Integrand size = 23, antiderivative size = 72 \[ \int e^{a+i b x} \sin (d+b x) \tan (d+b x) \, dx=\frac {i e^{a-i d+2 i (d+b x)}}{4 b}-\frac {1}{2} e^{a-i d} x-\frac {i e^{a-i d} \log \left (1+e^{2 i (d+b x)}\right )}{b} \] Output:
1/4*I*exp(a-I*d+2*I*(b*x+d))/b-1/2*exp(a-I*d)*x-I*exp(a-I*d)*ln(1+exp(2*I* (b*x+d)))/b
Time = 0.15 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.97 \[ \int e^{a+i b x} \sin (d+b x) \tan (d+b x) \, dx=-\frac {e^a \left (\cos (d) \left (-i e^{2 i b x}+2 b x+2 i \log \left (1+e^{4 i b x}+2 e^{2 i b x} \cos (2 d)\right )\right )-4 \arctan \left (\frac {\left (-1+e^{2 i b x}\right ) \tan (d)}{1+e^{2 i b x}}\right ) (\cos (d)-i \sin (d))+\left (e^{2 i b x}-2 i b x+2 \log \left (1+e^{4 i b x}+2 e^{2 i b x} \cos (2 d)\right )\right ) \sin (d)\right )}{4 b} \] Input:
Integrate[E^(a + I*b*x)*Sin[d + b*x]*Tan[d + b*x],x]
Output:
-1/4*(E^a*(Cos[d]*((-I)*E^((2*I)*b*x) + 2*b*x + (2*I)*Log[1 + E^((4*I)*b*x ) + 2*E^((2*I)*b*x)*Cos[2*d]]) - 4*ArcTan[((-1 + E^((2*I)*b*x))*Tan[d])/(1 + E^((2*I)*b*x))]*(Cos[d] - I*Sin[d]) + (E^((2*I)*b*x) - (2*I)*b*x + 2*Lo g[1 + E^((4*I)*b*x) + 2*E^((2*I)*b*x)*Cos[2*d]])*Sin[d]))/b
Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4974, 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.
\(\displaystyle \int e^{a+i b x} \sin (b x+d) \tan (b x+d) \, dx\) |
\(\Big \downarrow \) 4974 |
\(\displaystyle \int \left (-\frac {1}{2} e^{a+2 i (b x+d)-i d}-\frac {2 e^{a-i d}}{1+e^{2 i (b x+d)}}+\frac {3}{2} e^{a-i d}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i e^{a+2 i (b x+d)-i d}}{4 b}-\frac {i e^{a-i d} \log \left (1+e^{2 i (b x+d)}\right )}{b}-\frac {1}{2} x e^{a-i d}\) |
Input:
Int[E^(a + I*b*x)*Sin[d + b*x]*Tan[d + b*x],x]
Output:
((I/4)*E^(a - I*d + (2*I)*(d + b*x)))/b - (E^(a - I*d)*x)/2 - (I*E^(a - I* d)*Log[1 + E^((2*I)*(d + b*x))])/b
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[( d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] && IGtQ[ m, 0] && IGtQ[n, 0] && TrigQ[G] && TrigQ[H]
\[\int {\mathrm e}^{i b x +a} \sin \left (b x +d \right ) \tan \left (b x +d \right )d x\]
Input:
int(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d),x)
Output:
int(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d),x)
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.67 \[ \int e^{a+i b x} \sin (d+b x) \tan (d+b x) \, dx=-\frac {2 \, b x e^{\left (a - i \, d\right )} + 4 i \, e^{\left (a - i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - i \, e^{\left (2 i \, b x + a + i \, d\right )}}{4 \, b} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d),x, algorithm="fricas")
Output:
-1/4*(2*b*x*e^(a - I*d) + 4*I*e^(a - I*d)*log(e^(2*I*b*x + 2*I*d) + 1) - I *e^(2*I*b*x + a + I*d))/b
Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.35 \[ \int e^{a+i b x} \sin (d+b x) \tan (d+b x) \, dx=- \frac {x e^{a} e^{- i d}}{2} + \begin {cases} \frac {i e^{a} e^{i d} e^{2 i b x}}{4 b} & \text {for}\: b \neq 0 \\x \left (\frac {\left (- e^{a} e^{2 i d} - e^{a}\right ) e^{- i d}}{2} + \frac {e^{a} e^{- i d}}{2}\right ) & \text {otherwise} \end {cases} - \frac {i e^{a} e^{- i d} \log {\left (e^{2 i b x} + e^{- 2 i d} \right )}}{b} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d),x)
Output:
-x*exp(a)*exp(-I*d)/2 + Piecewise((I*exp(a)*exp(I*d)*exp(2*I*b*x)/(4*b), N e(b, 0)), (x*((-exp(a)*exp(2*I*d) - exp(a))*exp(-I*d)/2 + exp(a)*exp(-I*d) /2), True)) - I*exp(a)*exp(-I*d)*log(exp(2*I*b*x) + exp(-2*I*d))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (48) = 96\).
Time = 0.04 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.03 \[ \int e^{a+i b x} \sin (d+b x) \tan (d+b x) \, dx=-\frac {2 \, {\left (b \cos \left (d\right ) e^{a} - i \, b e^{a} \sin \left (d\right )\right )} x - 4 \, {\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \arctan \left (\sin \left (2 \, b x\right ) - \sin \left (2 \, d\right ), \cos \left (2 \, b x\right ) + \cos \left (2 \, d\right )\right ) - i \, \cos \left (2 \, b x + d\right ) e^{a} - 2 \, {\left (-i \, \cos \left (d\right ) e^{a} - e^{a} \sin \left (d\right )\right )} \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, d\right ) + \cos \left (2 \, d\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, d\right ) + \sin \left (2 \, d\right )^{2}\right ) + e^{a} \sin \left (2 \, b x + d\right )}{4 \, b} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d),x, algorithm="maxima")
Output:
-1/4*(2*(b*cos(d)*e^a - I*b*e^a*sin(d))*x - 4*(cos(d)*e^a - I*e^a*sin(d))* arctan2(sin(2*b*x) - sin(2*d), cos(2*b*x) + cos(2*d)) - I*cos(2*b*x + d)*e ^a - 2*(-I*cos(d)*e^a - e^a*sin(d))*log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2* d) + cos(2*d)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*sin(2*d) + sin(2*d)^2) + e^a *sin(2*b*x + d))/b
Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.71 \[ \int e^{a+i b x} \sin (d+b x) \tan (d+b x) \, dx=-\frac {2 \, {\left (b x + d\right )} e^{\left (a - i \, d\right )} + 4 i \, e^{\left (a - i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - i \, e^{\left (2 i \, b x + a + i \, d\right )}}{4 \, b} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d),x, algorithm="giac")
Output:
-1/4*(2*(b*x + d)*e^(a - I*d) + 4*I*e^(a - I*d)*log(e^(2*I*b*x + 2*I*d) + 1) - I*e^(2*I*b*x + a + I*d))/b
Time = 15.46 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int e^{a+i b x} \sin (d+b x) \tan (d+b x) \, dx=-\frac {x\,{\mathrm {e}}^{-d\,1{}\mathrm {i}}\,{\mathrm {e}}^a}{2}-\frac {{\mathrm {e}}^{-d\,1{}\mathrm {i}}\,{\mathrm {e}}^a\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{b}+\frac {{\mathrm {e}}^{d\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}\,{\mathrm {e}}^a\,1{}\mathrm {i}}{4\,b} \] Input:
int(exp(a + b*x*1i)*sin(d + b*x)*tan(d + b*x),x)
Output:
(exp(d*1i)*exp(b*x*2i)*exp(a)*1i)/(4*b) - (exp(-d*1i)*exp(a)*log(exp(2*a)* exp(b*x*2i) + exp(2*a)*exp(-d*2i))*1i)/b - (x*exp(-d*1i)*exp(a))/2
\[ \int e^{a+i b x} \sin (d+b x) \tan (d+b x) \, dx=\frac {2 e^{a} \left (-e^{b i x} \cos \left (b x +d \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} i -e^{b i x} \cos \left (b x +d \right ) i -2 e^{b i x} \sin \left (b x +d \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-2 e^{b i x} \sin \left (b x +d \right )+3 e^{b i x} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} i -4 e^{b i x} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )-e^{b i x} i -2 \left (\int \frac {e^{b i x}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}-1}d x \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} b -2 \left (\int \frac {e^{b i x}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}-1}d x \right ) b +2 \left (\int \frac {e^{b i x} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}-1}d x \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} b i +2 \left (\int \frac {e^{b i x} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}-1}d x \right ) b i -2 \left (\int \frac {e^{b i x} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}-1}d x \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} b i -2 \left (\int \frac {e^{b i x} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}-1}d x \right ) b i \right )}{b \left (\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}+1\right )} \] Input:
int(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d),x)
Output:
(2*e**a*( - e**(b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**2*i - e**(b*i*x)*cos (b*x + d)*i - 2*e**(b*i*x)*sin(b*x + d)*tan((b*x + d)/2)**2 - 2*e**(b*i*x) *sin(b*x + d) + 3*e**(b*i*x)*tan((b*x + d)/2)**2*i - 4*e**(b*i*x)*tan((b*x + d)/2) - e**(b*i*x)*i - 2*int(e**(b*i*x)/(tan((b*x + d)/2)**4 - 1),x)*ta n((b*x + d)/2)**2*b - 2*int(e**(b*i*x)/(tan((b*x + d)/2)**4 - 1),x)*b + 2* int((e**(b*i*x)*tan((b*x + d)/2)**3)/(tan((b*x + d)/2)**4 - 1),x)*tan((b*x + d)/2)**2*b*i + 2*int((e**(b*i*x)*tan((b*x + d)/2)**3)/(tan((b*x + d)/2) **4 - 1),x)*b*i - 2*int((e**(b*i*x)*tan((b*x + d)/2))/(tan((b*x + d)/2)**4 - 1),x)*tan((b*x + d)/2)**2*b*i - 2*int((e**(b*i*x)*tan((b*x + d)/2))/(ta n((b*x + d)/2)**4 - 1),x)*b*i))/(b*(tan((b*x + d)/2)**2 + 1))