Integrand size = 25, antiderivative size = 92 \[ \int e^{a+i b x} \cos ^2(d+b x) \sin (d+b x) \, dx=-\frac {e^{a-i d-2 i (d+b x)}}{16 b}-\frac {e^{a-i d+2 i (d+b x)}}{16 b}-\frac {e^{a-i d+4 i (d+b x)}}{32 b}+\frac {1}{8} i e^{a-i d} x \] Output:
-1/16*exp(a-I*d-2*I*(b*x+d))/b-1/16*exp(a-I*d+2*I*(b*x+d))/b-1/32*exp(a-I* d+4*I*(b*x+d))/b+1/8*I*exp(a-I*d)*x
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03 \[ \int e^{a+i b x} \cos ^2(d+b x) \sin (d+b x) \, dx=\frac {e^a \left (-2 e^{2 i b x} \cos (d)+4 i b x \cos (d)-2 i e^{2 i b x} \sin (d)+4 b x \sin (d)-e^{-2 i b x} \left (\left (2+e^{6 i b x}\right ) \cos (3 d)+i \left (-2+e^{6 i b x}\right ) \sin (3 d)\right )\right )}{32 b} \] Input:
Integrate[E^(a + I*b*x)*Cos[d + b*x]^2*Sin[d + b*x],x]
Output:
(E^a*(-2*E^((2*I)*b*x)*Cos[d] + (4*I)*b*x*Cos[d] - (2*I)*E^((2*I)*b*x)*Sin [d] + 4*b*x*Sin[d] - ((2 + E^((6*I)*b*x))*Cos[3*d] + I*(-2 + E^((6*I)*b*x) )*Sin[3*d])/E^((2*I)*b*x)))/(32*b)
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01, 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.
\(\displaystyle \int e^{a+i b x} \sin (b x+d) \cos ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \left (\frac {1}{4} e^{a+i b x} \sin (b x+d)+\frac {1}{4} e^{a+i b x} \sin (3 b x+3 d)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{a+2 i b x+i d}}{16 b}+\frac {i e^{a+i b x} \sin (3 b x+3 d)}{32 b}-\frac {3 e^{a+i b x} \cos (3 b x+3 d)}{32 b}+\frac {1}{8} i x e^{a-i d}\) |
Input:
Int[E^(a + I*b*x)*Cos[d + b*x]^2*Sin[d + b*x],x]
Output:
-1/16*E^(a + I*d + (2*I)*b*x)/b + (I/8)*E^(a - I*d)*x - (3*E^(a + I*b*x)*C os[3*d + 3*b*x])/(32*b) + ((I/32)*E^(a + I*b*x)*Sin[3*d + 3*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. 272 vs. \(2 (72 ) = 144\).
Time = 0.81 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.97
method | result | size |
norman | \(\frac {-\frac {i {\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{b}+\frac {x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{4}+\frac {x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{2}+\frac {x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}}{4}+\frac {i x \,{\mathrm e}^{i b x +a}}{8}+\frac {i x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{8}-\frac {i x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{8}-\frac {i x \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}}{8}-\frac {{\mathrm e}^{i b x +a}}{8 b}+\frac {11 \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}}{8 b}-\frac {11 \,{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}}{8 b}+\frac {{\mathrm e}^{i b x +a} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}}{8 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}\right )^{3}}\) | \(273\) |
orering | \(-\frac {\left (-4 b x +i\right ) {\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )}{4 b}+\frac {i \left (b x +i\right ) \left (i b \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )-2 \,{\mathrm e}^{i b x +a} \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} b +{\mathrm e}^{i b x +a} \cos \left (b x +d \right )^{3} b \right )}{4 b^{2}}-\frac {\left (-4 b x +i\right ) \left (-8 \cos \left (b x +d \right )^{2} \sin \left (b x +d \right ) {\mathrm e}^{i b x +a} b^{2}-4 i \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} {\mathrm e}^{i b x +a} b^{2}+2 i \cos \left (b x +d \right )^{3} {\mathrm e}^{i b x +a} b^{2}+2 \sin \left (b x +d \right )^{3} {\mathrm e}^{i b x +a} b^{2}\right )}{16 b^{3}}+\frac {i x \left (26 \cos \left (b x +d \right ) \sin \left (b x +d \right )^{2} {\mathrm e}^{i b x +a} b^{3}-10 \cos \left (b x +d \right )^{3} {\mathrm e}^{i b x +a} b^{3}-22 i \cos \left (b x +d \right )^{2} \sin \left (b x +d \right ) {\mathrm e}^{i b x +a} b^{3}+6 i b^{3} \sin \left (b x +d \right )^{3} {\mathrm e}^{i b x +a}\right )}{16 b^{3}}\) | \(335\) |
Input:
int(exp(a+I*b*x)*cos(b*x+d)^2*sin(b*x+d),x,method=_RETURNVERBOSE)
Output:
(-I/b*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^3+1/4*x*exp(a+I*b*x)*tan(1/2*b*x+1/2 *d)+1/2*x*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^3+1/4*x*exp(a+I*b*x)*tan(1/2*b*x +1/2*d)^5+1/8*I*x*exp(a+I*b*x)+1/8*I*x*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^2-1 /8*I*x*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^4-1/8*I*x*exp(a+I*b*x)*tan(1/2*b*x+ 1/2*d)^6-1/8*exp(a+I*b*x)/b+11/8/b*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^2-11/8/ b*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^4+1/8/b*exp(a+I*b*x)*tan(1/2*b*x+1/2*d)^ 6)/(1+tan(1/2*b*x+1/2*d)^2)^3
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int e^{a+i b x} \cos ^2(d+b x) \sin (d+b x) \, dx=\frac {{\left (4 i \, b x e^{\left (2 i \, b x + a + i \, d\right )} - e^{\left (6 i \, b x + a + 5 i \, d\right )} - 2 \, e^{\left (4 i \, b x + a + 3 i \, d\right )} - 2 \, e^{\left (a - i \, d\right )}\right )} e^{\left (-2 i \, b x - 2 i \, d\right )}}{32 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^2*sin(b*x+d),x, algorithm="fricas")
Output:
1/32*(4*I*b*x*e^(2*I*b*x + a + I*d) - e^(6*I*b*x + a + 5*I*d) - 2*e^(4*I*b *x + a + 3*I*d) - 2*e^(a - I*d))*e^(-2*I*b*x - 2*I*d)/b
Time = 0.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.76 \[ \int e^{a+i b x} \cos ^2(d+b x) \sin (d+b x) \, dx=\frac {i x e^{a} e^{- i d}}{8} + \begin {cases} \frac {\left (- 256 b^{2} e^{a} e^{6 i d} e^{4 i b x} - 512 b^{2} e^{a} e^{4 i d} e^{2 i b x} - 512 b^{2} e^{a} e^{- 2 i b x}\right ) e^{- 3 i d}}{8192 b^{3}} & \text {for}\: b^{3} e^{3 i d} \neq 0 \\x \left (\frac {\left (- i e^{a} e^{6 i d} - i e^{a} e^{4 i d} + i e^{a} e^{2 i d} + i e^{a}\right ) e^{- 3 i d}}{8} - \frac {i e^{a} e^{- i d}}{8}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)**2*sin(b*x+d),x)
Output:
I*x*exp(a)*exp(-I*d)/8 + Piecewise(((-256*b**2*exp(a)*exp(6*I*d)*exp(4*I*b *x) - 512*b**2*exp(a)*exp(4*I*d)*exp(2*I*b*x) - 512*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*((-I*exp(a)*exp( 6*I*d) - I*exp(a)*exp(4*I*d) + I*exp(a)*exp(2*I*d) + I*exp(a))*exp(-3*I*d) /8 - I*exp(a)*exp(-I*d)/8), True))
Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05 \[ \int e^{a+i b x} \cos ^2(d+b x) \sin (d+b x) \, dx=\frac {4 \, {\left (i \, b \cos \left (d\right ) e^{a} + b e^{a} \sin \left (d\right )\right )} x - \cos \left (4 \, b x + 3 \, d\right ) e^{a} - 2 \, \cos \left (2 \, b x + 3 \, d\right ) e^{a} - 2 \, \cos \left (2 \, b x + d\right ) e^{a} - i \, e^{a} \sin \left (4 \, b x + 3 \, d\right ) + 2 i \, e^{a} \sin \left (2 \, b x + 3 \, d\right ) - 2 i \, e^{a} \sin \left (2 \, b x + d\right )}{32 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^2*sin(b*x+d),x, algorithm="maxima")
Output:
1/32*(4*(I*b*cos(d)*e^a + b*e^a*sin(d))*x - cos(4*b*x + 3*d)*e^a - 2*cos(2 *b*x + 3*d)*e^a - 2*cos(2*b*x + d)*e^a - I*e^a*sin(4*b*x + 3*d) + 2*I*e^a* sin(2*b*x + 3*d) - 2*I*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. 141 vs. \(2 (55) = 110\).
Time = 0.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.53 \[ \int e^{a+i b x} \cos ^2(d+b x) \sin (d+b x) \, dx=-\frac {-8 i \, {\left (b x + d\right )} \cos \left (d\right ) e^{a} - 8 \, {\left (b x + d\right )} e^{a} \sin \left (d\right ) + {\left (e^{\left (4 i \, b x + 3 i \, d\right )} - e^{\left (-4 i \, b x - 3 i \, d\right )}\right )} e^{a} - 2 \, {\left (e^{\left (2 i \, b x + 3 i \, d\right )} - e^{\left (-2 i \, b x - 3 i \, d\right )}\right )} e^{a} + 2 \, {\left (e^{\left (2 i \, b x + i \, d\right )} - e^{\left (-2 i \, b x - i \, d\right )}\right )} e^{a} + 4 \, \cos \left (2 \, b x + 3 \, d\right ) e^{a} + 4 \, \cos \left (-2 \, b x - d\right ) e^{a} + 2 \, \cos \left (-4 \, b x - 3 \, d\right ) e^{a}}{64 \, b} \] Input:
integrate(exp(a+I*b*x)*cos(b*x+d)^2*sin(b*x+d),x, algorithm="giac")
Output:
-1/64*(-8*I*(b*x + d)*cos(d)*e^a - 8*(b*x + d)*e^a*sin(d) + (e^(4*I*b*x + 3*I*d) - e^(-4*I*b*x - 3*I*d))*e^a - 2*(e^(2*I*b*x + 3*I*d) - e^(-2*I*b*x - 3*I*d))*e^a + 2*(e^(2*I*b*x + I*d) - e^(-2*I*b*x - I*d))*e^a + 4*cos(2*b *x + 3*d)*e^a + 4*cos(-2*b*x - d)*e^a + 2*cos(-4*b*x - 3*d)*e^a)/b
Time = 16.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.20 \[ \int e^{a+i b x} \cos ^2(d+b x) \sin (d+b x) \, dx=\frac {x\,{\mathrm {e}}^a\,\left (\cos \left (d\right )-\sin \left (d\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8}-\frac {{\mathrm {e}}^a\,\left (\cos \left (2\,b\,x\right )-\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )-\sin \left (3\,d\right )\,1{}\mathrm {i}\right )}{16\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (4\,b\,x\right )+\sin \left (4\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,d\right )+\sin \left (3\,d\right )\,1{}\mathrm {i}\right )}{32\,b}-\frac {{\mathrm {e}}^a\,\left (\cos \left (2\,b\,x\right )+\sin \left (2\,b\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (d\right )+\sin \left (d\right )\,1{}\mathrm {i}\right )}{16\,b} \] Input:
int(cos(d + b*x)^2*exp(a + b*x*1i)*sin(d + b*x),x)
Output:
(x*exp(a)*(cos(d) - sin(d)*1i)*1i)/8 - (exp(a)*(cos(2*b*x) - sin(2*b*x)*1i )*(cos(3*d) - sin(3*d)*1i))/(16*b) - (exp(a)*(cos(4*b*x) + sin(4*b*x)*1i)* (cos(3*d) + sin(3*d)*1i))/(32*b) - (exp(a)*(cos(2*b*x) + sin(2*b*x)*1i)*(c os(d) + sin(d)*1i))/(16*b)
\[ \int e^{a+i b x} \cos ^2(d+b x) \sin (d+b x) \, dx=e^{a} \left (\int e^{b i x} \cos \left (b x +d \right )^{2} \sin \left (b x +d \right )d x \right ) \] Input:
int(exp(a+I*b*x)*cos(b*x+d)^2*sin(b*x+d),x)
Output:
e**a*int(e**(b*i*x)*cos(b*x + d)**2*sin(b*x + d),x)