Integrand size = 25, antiderivative size = 96 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan (d+b x) \, dx=-\frac {2 e^{2 (a-i d)+i (d+b x)}}{b}-\frac {2 e^{2 (a-i d)+i (d+b x)}}{b \left (1+e^{2 i (d+b x)}\right )}+\frac {4 e^{2 a-2 i d} \arctan \left (e^{i (d+b x)}\right )}{b} \] Output:
-2*exp(2*a-2*I*d+I*(b*x+d))/b-2*exp(2*a-2*I*d+I*(b*x+d))/b/(1+exp(2*I*(b*x +d)))+4*exp(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. \(435\) vs. \(2(96)=192\).
Time = 0.48 (sec) , antiderivative size = 435, normalized size of antiderivative = 4.53 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan (d+b x) \, dx=\frac {e^{2 a} \left (-2 e^{i b x} \cos (d)+2 \arctan \left (\frac {-\cos \left (\frac {d}{2}\right )+e^{i b x} \sin \left (\frac {d}{2}\right )}{e^{i b x} \cos \left (\frac {d}{2}\right )-\sin \left (\frac {d}{2}\right )}\right ) \cos (2 d)-2 \arctan \left (\frac {\cos \left (\frac {d}{2}\right )+e^{i b x} \sin \left (\frac {d}{2}\right )}{e^{i b x} \cos \left (\frac {d}{2}\right )+\sin \left (\frac {d}{2}\right )}\right ) \cos (2 d)-i \cos (2 d) \log \left (1+e^{2 i b x}-2 e^{i b x} \sin (d)\right )+i \cos (2 d) \log \left (1+e^{2 i b x}+2 e^{i b x} \sin (d)\right )+2 i e^{i b x} \sin (d)+4 i \arctan \left (\frac {\cos \left (\frac {d}{2}\right )-e^{i b x} \sin \left (\frac {d}{2}\right )}{e^{i b x} \cos \left (\frac {d}{2}\right )-\sin \left (\frac {d}{2}\right )}\right ) \cos (d) \sin (d)-\frac {2 e^{i b x} (\cos (d)-i \sin (d))^2}{\left (1+e^{2 i b x}\right ) \cos (d)+i \left (-1+e^{2 i b x}\right ) \sin (d)}+2 i \arctan \left (\frac {\cos \left (\frac {d}{2}\right )+e^{i b x} \sin \left (\frac {d}{2}\right )}{e^{i b x} \cos \left (\frac {d}{2}\right )+\sin \left (\frac {d}{2}\right )}\right ) \sin (2 d)-\log \left (1+e^{2 i b x}-2 e^{i b x} \sin (d)\right ) \sin (2 d)+\log \left (1+e^{2 i b x}+2 e^{i b x} \sin (d)\right ) \sin (2 d)\right )}{b} \] Input:
Integrate[E^(2*(a + I*b*x))*Sec[d + b*x]*Tan[d + b*x],x]
Output:
(E^(2*a)*(-2*E^(I*b*x)*Cos[d] + 2*ArcTan[(-Cos[d/2] + E^(I*b*x)*Sin[d/2])/ (E^(I*b*x)*Cos[d/2] - Sin[d/2])]*Cos[2*d] - 2*ArcTan[(Cos[d/2] + E^(I*b*x) *Sin[d/2])/(E^(I*b*x)*Cos[d/2] + Sin[d/2])]*Cos[2*d] - I*Cos[2*d]*Log[1 + E^((2*I)*b*x) - 2*E^(I*b*x)*Sin[d]] + I*Cos[2*d]*Log[1 + E^((2*I)*b*x) + 2 *E^(I*b*x)*Sin[d]] + (2*I)*E^(I*b*x)*Sin[d] + (4*I)*ArcTan[(Cos[d/2] - E^( I*b*x)*Sin[d/2])/(E^(I*b*x)*Cos[d/2] - Sin[d/2])]*Cos[d]*Sin[d] - (2*E^(I* b*x)*(Cos[d] - I*Sin[d])^2)/((1 + E^((2*I)*b*x))*Cos[d] + I*(-1 + E^((2*I) *b*x))*Sin[d]) + (2*I)*ArcTan[(Cos[d/2] + E^(I*b*x)*Sin[d/2])/(E^(I*b*x)*C os[d/2] + Sin[d/2])]*Sin[2*d] - Log[1 + E^((2*I)*b*x) - 2*E^(I*b*x)*Sin[d] ]*Sin[2*d] + Log[1 + E^((2*I)*b*x) + 2*E^(I*b*x)*Sin[d]]*Sin[2*d]))/b
Time = 0.35 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.
\(\displaystyle \int e^{2 (a+i b x)} \tan (b x+d) \sec (b x+d) \, dx\) |
\(\Big \downarrow \) 4974 |
\(\displaystyle \int \left (\frac {4 i e^{2 a+3 i b x+i d}}{\left (1+e^{2 i (b x+d)}\right )^2}-\frac {2 i e^{2 a+3 i b x+i d}}{1+e^{2 i (b x+d)}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 e^{2 a-2 i d} \arctan \left (e^{i (b x+d)}\right )}{b}-\frac {2 e^{2 (a-i d)+i (b x+d)}}{b}-\frac {2 e^{2 (a-i d)+i (b x+d)}}{b \left (1+e^{2 i (b x+d)}\right )}\) |
Input:
Int[E^(2*(a + I*b*x))*Sec[d + b*x]*Tan[d + b*x],x]
Output:
(-2*E^(2*(a - I*d) + I*(d + b*x)))/b - (2*E^(2*(a - I*d) + I*(d + b*x)))/( b*(1 + E^((2*I)*(d + b*x)))) + (4*E^(2*a - (2*I)*d)*ArcTan[E^(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]
Time = 0.58 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.82
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{2 a} {\mathrm e}^{3 i b x} {\mathrm e}^{i d}}{\left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right ) b}+\frac {4 \,{\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \arctan \left ({\mathrm e}^{i \left (b x +d \right )}\right )}{b}-\frac {4 \,{\mathrm e}^{2 a} {\mathrm e}^{-i d} {\mathrm e}^{i b x}}{b}\) | \(79\) |
Input:
int(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d),x,method=_RETURNVERBOSE)
Output:
2/(1+exp(2*I*(b*x+d)))/b*exp(2*a)*exp(3*I*b*x)*exp(I*d)+4/b*exp(2*a)*exp(- 2*I*d)*arctan(exp(I*(b*x+d)))-4/b*exp(2*a)*exp(-I*d)*exp(I*b*x)
Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.19 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan (d+b x) \, dx=-\frac {2 \, {\left ({\left (-i \, e^{\left (2 i \, b x + 2 \, a\right )} - i \, e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) + {\left (i \, e^{\left (2 i \, b x + 2 \, a\right )} + i \, e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + e^{\left (3 i \, b x + 2 \, a + i \, d\right )} + 2 \, e^{\left (i \, b x + 2 \, a - i \, d\right )}\right )}}{b e^{\left (2 i \, b x + 2 i \, d\right )} + b} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d),x, algorithm="fricas")
Output:
-2*((-I*e^(2*I*b*x + 2*a) - I*e^(2*a - 2*I*d))*log(e^(I*b*x + I*d) + I) + (I*e^(2*I*b*x + 2*a) + I*e^(2*a - 2*I*d))*log(e^(I*b*x + I*d) - I) + e^(3* I*b*x + 2*a + I*d) + 2*e^(I*b*x + 2*a - I*d))/(b*e^(2*I*b*x + 2*I*d) + b)
\[ \int e^{2 (a+i b x)} \sec (d+b x) \tan (d+b x) \, dx=e^{2 a} \int e^{2 i b x} \tan {\left (b x + d \right )} \sec {\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d),x)
Output:
exp(2*a)*Integral(exp(2*I*b*x)*tan(b*x + d)*sec(b*x + d), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 579 vs. \(2 (71) = 142\).
Time = 0.17 (sec) , antiderivative size = 579, normalized size of antiderivative = 6.03 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan (d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d),x, algorithm="maxima")
Output:
(2*((I*cos(2*d)*e^(2*a) + e^(2*a)*sin(2*d))*cos(2*b*x + 3*d) + (I*cos(d)*e ^(2*a) - e^(2*a)*sin(d))*cos(2*d) - (cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d) )*sin(2*b*x + 3*d) + (cos(d)*e^(2*a) + I*e^(2*a)*sin(d))*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)*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 + 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)) + 2*I*cos(3*b*x + 2*d)*e^(2*a) + 4*I*cos(b*x)*e^(2*a) - ((cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*cos(2*b*x + 3*d) + (cos(d)*e^(2*a) + I*e^(2*a)*sin(d))*cos(2*d) + (I*cos(2*d)*e^(2*a ) + e^(2*a)*sin(2*d))*sin(2*b*x + 3*d) + (-I*cos(d)*e^(2*a) + e^(2*a)*sin( d))*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)) - 2*e^(2*a)*sin(3*b*x + 2*d) - 4*e^(2*a)*sin(b*x))/(- I*b*cos(2*b*x + 3*d) - I*b*cos(d) + b*sin(2*b*x + 3*d) + b*sin(d))
Time = 0.19 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.41 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan (d+b x) \, dx=-\frac {2 \, {\left (-i \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) - i \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) + i \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + i \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + e^{\left (3 i \, b x + 2 \, a + i \, d\right )} + 2 \, e^{\left (i \, b x + 2 \, a - i \, d\right )}\right )}}{b {\left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d),x, algorithm="giac")
Output:
-2*(-I*e^(2*I*b*x + 2*a)*log(e^(I*b*x + I*d) + I) - I*e^(2*a - 2*I*d)*log( e^(I*b*x + I*d) + I) + I*e^(2*I*b*x + 2*a)*log(e^(I*b*x + I*d) - I) + I*e^ (2*a - 2*I*d)*log(e^(I*b*x + I*d) - I) + e^(3*I*b*x + 2*a + I*d) + 2*e^(I* b*x + 2*a - I*d))/(b*(e^(2*I*b*x + 2*I*d) + 1))
Time = 1.07 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.95 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan (d+b x) \, dx=-\frac {2\,{\mathrm {e}}^{2\,a-d\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{b}-\frac {2\,{\mathrm {e}}^{4\,a-d\,3{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{b\,\left ({\mathrm {e}}^{2\,a-d\,2{}\mathrm {i}}+{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\right )}+\frac {2\,\sqrt {-{\mathrm {e}}^{4\,a-d\,4{}\mathrm {i}}}\,\ln \left (-{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,3{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,4{}\mathrm {i}-{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {-{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,4{}\mathrm {i}}}\,4{}\mathrm {i}\right )}{b}-\frac {2\,\sqrt {-{\mathrm {e}}^{4\,a-d\,4{}\mathrm {i}}}\,\ln \left (-{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,3{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,4{}\mathrm {i}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {-{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,4{}\mathrm {i}}}\,4{}\mathrm {i}\right )}{b} \] Input:
int((exp(2*a + b*x*2i)*tan(d + b*x))/cos(d + b*x),x)
Output:
(2*(-exp(4*a - d*4i))^(1/2)*log(- exp(4*a)*exp(-d*3i)*exp(b*x*1i)*4i - exp (2*a)*exp(-d*2i)*(-exp(4*a)*exp(-d*4i))^(1/2)*4i))/b - (2*exp(4*a - d*3i + b*x*1i))/(b*(exp(2*a - d*2i) + exp(2*a + b*x*2i))) - (2*exp(2*a - d*1i + b*x*1i))/b - (2*(-exp(4*a - d*4i))^(1/2)*log(exp(2*a)*exp(-d*2i)*(-exp(4*a )*exp(-d*4i))^(1/2)*4i - exp(4*a)*exp(-d*3i)*exp(b*x*1i)*4i))/b
\[ \int e^{2 (a+i b x)} \sec (d+b x) \tan (d+b x) \, dx=\frac {2 e^{2 a} \left (-e^{2 b i x}+2 \left (\int \frac {e^{2 b i x}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-1}d x \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} b i -2 \left (\int \frac {e^{2 b i x}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-1}d x \right ) b i \right )}{b \left (\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-1\right )} \] Input:
int(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d),x)
Output:
(2*e**(2*a)*( - e**(2*b*i*x) + 2*int(e**(2*b*i*x)/(tan((b*x + d)/2)**2 - 1 ),x)*tan((b*x + d)/2)**2*b*i - 2*int(e**(2*b*i*x)/(tan((b*x + d)/2)**2 - 1 ),x)*b*i))/(b*(tan((b*x + d)/2)**2 - 1))