + sin(8*b*x)*1i)*(cos(6*d) + sin(6*d)*1i))/(512*b)
\[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \sin ^3(d+b x) \, dx=e^{2 a} \left (\int e^{2 b i x} \cos \left (b x +d \right )^{3} \sin \left (b x +d \right )^{3}d x \right ) \] Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)^3*sin(b*x+d)^3,x)
Output:
e**(2*a)*int(e**(2*b*i*x)*cos(b*x + d)**3*sin(b*x + d)**3,x)