Integrand size = 21, antiderivative size = 96 \[ \int e^{2 (a+i b x)} \tan ^2(d+b x) \, dx=\frac {i e^{2 (a-i d)+2 i (d+b x)}}{2 b}-\frac {2 i e^{2 a-2 i d}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {2 i e^{2 a-2 i d} \log \left (1+e^{2 i (d+b x)}\right )}{b} \] Output:
1/2*I*exp(2*a-2*I*d+2*I*(b*x+d))/b-2*I*exp(2*a-2*I*d)/b/(1+exp(2*I*(b*x+d) ))-2*I*exp(2*a-2*I*d)*ln(1+exp(2*I*(b*x+d)))/b
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(201\) vs. \(2(96)=192\).
Time = 0.25 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.09 \[ \int e^{2 (a+i b x)} \tan ^2(d+b x) \, dx=\frac {e^{2 a} \left (i e^{2 i b x}+4 \arctan \left (\frac {\left (-1+e^{2 i b x}\right ) \tan (d)}{1+e^{2 i b x}}\right ) \cos (2 d)-2 i \cos (2 d) \log \left (1+e^{4 i b x}+2 e^{2 i b x} \cos (2 d)\right )+\frac {4 (\cos (d)-i \sin (d))^3}{i \left (1+e^{2 i b x}\right ) \cos (d)-\left (-1+e^{2 i b x}\right ) \sin (d)}-4 i \arctan \left (\frac {\left (-1+e^{2 i b x}\right ) \tan (d)}{1+e^{2 i b x}}\right ) \sin (2 d)-2 \log \left (1+e^{4 i b x}+2 e^{2 i b x} \cos (2 d)\right ) \sin (2 d)\right )}{2 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Tan[d + b*x]^2,x]
Output:
(E^(2*a)*(I*E^((2*I)*b*x) + 4*ArcTan[((-1 + E^((2*I)*b*x))*Tan[d])/(1 + E^ ((2*I)*b*x))]*Cos[2*d] - (2*I)*Cos[2*d]*Log[1 + E^((4*I)*b*x) + 2*E^((2*I) *b*x)*Cos[2*d]] + (4*(Cos[d] - I*Sin[d])^3)/(I*(1 + E^((2*I)*b*x))*Cos[d] - (-1 + E^((2*I)*b*x))*Sin[d]) - (4*I)*ArcTan[((-1 + E^((2*I)*b*x))*Tan[d] )/(1 + E^((2*I)*b*x))]*Sin[2*d] - 2*Log[1 + E^((4*I)*b*x) + 2*E^((2*I)*b*x )*Cos[2*d]]*Sin[2*d]))/(2*b)
Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4942, 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.
\(\displaystyle \int e^{2 (a+i b x)} \tan ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 4942 |
\(\displaystyle -\int \left (e^{2 (a+i b x)}-\frac {4 e^{2 (a+i b x)}}{1+e^{2 i (d+b x)}}+\frac {4 e^{2 (a+i b x)}}{\left (1+e^{2 i (d+b x)}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 i e^{2 a-2 i d}}{b \left (1+e^{2 i (b x+d)}\right )}-\frac {2 i e^{2 a-2 i d} \log \left (1+e^{2 i (b x+d)}\right )}{b}+\frac {i e^{2 (a+i b x)}}{2 b}\) |
Input:
Int[E^(2*(a + I*b*x))*Tan[d + b*x]^2,x]
Output:
((I/2)*E^(2*(a + I*b*x)))/b - ((2*I)*E^(2*a - (2*I)*d))/(b*(1 + E^((2*I)*( d + b*x)))) - ((2*I)*E^(2*a - (2*I)*d)*Log[1 + E^((2*I)*(d + b*x))])/b
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tan[(d_.) + (e_.)*(x_)]^(n_.), x_Symb ol] :> Simp[I^n Int[ExpandIntegrand[F^(c*(a + b*x))*((1 - E^(2*I*(d + e*x )))^n/(1 + E^(2*I*(d + e*x)))^n), x], x], x] /; FreeQ[{F, a, b, c, d, e}, x ] && IntegerQ[n]
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {i {\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{2 b}+\frac {2 i {\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{b \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}-\frac {2 i {\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \ln \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}\) | \(74\) |
Input:
int(exp(2*a+2*I*b*x)*tan(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
1/2*I/b*exp(2*a)*exp(2*I*b*x)+2*I*exp(2*a)*exp(2*I*b*x)/b/(1+exp(2*I*(b*x+ d)))-2*I*exp(2*a)*exp(-2*I*d)*ln(1+exp(2*I*(b*x+d)))/b
Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.93 \[ \int e^{2 (a+i b x)} \tan ^2(d+b x) \, dx=-\frac {4 \, {\left (i \, e^{\left (2 i \, b x + 2 \, a\right )} + i \, e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - i \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - i \, e^{\left (2 i \, b x + 2 \, a\right )} + 4 i \, e^{\left (2 \, a - 2 i \, d\right )}}{2 \, {\left (b e^{\left (2 i \, b x + 2 i \, d\right )} + b\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*tan(b*x+d)^2,x, algorithm="fricas")
Output:
-1/2*(4*(I*e^(2*I*b*x + 2*a) + I*e^(2*a - 2*I*d))*log(e^(2*I*b*x + 2*I*d) + 1) - I*e^(4*I*b*x + 2*a + 2*I*d) - I*e^(2*I*b*x + 2*a) + 4*I*e^(2*a - 2* I*d))/(b*e^(2*I*b*x + 2*I*d) + b)
Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99 \[ \int e^{2 (a+i b x)} \tan ^2(d+b x) \, dx=\begin {cases} \frac {i e^{2 a} e^{2 i b x}}{2 b} & \text {for}\: b \neq 0 \\- x e^{2 a} & \text {otherwise} \end {cases} - \frac {2 i e^{2 a}}{b e^{4 i d} e^{2 i b x} + b e^{2 i d}} - \frac {2 i e^{2 a} e^{- 2 i d} \log {\left (e^{2 i b x} + e^{- 2 i d} \right )}}{b} \] Input:
integrate(exp(2*a+2*I*b*x)*tan(b*x+d)**2,x)
Output:
Piecewise((I*exp(2*a)*exp(2*I*b*x)/(2*b), Ne(b, 0)), (-x*exp(2*a), True)) - 2*I*exp(2*a)/(b*exp(4*I*d)*exp(2*I*b*x) + b*exp(2*I*d)) - 2*I*exp(2*a)*e xp(-2*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. 355 vs. \(2 (66) = 132\).
Time = 0.05 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.70 \[ \int e^{2 (a+i b x)} \tan ^2(d+b x) \, dx=\frac {4 \, {\left (-i \, \cos \left (2 \, d\right )^{2} e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (2 \, d\right )^{2} + {\left (-i \, \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \cos \left (2 \, b x + 4 \, d\right ) + {\left (\cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \sin \left (2 \, b x + 4 \, d\right )\right )} \arctan \left (\sin \left (2 \, b x\right ) - \sin \left (2 \, d\right ), \cos \left (2 \, b x\right ) + \cos \left (2 \, d\right )\right ) + \cos \left (4 \, b x + 4 \, d\right ) e^{\left (2 \, a\right )} + \cos \left (2 \, b x + 2 \, d\right ) e^{\left (2 \, a\right )} - 2 \, {\left (\cos \left (2 \, d\right )^{2} e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )} \sin \left (2 \, d\right )^{2} + {\left (\cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \cos \left (2 \, b x + 4 \, d\right ) - {\left (-i \, \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \sin \left (2 \, b x + 4 \, d\right )\right )} \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, d\right ) + \cos \left (2 \, d\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, d\right ) + \sin \left (2 \, d\right )^{2}\right ) + i \, e^{\left (2 \, a\right )} \sin \left (4 \, b x + 4 \, d\right ) + i \, e^{\left (2 \, a\right )} \sin \left (2 \, b x + 2 \, d\right ) - 4 \, e^{\left (2 \, a\right )}}{-2 i \, b \cos \left (2 \, b x + 4 \, d\right ) - 2 i \, b \cos \left (2 \, d\right ) + 2 \, b \sin \left (2 \, b x + 4 \, d\right ) + 2 \, b \sin \left (2 \, d\right )} \] Input:
integrate(exp(2*a+2*I*b*x)*tan(b*x+d)^2,x, algorithm="maxima")
Output:
(4*(-I*cos(2*d)^2*e^(2*a) - I*e^(2*a)*sin(2*d)^2 + (-I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*cos(2*b*x + 4*d) + (cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d ))*sin(2*b*x + 4*d))*arctan2(sin(2*b*x) - sin(2*d), cos(2*b*x) + cos(2*d)) + cos(4*b*x + 4*d)*e^(2*a) + cos(2*b*x + 2*d)*e^(2*a) - 2*(cos(2*d)^2*e^( 2*a) + e^(2*a)*sin(2*d)^2 + (cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*cos(2* b*x + 4*d) - (-I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*sin(2*b*x + 4*d))*lo g(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) + I*e^(2*a)*sin(4*b*x + 4*d) + I*e^(2*a)*si n(2*b*x + 2*d) - 4*e^(2*a))/(-2*I*b*cos(2*b*x + 4*d) - 2*I*b*cos(2*d) + 2* b*sin(2*b*x + 4*d) + 2*b*sin(2*d))
Time = 0.53 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.03 \[ \int e^{2 (a+i b x)} \tan ^2(d+b x) \, dx=\frac {-4 i \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - 4 i \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + i \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + i \, e^{\left (2 i \, b x + 2 \, a\right )} - 4 i \, e^{\left (2 \, a - 2 i \, d\right )}}{2 \, b {\left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*tan(b*x+d)^2,x, algorithm="giac")
Output:
1/2*(-4*I*e^(2*I*b*x + 2*a)*log(e^(2*I*b*x + 2*I*d) + 1) - 4*I*e^(2*a - 2* I*d)*log(e^(2*I*b*x + 2*I*d) + 1) + I*e^(4*I*b*x + 2*a + 2*I*d) + I*e^(2*I *b*x + 2*a) - 4*I*e^(2*a - 2*I*d))/(b*(e^(2*I*b*x + 2*I*d) + 1))
Time = 0.00 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int e^{2 (a+i b x)} \tan ^2(d+b x) \, dx=\frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2\,b}-\frac {{\mathrm {e}}^{2\,a-d\,2{}\mathrm {i}}\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\right )\,2{}\mathrm {i}}{b}-\frac {{\mathrm {e}}^{4\,a-d\,4{}\mathrm {i}}\,2{}\mathrm {i}}{b\,\left ({\mathrm {e}}^{2\,a-d\,2{}\mathrm {i}}+{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\right )} \] Input:
int(exp(2*a + b*x*2i)*tan(d + b*x)^2,x)
Output:
(exp(2*a + b*x*2i)*1i)/(2*b) - (exp(2*a - d*2i)*log(exp(2*a)*exp(b*x*2i) + exp(2*a)*exp(-d*2i))*2i)/b - (exp(4*a - d*4i)*2i)/(b*(exp(2*a - d*2i) + e xp(2*a + b*x*2i)))
\[ \int e^{2 (a+i b x)} \tan ^2(d+b x) \, dx=e^{2 a} \left (\int e^{2 b i x} \tan \left (b x +d \right )^{2}d x \right ) \] Input:
int(exp(2*a+2*I*b*x)*tan(b*x+d)^2,x)
Output:
e**(2*a)*int(e**(2*b*i*x)*tan(b*x + d)**2,x)