Integrand size = 25, antiderivative size = 91 \[ \int e^{2 (a+i b x)} \sin (d+b x) \tan (d+b x) \, dx=-\frac {3 i e^{2 (a-i d)+i (d+b x)}}{2 b}+\frac {i e^{2 (a-i d)+3 i (d+b x)}}{6 b}+\frac {2 i e^{2 a-2 i d} \arctan \left (e^{i (d+b x)}\right )}{b} \] Output:
-3/2*I*exp(2*a-2*I*d+I*(b*x+d))/b+1/6*I*exp(2*a-2*I*d+3*I*(b*x+d))/b+2*I*e xp(2*a-2*I*d)*arctan(exp(I*(b*x+d)))/b
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(316\) vs. \(2(91)=182\).
Time = 0.24 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.47 \[ \int e^{2 (a+i b x)} \sin (d+b x) \tan (d+b x) \, dx=\frac {e^{2 a} \left (-9 i e^{i b x} \cos (d)+i e^{3 i b x} \cos (d)+3 \cos (2 d) \log \left (1+e^{2 i b x}-2 e^{i b x} \sin (d)\right )-3 \cos (2 d) \log \left (1+e^{2 i b x}+2 e^{i b x} \sin (d)\right )-9 e^{i b x} \sin (d)-e^{3 i b x} \sin (d)+\arctan \left (\frac {\cos \left (\frac {d}{2}\right )+e^{i b x} \sin \left (\frac {d}{2}\right )}{e^{i b x} \cos \left (\frac {d}{2}\right )+\sin \left (\frac {d}{2}\right )}\right ) (-6 i \cos (2 d)-6 \sin (2 d))-3 i \log \left (1+e^{2 i b x}-2 e^{i b x} \sin (d)\right ) \sin (2 d)+3 i \log \left (1+e^{2 i b x}+2 e^{i b x} \sin (d)\right ) \sin (2 d)+6 \arctan \left (\frac {-\cos \left (\frac {d}{2}\right )+e^{i b x} \sin \left (\frac {d}{2}\right )}{e^{i b x} \cos \left (\frac {d}{2}\right )-\sin \left (\frac {d}{2}\right )}\right ) (i \cos (2 d)+\sin (2 d))\right )}{6 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Sin[d + b*x]*Tan[d + b*x],x]
Output:
(E^(2*a)*((-9*I)*E^(I*b*x)*Cos[d] + I*E^((3*I)*b*x)*Cos[d] + 3*Cos[2*d]*Lo g[1 + E^((2*I)*b*x) - 2*E^(I*b*x)*Sin[d]] - 3*Cos[2*d]*Log[1 + E^((2*I)*b* x) + 2*E^(I*b*x)*Sin[d]] - 9*E^(I*b*x)*Sin[d] - E^((3*I)*b*x)*Sin[d] + Arc Tan[(Cos[d/2] + E^(I*b*x)*Sin[d/2])/(E^(I*b*x)*Cos[d/2] + Sin[d/2])]*((-6* I)*Cos[2*d] - 6*Sin[2*d]) - (3*I)*Log[1 + E^((2*I)*b*x) - 2*E^(I*b*x)*Sin[ d]]*Sin[2*d] + (3*I)*Log[1 + E^((2*I)*b*x) + 2*E^(I*b*x)*Sin[d]]*Sin[2*d] + 6*ArcTan[(-Cos[d/2] + E^(I*b*x)*Sin[d/2])/(E^(I*b*x)*Cos[d/2] - Sin[d/2] )]*(I*Cos[2*d] + Sin[2*d])))/(6*b)
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.
\(\displaystyle \int e^{2 (a+i b x)} \sin (b x+d) \tan (b x+d) \, dx\) |
\(\Big \downarrow \) 4974 |
\(\displaystyle \int \left (\frac {3}{2} e^{2 a+i b x-i d}-\frac {1}{2} e^{2 a+2 i (b x+d)+i b x-i d}-\frac {2 e^{2 a+i b x-i d}}{1+e^{2 i (b x+d)}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 i e^{2 a-2 i d} \arctan \left (e^{i b x+i d}\right )}{b}-\frac {3 i e^{2 a+i b x-i d}}{2 b}+\frac {i e^{2 a+2 i (b x+d)+i b x-i d}}{6 b}\) |
Input:
Int[E^(2*(a + I*b*x))*Sin[d + b*x]*Tan[d + b*x],x]
Output:
(((-3*I)/2)*E^(2*a - I*d + I*b*x))/b + ((I/6)*E^(2*a - I*d + I*b*x + (2*I) *(d + b*x)))/b + ((2*I)*E^(2*a - (2*I)*d)*ArcTan[E^(I*d + I*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]
\[\int {\mathrm e}^{2 i b x +2 a} \sin \left (b x +d \right ) \tan \left (b x +d \right )d x\]
Input:
int(exp(2*a+2*I*b*x)*sin(b*x+d)*tan(b*x+d),x)
Output:
int(exp(2*a+2*I*b*x)*sin(b*x+d)*tan(b*x+d),x)
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.86 \[ \int e^{2 (a+i b x)} \sin (d+b x) \tan (d+b x) \, dx=-\frac {6 \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) - 6 \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) - i \, e^{\left (3 i \, b x + 2 \, a + i \, d\right )} + 9 i \, e^{\left (i \, b x + 2 \, a - i \, d\right )}}{6 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*sin(b*x+d)*tan(b*x+d),x, algorithm="fricas")
Output:
-1/6*(6*e^(2*a - 2*I*d)*log(e^(I*b*x + I*d) + I) - 6*e^(2*a - 2*I*d)*log(e ^(I*b*x + I*d) - I) - I*e^(3*I*b*x + 2*a + I*d) + 9*I*e^(I*b*x + 2*a - I*d ))/b
Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.42 \[ \int e^{2 (a+i b x)} \sin (d+b x) \tan (d+b x) \, dx=\begin {cases} \frac {\left (2 i b e^{2 a} e^{2 i d} e^{3 i b x} - 18 i b e^{2 a} e^{i b x}\right ) e^{- i d}}{12 b^{2}} & \text {for}\: b^{2} e^{i d} \neq 0 \\\frac {x \left (- e^{2 a} e^{2 i d} + 3 e^{2 a}\right ) e^{- i d}}{2} & \text {otherwise} \end {cases} + \frac {\left (\log {\left (e^{i b x} - i e^{- i d} \right )} - \log {\left (e^{i b x} + i e^{- i d} \right )}\right ) e^{2 a} e^{- 2 i d}}{b} \] Input:
integrate(exp(2*a+2*I*b*x)*sin(b*x+d)*tan(b*x+d),x)
Output:
Piecewise(((2*I*b*exp(2*a)*exp(2*I*d)*exp(3*I*b*x) - 18*I*b*exp(2*a)*exp(I *b*x))*exp(-I*d)/(12*b**2), Ne(b**2*exp(I*d), 0)), (x*(-exp(2*a)*exp(2*I*d ) + 3*exp(2*a))*exp(-I*d)/2, True)) + (log(exp(I*b*x) - I*exp(-I*d)) - log (exp(I*b*x) + I*exp(-I*d)))*exp(2*a)*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. 385 vs. \(2 (58) = 116\).
Time = 0.16 (sec) , antiderivative size = 385, normalized size of antiderivative = 4.23 \[ \int e^{2 (a+i b x)} \sin (d+b x) \tan (d+b x) \, dx=\frac {6 \, {\left (-i \, \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \arctan \left (\frac {2 \, {\left (\cos \left (b x + 2 \, d\right ) \cos \left (d\right ) + \sin \left (b x + 2 \, d\right ) \sin \left (d\right )\right )}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}, \frac {\cos \left (b x + 2 \, d\right )^{2} - \cos \left (d\right )^{2} + \sin \left (b x + 2 \, d\right )^{2} - \sin \left (d\right )^{2}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}\right ) + i \, \cos \left (3 \, b x + d\right ) e^{\left (2 \, a\right )} - 9 i \, \cos \left (b x - d\right ) e^{\left (2 \, a\right )} + 3 \, {\left (\cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \log \left (\frac {\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} - 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} + 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}\right ) - e^{\left (2 \, a\right )} \sin \left (3 \, b x + d\right ) + 9 \, e^{\left (2 \, a\right )} \sin \left (b x - d\right )}{6 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*sin(b*x+d)*tan(b*x+d),x, algorithm="maxima")
Output:
1/6*(6*(-I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*arctan2(2*(cos(b*x + 2*d)* cos(d) + sin(b*x + 2*d)*sin(d))/(cos(b*x + 2*d)^2 + cos(d)^2 + 2*cos(d)*si n(b*x + 2*d) + sin(b*x + 2*d)^2 - 2*cos(b*x + 2*d)*sin(d) + sin(d)^2), (co s(b*x + 2*d)^2 - cos(d)^2 + sin(b*x + 2*d)^2 - sin(d)^2)/(cos(b*x + 2*d)^2 + cos(d)^2 + 2*cos(d)*sin(b*x + 2*d) + sin(b*x + 2*d)^2 - 2*cos(b*x + 2*d )*sin(d) + sin(d)^2)) + I*cos(3*b*x + d)*e^(2*a) - 9*I*cos(b*x - d)*e^(2*a ) + 3*(cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*log((cos(b*x + 2*d)^2 + cos( d)^2 - 2*cos(d)*sin(b*x + 2*d) + sin(b*x + 2*d)^2 + 2*cos(b*x + 2*d)*sin(d ) + sin(d)^2)/(cos(b*x + 2*d)^2 + cos(d)^2 + 2*cos(d)*sin(b*x + 2*d) + sin (b*x + 2*d)^2 - 2*cos(b*x + 2*d)*sin(d) + sin(d)^2)) - e^(2*a)*sin(3*b*x + d) + 9*e^(2*a)*sin(b*x - d))/b
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int e^{2 (a+i b x)} \sin (d+b x) \tan (d+b x) \, dx=-\frac {6 \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) - 6 \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) - i \, e^{\left (3 i \, b x + 2 \, a + i \, d\right )} + 9 i \, e^{\left (i \, b x + 2 \, a - i \, d\right )}}{6 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*sin(b*x+d)*tan(b*x+d),x, algorithm="giac")
Output:
-1/6*(6*e^(2*a - 2*I*d)*log(I*e^(I*b*x + I*d) - 1) - 6*e^(2*a - 2*I*d)*log (-I*e^(I*b*x + I*d) - 1) - I*e^(3*I*b*x + 2*a + I*d) + 9*I*e^(I*b*x + 2*a - I*d))/b
Time = 15.49 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.74 \[ \int e^{2 (a+i b x)} \sin (d+b x) \tan (d+b x) \, dx=-\frac {{\mathrm {e}}^{2\,a-d\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,3{}\mathrm {i}}{2\,b}+\frac {{\mathrm {e}}^{2\,a+d\,1{}\mathrm {i}+b\,x\,3{}\mathrm {i}}\,1{}\mathrm {i}}{6\,b}+\frac {\sqrt {{\mathrm {e}}^{4\,a-d\,4{}\mathrm {i}}}\,\ln \left (2\,{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,3{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}-{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,4{}\mathrm {i}}}\,2{}\mathrm {i}\right )}{b}-\frac {\sqrt {{\mathrm {e}}^{4\,a-d\,4{}\mathrm {i}}}\,\ln \left (2\,{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,3{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,4{}\mathrm {i}}}\,2{}\mathrm {i}\right )}{b} \] Input:
int(exp(2*a + b*x*2i)*sin(d + b*x)*tan(d + b*x),x)
Output:
(exp(2*a + d*1i + b*x*3i)*1i)/(6*b) - (exp(2*a - d*1i + b*x*1i)*3i)/(2*b) + (exp(4*a - d*4i)^(1/2)*log(2*exp(4*a)*exp(-d*3i)*exp(b*x*1i) - exp(2*a)* exp(-d*2i)*(exp(4*a)*exp(-d*4i))^(1/2)*2i))/b - (exp(4*a - d*4i)^(1/2)*log (2*exp(4*a)*exp(-d*3i)*exp(b*x*1i) + exp(2*a)*exp(-d*2i)*(exp(4*a)*exp(-d* 4i))^(1/2)*2i))/b
\[ \int e^{2 (a+i b x)} \sin (d+b x) \tan (d+b x) \, dx=\frac {e^{2 a} \left (4 e^{2 b i x} \cos \left (b x +d \right ) i +2 e^{2 b i x} \sin \left (b x +d \right )+3 e^{2 b i x} i -12 \left (\int \frac {e^{2 b i x}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}-1}d x \right ) b \right )}{3 b} \] Input:
int(exp(2*a+2*I*b*x)*sin(b*x+d)*tan(b*x+d),x)
Output:
(e**(2*a)*(4*e**(2*b*i*x)*cos(b*x + d)*i + 2*e**(2*b*i*x)*sin(b*x + d) + 3 *e**(2*b*i*x)*i - 12*int(e**(2*b*i*x)/(tan((b*x + d)/2)**4 - 1),x)*b))/(3* b)