Integrand size = 21, antiderivative size = 121 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \, dx=\frac {i e^{2 (a-i d)-i (d+b x)}}{8 b}-\frac {3 i e^{2 (a-i d)+i (d+b x)}}{8 b}-\frac {i e^{2 (a-i d)+3 i (d+b x)}}{8 b}-\frac {i e^{2 (a-i d)+5 i (d+b x)}}{40 b} \] Output:
1/8*I*exp(2*a-2*I*d-I*(b*x+d))/b-3/8*I*exp(2*a-2*I*d+I*(b*x+d))/b-1/8*I*ex p(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 = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.87 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \, dx=\frac {e^{2 a-i b x} \left (-5 i e^{2 i b x} \left (3+e^{2 i b x}\right ) \cos (d)-i \left (-5+e^{6 i b x}\right ) \cos (3 d)-15 e^{2 i b x} \sin (d)+5 e^{4 i b x} \sin (d)+5 \sin (3 d)+e^{6 i b x} \sin (3 d)\right )}{40 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Cos[d + b*x]^3,x]
Output:
(E^(2*a - I*b*x)*((-5*I)*E^((2*I)*b*x)*(3 + E^((2*I)*b*x))*Cos[d] - I*(-5 + E^((6*I)*b*x))*Cos[3*d] - 15*E^((2*I)*b*x)*Sin[d] + 5*E^((4*I)*b*x)*Sin[ d] + 5*Sin[3*d] + E^((6*I)*b*x)*Sin[3*d]))/(40*b)
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4935, 4933}
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.
\(\displaystyle \int e^{2 (a+i b x)} \cos ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 4935 |
\(\displaystyle \frac {6}{5} \int e^{2 (a+i b x)} \cos (d+b x)dx+\frac {2 i e^{2 (a+i b x)} \cos ^3(b x+d)}{5 b}+\frac {3 e^{2 (a+i b x)} \sin (b x+d) \cos ^2(b x+d)}{5 b}\) |
\(\Big \downarrow \) 4933 |
\(\displaystyle \frac {2 i e^{2 (a+i b x)} \cos ^3(b x+d)}{5 b}+\frac {3 e^{2 (a+i b x)} \sin (b x+d) \cos ^2(b x+d)}{5 b}+\frac {6}{5} \left (-\frac {e^{2 (a+i b x)} \sin (b x+d)}{3 b}-\frac {2 i e^{2 (a+i b x)} \cos (b x+d)}{3 b}\right )\) |
Input:
Int[E^(2*(a + I*b*x))*Cos[d + b*x]^3,x]
Output:
(((2*I)/5)*E^(2*(a + I*b*x))*Cos[d + b*x]^3)/b + (3*E^(2*(a + I*b*x))*Cos[ d + b*x]^2*Sin[d + b*x])/(5*b) + (6*((((-2*I)/3)*E^(2*(a + I*b*x))*Cos[d + b*x])/b - (E^(2*(a + I*b*x))*Sin[d + b*x])/(3*b)))/5
Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Cos[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x ] + Simp[e*F^(c*(a + b*x))*(Sin[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] /; F reeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]
Int[Cos[(d_.) + (e_.)*(x_)]^(m_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbo l] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Cos[d + e*x]^m/(e^2*m^2 + b^2*c^2*Lo g[F]^2)), x] + (Simp[e*m*F^(c*(a + b*x))*Sin[d + e*x]*(Cos[d + e*x]^(m - 1) /(e^2*m^2 + b^2*c^2*Log[F]^2)), x] + Simp[(m*(m - 1)*e^2)/(e^2*m^2 + b^2*c^ 2*Log[F]^2) Int[F^(c*(a + b*x))*Cos[d + e*x]^(m - 2), x], x]) /; FreeQ[{F , a, b, c, d, e}, x] && NeQ[e^2*m^2 + b^2*c^2*Log[F]^2, 0] && GtQ[m, 1]
Time = 0.77 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {i {\mathrm e}^{2 i b x +2 a} \cos \left (3 b x +3 d \right )}{10 b}+\frac {3 \,{\mathrm e}^{2 i b x +2 a} \sin \left (3 b x +3 d \right )}{20 b}-\frac {i {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )}{2 b}-\frac {{\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )}{4 b}\) | \(94\) |
parallelrisch | \(\frac {4 \tan \left (\frac {b x}{2}+\frac {d}{2}\right ) {\mathrm e}^{2 i b x +2 a} \left (i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}-\frac {3 \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{2}-5 i \tan \left (\frac {b x}{2}+\frac {d}{2}\right )-5 \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}+\frac {5}{2}\right )}{5 b \left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{3}}\) | \(94\) |
norman | \(\frac {-\frac {4 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{b}-\frac {4 i {\mathrm e}^{2 i b x +2 a}}{5 b}+\frac {4 i {\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{b}+\frac {2 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}}{b}-\frac {6 \,{\mathrm e}^{2 i b x +2 a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{5 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{3}}\) | \(137\) |
orering | \(-\frac {8 i {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3}}{15 b}-\frac {14 \left (2 i b \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3}-3 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} b \sin \left (b x +d \right )\right )}{15 b^{2}}-\frac {8 i \left (-7 \,{\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3} b^{2}-12 i b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )+6 b^{2} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}\right )}{15 b^{3}}+\frac {-26 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{3}+57 b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )+36 i b^{3} {\mathrm e}^{2 i b x +2 a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}-6 b^{3} {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right )^{3}}{15 b^{4}}\) | \(282\) |
Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
1/10*I/b*exp(2*a+2*I*b*x)*cos(3*b*x+3*d)+3/20/b*exp(2*a+2*I*b*x)*sin(3*b*x +3*d)-1/2*I/b*exp(2*a+2*I*b*x)*cos(b*x+d)-1/4/b*exp(2*a+2*I*b*x)*sin(b*x+d )
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.53 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \, dx=\frac {{\left (-i \, e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} - 5 i \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - 15 i \, e^{\left (2 i \, b x + 2 \, a\right )} + 5 i \, e^{\left (2 \, a - 2 i \, d\right )}\right )} e^{\left (-i \, b x - i \, d\right )}}{40 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^3,x, algorithm="fricas")
Output:
1/40*(-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) + 5*I*e^(2*a - 2*I*d))*e^(-I*b*x - I*d)/b
Time = 0.23 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.46 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \, dx=\begin {cases} \frac {\left (- 512 i b^{3} e^{2 a} e^{7 i d} e^{5 i b x} - 2560 i b^{3} e^{2 a} e^{5 i d} e^{3 i b x} - 7680 i b^{3} e^{2 a} e^{3 i d} e^{i b x} + 2560 i b^{3} e^{2 a} e^{i d} e^{- i b x}\right ) e^{- 4 i d}}{20480 b^{4}} & \text {for}\: b^{4} e^{4 i d} \neq 0 \\\frac {x \left (e^{2 a} e^{6 i d} + 3 e^{2 a} e^{4 i d} + 3 e^{2 a} e^{2 i d} + e^{2 a}\right ) e^{- 3 i d}}{8} & \text {otherwise} \end {cases} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)**3,x)
Output:
Piecewise(((-512*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) + 2560*I*b**3*exp(2*a)*exp(I*d)*exp(-I*b*x))*exp(-4*I*d)/(20480*b**4), Ne (b**4*exp(4*I*d), 0)), (x*(exp(2*a)*exp(6*I*d) + 3*exp(2*a)*exp(4*I*d) + 3 *exp(2*a)*exp(2*I*d) + exp(2*a))*exp(-3*I*d)/8, True))
Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.97 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \, dx=\frac {-i \, \cos \left (5 \, b x + 3 \, d\right ) e^{\left (2 \, a\right )} - 5 i \, \cos \left (3 \, b x + d\right ) e^{\left (2 \, a\right )} + 5 i \, \cos \left (b x + 3 \, d\right ) e^{\left (2 \, a\right )} - 15 i \, \cos \left (b x - d\right ) e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )} \sin \left (5 \, b x + 3 \, d\right ) + 5 \, e^{\left (2 \, a\right )} \sin \left (3 \, b x + d\right ) + 5 \, e^{\left (2 \, a\right )} \sin \left (b x + 3 \, d\right ) + 15 \, e^{\left (2 \, a\right )} \sin \left (b x - d\right )}{40 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^3,x, algorithm="maxima")
Output:
1/40*(-I*cos(5*b*x + 3*d)*e^(2*a) - 5*I*cos(3*b*x + d)*e^(2*a) + 5*I*cos(b *x + 3*d)*e^(2*a) - 15*I*cos(b*x - d)*e^(2*a) + e^(2*a)*sin(5*b*x + 3*d) + 5*e^(2*a)*sin(3*b*x + d) + 5*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 = 0.16 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.35 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \, dx=-\frac {i \, {\left (e^{\left (5 i \, b x + 3 i \, d\right )} + e^{\left (-5 i \, b x - 3 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 5 i \, {\left (e^{\left (3 i \, b x + i \, d\right )} + e^{\left (-3 i \, b x - i \, d\right )}\right )} e^{\left (2 \, a\right )} - 5 i \, {\left (e^{\left (i \, b x + 3 i \, d\right )} + e^{\left (-i \, b x - 3 i \, d\right )}\right )} e^{\left (2 \, a\right )} + 15 i \, {\left (e^{\left (i \, b x - i \, d\right )} + e^{\left (-i \, b x + i \, d\right )}\right )} e^{\left (2 \, a\right )} - 10 \, e^{\left (2 \, a\right )} \sin \left (b x + 3 \, d\right ) + 30 \, e^{\left (2 \, a\right )} \sin \left (-b x + d\right ) + 10 \, e^{\left (2 \, a\right )} \sin \left (-3 \, b x - d\right ) + 2 \, e^{\left (2 \, a\right )} \sin \left (-5 \, b x - 3 \, d\right )}{80 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*cos(b*x+d)^3,x, algorithm="giac")
Output:
-1/80*(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) - 5*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 ) - 10*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) + 2*e^(2*a)*sin(-5*b*x - 3*d))/b
Time = 0.00 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \, dx=\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (b\,x\right )-\sin \left (b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,b}-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (5\,b\,x\right )+\sin \left (5\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{40\,b}-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (b\,x\right )+\sin \left (b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{8\,b}-\frac {{\mathrm {e}}^{2\,a}\,\left (\cos \left (3\,b\,x\right )+\sin \left (3\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,b} \] Input:
int(cos(d + b*x)^3*exp(2*a + b*x*2i),x)
Output:
(exp(2*a)*(cos(b*x) - sin(b*x)*1i)*(cos(3*d) - sin(3*d)*1i)*1i)/(8*b) - (e xp(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(d) - sin(d)*1i)*3i)/(8*b) - (exp( 2*a)*(cos(3*b*x) + sin(3*b*x)*1i)*(cos(d) + sin(d)*1i)*1i)/(8*b)
Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74 \[ \int e^{2 (a+i b x)} \cos ^3(d+b x) \, dx=\frac {e^{2 b i x +2 a} \left (-4 \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} i -2 \cos \left (b x +d \right ) \sin \left (b x +d \right )-4 \cos \left (b x +d \right ) i -6 \sin \left (b x +d \right )^{3}+2 \sin \left (b x +d \right )^{2} i +2 \sin \left (b x +d \right )-i \right )}{10 b} \] Input:
int(exp(2*a+2*I*b*x)*cos(b*x+d)^3,x)
Output:
(e**(2*a + 2*b*i*x)*( - 4*cos(b*x + d)*sin(b*x + d)**2*i - 2*cos(b*x + d)* sin(b*x + d) - 4*cos(b*x + d)*i - 6*sin(b*x + d)**3 + 2*sin(b*x + d)**2*i + 2*sin(b*x + d) - i))/(10*b)