Integrand size = 19, antiderivative size = 96 \[ \int e^{a+i b x} \cos ^3(d+b x) \, dx=\frac {i e^{a-i d-2 i (d+b x)}}{16 b}-\frac {3 i e^{a-i d+2 i (d+b x)}}{16 b}-\frac {i e^{a-i d+4 i (d+b x)}}{32 b}+\frac {3}{8} e^{a-i d} x \] Output:
1/16*I*exp(a-I*d-2*I*(b*x+d))/b-3/16*I*exp(a-I*d+2*I*(b*x+d))/b-1/32*I*exp (a-I*d+4*I*(b*x+d))/b+3/8*exp(a-I*d)*x
Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17 \[ \int e^{a+i b x} \cos ^3(d+b x) \, dx=\frac {e^{a-2 i b x} \left (6 e^{2 i b x} \left (-i e^{2 i b x}+2 b x\right ) \cos (d)-i \left (-2+e^{6 i b x}\right ) \cos (3 d)+6 e^{4 i b x} \sin (d)-12 i b e^{2 i b x} x \sin (d)+2 \sin (3 d)+e^{6 i b x} \sin (3 d)\right )}{32 b} \] Input:
Integrate[E^(a + I*b*x)*Cos[d + b*x]^3,x]
Output:
(E^(a - (2*I)*b*x)*(6*E^((2*I)*b*x)*((-I)*E^((2*I)*b*x) + 2*b*x)*Cos[d] - I*(-2 + E^((6*I)*b*x))*Cos[3*d] + 6*E^((4*I)*b*x)*Sin[d] - (12*I)*b*E^((2* I)*b*x)*x*Sin[d] + 2*Sin[3*d] + E^((6*I)*b*x)*Sin[3*d]))/(32*b)
Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4935, 4976, 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.
\(\displaystyle \int e^{a+i b x} \cos ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 4935 |
\(\displaystyle \frac {3}{4} \int e^{a+i b x} \cos (d+b x)dx+\frac {i e^{a+i b x} \cos ^3(b x+d)}{8 b}+\frac {3 e^{a+i b x} \sin (b x+d) \cos ^2(b x+d)}{8 b}\) |
\(\Big \downarrow \) 4976 |
\(\displaystyle \frac {3}{4} \int \left (\frac {1}{2} e^{a-i d}+\frac {1}{2} e^{a+i d+2 i b x}\right )dx+\frac {i e^{a+i b x} \cos ^3(b x+d)}{8 b}+\frac {3 e^{a+i b x} \sin (b x+d) \cos ^2(b x+d)}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} x e^{a-i d}-\frac {i e^{a+2 i b x+i d}}{4 b}\right )+\frac {i e^{a+i b x} \cos ^3(b x+d)}{8 b}+\frac {3 e^{a+i b x} \sin (b x+d) \cos ^2(b x+d)}{8 b}\) |
Input:
Int[E^(a + I*b*x)*Cos[d + b*x]^3,x]
Output:
(3*(((-1/4*I)*E^(a + I*d + (2*I)*b*x))/b + (E^(a - I*d)*x)/2))/4 + ((I/8)* E^(a + I*b*x)*Cos[d + b*x]^3)/b + (3*E^(a + I*b*x)*Cos[d + b*x]^2*Sin[d + b*x])/(8*b)
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]
Int[Cos[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cos[v]^n , x], x] /; FreeQ[F, x] && (LinearQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]
Time = 0.79 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.79
method | result | size |
parallelrisch | \(-\frac {\left (12 i b x \sin \left (b x +d \right )-12 b x \cos \left (b x +d \right )+i \cos \left (b x +d \right )-i \cos \left (3 b x +3 d \right )-11 \sin \left (b x +d \right )-3 \sin \left (3 b x +3 d \right )\right ) {\mathrm e}^{i b x +a}}{32 b}\) | \(76\) |
norman | \(\frac {-\frac {3 \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{b}+\frac {3 x \,{\mathrm e}^{i b x +a}}{8}+\frac {3 x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{8}-\frac {3 x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{8}-\frac {3 x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}}{8}-\frac {3 i x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{4}-\frac {3 i x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{2}-\frac {3 i x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}}{4}-\frac {5 i {\mathrm e}^{i b x +a}}{8 b}-\frac {9 i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{8 b}+\frac {9 i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{8 b}+\frac {5 i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}}{8 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{3}}\) | \(275\) |
orering | \(-\frac {\left (-4 b x +i\right ) {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3}}{4 b}+\frac {i \left (b x +i\right ) \left (i b \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3}-3 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} b \sin \left (b x +d \right )\right )}{4 b^{2}}-\frac {\left (-4 b x +i\right ) \left (-4 \cos \left (b x +d \right )^{3} {\mathrm e}^{i b x +a} b^{2}-6 i b^{2} {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )+6 \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} {\mathrm e}^{i b x +a} b^{2}\right )}{16 b^{3}}+\frac {i x \left (30 \cos \left (b x +d \right )^{2} \sin \left (b x +d \right ) {\mathrm e}^{i b x +a} b^{3}-10 i \cos \left (b x +d \right )^{3} b^{3} {\mathrm e}^{i b x +a}+18 i b^{3} {\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2}-6 \sin \left (b x +d \right )^{3} {\mathrm e}^{i b x +a} b^{3}\right )}{16 b^{3}}\) | \(283\) |
Input:
int(exp(a+I*b*x)*cos(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-1/32*(12*I*b*x*sin(b*x+d)-12*b*x*cos(b*x+d)+I*cos(b*x+d)-I*cos(3*b*x+3*d) -11*sin(b*x+d)-3*sin(3*b*x+3*d))*exp(a+I*b*x)/b
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.64 \[ \int e^{a+i b x} \cos ^3(d+b x) \, dx=\frac {{\left (12 \, b x e^{\left (2 i \, b x + a + i \, d\right )} - i \, e^{\left (6 i \, b x + a + 5 i \, d\right )} - 6 i \, e^{\left (4 i \, b x + a + 3 i \, d\right )} + 2 i \, e^{\left (a - i \, d\right )}\right )} e^{\left (-2 i \, b x - 2 i \, d\right )}}{32 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^3,x, algorithm="fricas")
Output:
1/32*(12*b*x*e^(2*I*b*x + a + I*d) - I*e^(6*I*b*x + a + 5*I*d) - 6*I*e^(4* I*b*x + a + 3*I*d) + 2*I*e^(a - I*d))*e^(-2*I*b*x - 2*I*d)/b
Time = 0.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.69 \[ \int e^{a+i b x} \cos ^3(d+b x) \, dx=\frac {3 x e^{a} e^{- i d}}{8} + \begin {cases} \frac {\left (- 256 i b^{2} e^{a} e^{6 i d} e^{4 i b x} - 1536 i b^{2} e^{a} e^{4 i d} e^{2 i b x} + 512 i b^{2} e^{a} e^{- 2 i b x}\right ) e^{- 3 i d}}{8192 b^{3}} & \text {for}\: b^{3} e^{3 i d} \neq 0 \\x \left (\frac {\left (e^{a} e^{6 i d} + 3 e^{a} e^{4 i d} + 3 e^{a} e^{2 i d} + e^{a}\right ) e^{- 3 i d}}{8} - \frac {3 e^{a} e^{- i d}}{8}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)**3,x)
Output:
3*x*exp(a)*exp(-I*d)/8 + Piecewise(((-256*I*b**2*exp(a)*exp(6*I*d)*exp(4*I *b*x) - 1536*I*b**2*exp(a)*exp(4*I*d)*exp(2*I*b*x) + 512*I*b**2*exp(a)*exp (-2*I*b*x))*exp(-3*I*d)/(8192*b**3), Ne(b**3*exp(3*I*d), 0)), (x*((exp(a)* exp(6*I*d) + 3*exp(a)*exp(4*I*d) + 3*exp(a)*exp(2*I*d) + exp(a))*exp(-3*I* d)/8 - 3*exp(a)*exp(-I*d)/8), True))
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int e^{a+i b x} \cos ^3(d+b x) \, dx=\frac {12 \, {\left (b \cos \left (d\right ) e^{a} - i \, b e^{a} \sin \left (d\right )\right )} x - i \, \cos \left (4 \, b x + 3 \, d\right ) e^{a} + 2 i \, \cos \left (2 \, b x + 3 \, d\right ) e^{a} - 6 i \, \cos \left (2 \, b x + d\right ) e^{a} + e^{a} \sin \left (4 \, b x + 3 \, d\right ) + 2 \, e^{a} \sin \left (2 \, b x + 3 \, d\right ) + 6 \, e^{a} \sin \left (2 \, b x + d\right )}{32 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^3,x, algorithm="maxima")
Output:
1/32*(12*(b*cos(d)*e^a - I*b*e^a*sin(d))*x - I*cos(4*b*x + 3*d)*e^a + 2*I* cos(2*b*x + 3*d)*e^a - 6*I*cos(2*b*x + d)*e^a + e^a*sin(4*b*x + 3*d) + 2*e ^a*sin(2*b*x + 3*d) + 6*e^a*sin(2*b*x + d))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (55) = 110\).
Time = 0.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.42 \[ \int e^{a+i b x} \cos ^3(d+b x) \, dx=\frac {24 \, {\left (b x + d\right )} \cos \left (d\right ) e^{a} - 24 i \, {\left (b x + d\right )} e^{a} \sin \left (d\right ) - i \, {\left (e^{\left (4 i \, b x + 3 i \, d\right )} + e^{\left (-4 i \, b x - 3 i \, d\right )}\right )} e^{a} + 2 i \, {\left (e^{\left (2 i \, b x + 3 i \, d\right )} + e^{\left (-2 i \, b x - 3 i \, d\right )}\right )} e^{a} - 6 i \, {\left (e^{\left (2 i \, b x + i \, d\right )} + e^{\left (-2 i \, b x - i \, d\right )}\right )} e^{a} + 4 \, e^{a} \sin \left (2 \, b x + 3 \, d\right ) - 12 \, e^{a} \sin \left (-2 \, b x - d\right ) - 2 \, e^{a} \sin \left (-4 \, b x - 3 \, d\right )}{64 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^3,x, algorithm="giac")
Output:
1/64*(24*(b*x + d)*cos(d)*e^a - 24*I*(b*x + d)*e^a*sin(d) - I*(e^(4*I*b*x + 3*I*d) + e^(-4*I*b*x - 3*I*d))*e^a + 2*I*(e^(2*I*b*x + 3*I*d) + e^(-2*I* b*x - 3*I*d))*e^a - 6*I*(e^(2*I*b*x + I*d) + e^(-2*I*b*x - I*d))*e^a + 4*e ^a*sin(2*b*x + 3*d) - 12*e^a*sin(-2*b*x - d) - 2*e^a*sin(-4*b*x - 3*d))/b
Time = 0.00 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17 \[ \int e^{a+i b x} \cos ^3(d+b x) \, dx=\frac {3\,x\,{\mathrm {e}}^a\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )}{8}+\frac {{\mathrm {e}}^a\,\left (\cos \left (2\,b\,x\right )-\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (4\,b\,x\right )+\sin \left (4\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (2\,b\,x\right )+\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{16\,b} \] Input:
int(cos(d + b*x)^3*exp(a + b*x*1i),x)
Output:
(3*x*exp(a)*(cos(d) - sin(d)*1i))/8 + (exp(a)*(cos(2*b*x) - sin(2*b*x)*1i) *(cos(3*d) - sin(3*d)*1i)*1i)/(16*b) - (exp(a)*(cos(4*b*x) + sin(4*b*x)*1i )*(cos(3*d) + sin(3*d)*1i)*1i)/(32*b) - (exp(a)*(cos(2*b*x) + sin(2*b*x)*1 i)*(cos(d) + sin(d)*1i)*3i)/(16*b)
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.74 \[ \int e^{a+i b x} \cos ^3(d+b x) \, dx=\frac {e^{b i x +a} \left (-\cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} i +3 \cos \left (b x +d \right ) b x -5 \cos \left (b x +d \right ) i -3 \sin \left (b x +d \right )^{3}-3 \sin \left (b x +d \right ) b i x \right )}{8 b} \] Input:
int(exp(a+I*b*x)*cos(b*x+d)^3,x)
Output:
(e**(a + b*i*x)*( - cos(b*x + d)*sin(b*x + d)**2*i + 3*cos(b*x + d)*b*x - 5*cos(b*x + d)*i - 3*sin(b*x + d)**3 - 3*sin(b*x + d)*b*i*x))/(8*b)