\(\int e^{2 (a+i b x)} \sec (d+b x) \tan ^2(d+b x) \, dx\) [79]

Optimal result

Integrand size = 27, antiderivative size = 145 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan ^2(d+b x) \, dx=\frac {2 i e^{2 (a-i d)+i (d+b x)}}{b}-\frac {2 i e^{2 (a-i d)+i (d+b x)}}{b \left (1+e^{2 i (d+b x)}\right )^2}+\frac {5 i e^{2 (a-i d)+i (d+b x)}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {5 i e^{2 a-2 i d} \arctan \left (e^{i (d+b x)}\right )}{b} \] Output:

2*I*exp(2*a-2*I*d+I*(b*x+d))/b-2*I*exp(2*a-2*I*d+I*(b*x+d))/b/(1+exp(2*I*( 
b*x+d)))^2+5*I*exp(2*a-2*I*d+I*(b*x+d))/b/(1+exp(2*I*(b*x+d)))-5*I*exp(2*a 
-2*I*d)*arctan(exp(I*(b*x+d)))/b
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.71 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan ^2(d+b x) \, dx=\frac {i e^{2 a-2 i d} \left (e^{i (d+b x)} \left (5+9 e^{2 i (d+b x)}+2 e^{4 i (d+b x)}\right )-5 \left (1+e^{2 i (d+b x)}\right )^2 \arctan \left (e^{i (d+b x)}\right )\right )}{b \left (1+e^{2 i (d+b x)}\right )^2} \] Input:

Integrate[E^(2*(a + I*b*x))*Sec[d + b*x]*Tan[d + b*x]^2,x]
 

Output:

(I*E^(2*a - (2*I)*d)*(E^(I*(d + b*x))*(5 + 9*E^((2*I)*(d + b*x)) + 2*E^((4 
*I)*(d + b*x))) - 5*(1 + E^((2*I)*(d + b*x)))^2*ArcTan[E^(I*(d + b*x))]))/ 
(b*(1 + E^((2*I)*(d + b*x)))^2)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, 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.

\(\displaystyle \int e^{2 (a+i b x)} \tan ^2(b x+d) \sec (b x+d) \, dx\)

\(\Big \downarrow \) 4974

\(\displaystyle \int \left (\frac {8 e^{2 a+3 i b x+i d}}{\left (1+e^{2 i (b x+d)}\right )^2}-\frac {8 e^{2 a+3 i b x+i d}}{\left (1+e^{2 i (b x+d)}\right )^3}+\frac {2 e^{2 a+3 i b x+i d}}{-1-e^{2 i (b x+d)}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {5 i e^{2 a-2 i d} \arctan \left (e^{i (b x+d)}\right )}{b}+\frac {2 i e^{2 (a-i d)+i (b x+d)}}{b}+\frac {5 i e^{2 (a-i d)+i (b x+d)}}{b \left (1+e^{2 i (b x+d)}\right )}-\frac {2 i e^{2 (a-i d)+i (b x+d)}}{b \left (1+e^{2 i (b x+d)}\right )^2}\)

Input:

Int[E^(2*(a + I*b*x))*Sec[d + b*x]*Tan[d + b*x]^2,x]
 

Output:

((2*I)*E^(2*(a - I*d) + I*(d + b*x)))/b - ((2*I)*E^(2*(a - I*d) + I*(d + b 
*x)))/(b*(1 + E^((2*I)*(d + b*x)))^2) + ((5*I)*E^(2*(a - I*d) + I*(d + b*x 
)))/(b*(1 + E^((2*I)*(d + b*x)))) - ((5*I)*E^(2*a - (2*I)*d)*ArcTan[E^(I*( 
d + b*x))])/b
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4974

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]
 

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.70

method	result	size
risch	\(-\frac {i \left (3 \,{\mathrm e}^{5 i b x} {\mathrm e}^{3 i d} {\mathrm e}^{2 a}+{\mathrm e}^{3 i b x} {\mathrm e}^{i d} {\mathrm e}^{2 a}\right )}{\left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{2} b}-\frac {5 i {\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \arctan \left ({\mathrm e}^{i \left (b x +d \right )}\right )}{b}+\frac {5 i {\mathrm e}^{2 a} {\mathrm e}^{i b x} {\mathrm e}^{-i d}}{b}\)	\(101\)

Input:

int(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

-I/(1+exp(2*I*(b*x+d)))^2/b*(3*exp(5*I*b*x)*exp(3*I*d)*exp(2*a)+exp(3*I*b* 
x)*exp(I*d)*exp(2*a))-5*I/b*exp(2*a)*exp(-2*I*d)*arctan(exp(I*(b*x+d)))+5* 
I/b*exp(2*a)*exp(I*b*x)*exp(-I*d)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.13 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan ^2(d+b x) \, dx=\frac {5 \, {\left (e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + 2 \, e^{\left (2 i \, b x + 2 \, a\right )} + e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) - 5 \, {\left (e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + 2 \, e^{\left (2 i \, b x + 2 \, a\right )} + e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + 4 i \, e^{\left (5 i \, b x + 2 \, a + 3 i \, d\right )} + 18 i \, e^{\left (3 i \, b x + 2 \, a + i \, d\right )} + 10 i \, e^{\left (i \, b x + 2 \, a - i \, d\right )}}{2 \, {\left (b e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, b e^{\left (2 i \, b x + 2 i \, d\right )} + b\right )}} \] Input:

integrate(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d)^2,x, algorithm="fricas")
 

Output:

1/2*(5*(e^(4*I*b*x + 2*a + 2*I*d) + 2*e^(2*I*b*x + 2*a) + e^(2*a - 2*I*d)) 
*log(e^(I*b*x + I*d) + I) - 5*(e^(4*I*b*x + 2*a + 2*I*d) + 2*e^(2*I*b*x + 
2*a) + e^(2*a - 2*I*d))*log(e^(I*b*x + I*d) - I) + 4*I*e^(5*I*b*x + 2*a + 
3*I*d) + 18*I*e^(3*I*b*x + 2*a + I*d) + 10*I*e^(I*b*x + 2*a - I*d))/(b*e^( 
4*I*b*x + 4*I*d) + 2*b*e^(2*I*b*x + 2*I*d) + b)
 

Sympy [F]

\[ \int e^{2 (a+i b x)} \sec (d+b x) \tan ^2(d+b x) \, dx=e^{2 a} \int e^{2 i b x} \tan ^{2}{\left (b x + d \right )} \sec {\left (b x + d \right )}\, dx \] Input:

integrate(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d)**2,x)
 

Output:

exp(2*a)*Integral(exp(2*I*b*x)*tan(b*x + d)**2*sec(b*x + d), x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 765 vs. \(2 (101) = 202\).

Time = 0.23 (sec) , antiderivative size = 765, normalized size of antiderivative = 5.28 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan ^2(d+b x) \, dx=\text {Too large to display} \] Input:

integrate(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d)^2,x, algorithm="maxima")
 

Output:

(10*((cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*cos(4*b*x + 5*d) + 2*(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(4*b*x 
 + 5*d) - 2*(-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))*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 + c 
os(d)^2 + 2*cos(d)*sin(b*x + 2*d) + sin(b*x + 2*d)^2 - 2*cos(b*x + 2*d)*si 
n(d) + sin(d)^2)) + 8*cos(5*b*x + 4*d)*e^(2*a) + 36*cos(3*b*x + 2*d)*e^(2* 
a) + 20*cos(b*x)*e^(2*a) - 5*((-I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*cos 
(4*b*x + 5*d) + 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(4*b*x + 5*d) + 2*(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))*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 
)) + 8*I*e^(2*a)*sin(5*b*x + 4*d) + 36*I*e^(2*a)*sin(3*b*x + 2*d) + 20*I*e 
^(2*a)*sin(b*x))/(-4*I*b*cos(4*b*x + 5*d) - 8*I*b*cos(2*b*x + 3*d) - 4*...
 

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (101) = 202\).

Time = 0.67 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.61 \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan ^2(d+b x) \, dx=\frac {e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} \log \left (i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) + 2 \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) + e^{\left (2 \, a - 2 i \, d\right )} \log \left (i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) + 21 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} \log \left (i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) + 42 \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) + 21 \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) - e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) - 2 \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) - e^{\left (2 \, a - 2 i \, d\right )} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) - 21 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) - 42 \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) - 21 \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) + 16 i \, e^{\left (5 i \, b x + 2 \, a + 3 i \, d\right )} + 72 i \, e^{\left (3 i \, b x + 2 \, a + i \, d\right )} + 40 i \, e^{\left (i \, b x + 2 \, a - i \, d\right )}}{8 \, b {\left (e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:

integrate(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d)^2,x, algorithm="giac")
 

Output:

1/8*(e^(4*I*b*x + 2*a + 2*I*d)*log(I*e^(I*b*x + I*d) + 1) + 2*e^(2*I*b*x + 
 2*a)*log(I*e^(I*b*x + I*d) + 1) + e^(2*a - 2*I*d)*log(I*e^(I*b*x + I*d) + 
 1) + 21*e^(4*I*b*x + 2*a + 2*I*d)*log(I*e^(I*b*x + I*d) - 1) + 42*e^(2*I* 
b*x + 2*a)*log(I*e^(I*b*x + I*d) - 1) + 21*e^(2*a - 2*I*d)*log(I*e^(I*b*x 
+ I*d) - 1) - e^(4*I*b*x + 2*a + 2*I*d)*log(-I*e^(I*b*x + I*d) + 1) - 2*e^ 
(2*I*b*x + 2*a)*log(-I*e^(I*b*x + I*d) + 1) - e^(2*a - 2*I*d)*log(-I*e^(I* 
b*x + I*d) + 1) - 21*e^(4*I*b*x + 2*a + 2*I*d)*log(-I*e^(I*b*x + I*d) - 1) 
 - 42*e^(2*I*b*x + 2*a)*log(-I*e^(I*b*x + I*d) - 1) - 21*e^(2*a - 2*I*d)*l 
og(-I*e^(I*b*x + I*d) - 1) + 16*I*e^(5*I*b*x + 2*a + 3*I*d) + 72*I*e^(3*I* 
b*x + 2*a + I*d) + 40*I*e^(I*b*x + 2*a - I*d))/(b*(e^(4*I*b*x + 4*I*d) + 2 
*e^(2*I*b*x + 2*I*d) + 1))
 

Mupad [F(-1)]

Timed out. \[ \int e^{2 (a+i b x)} \sec (d+b x) \tan ^2(d+b x) \, dx=\int \frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\,{\mathrm {tan}\left (d+b\,x\right )}^2}{\cos \left (d+b\,x\right )} \,d x \] Input:

int((exp(2*a + b*x*2i)*tan(d + b*x)^2)/cos(d + b*x),x)
 

Output:

int((exp(2*a + b*x*2i)*tan(d + b*x)^2)/cos(d + b*x), x)
 

Reduce [F]

\[ \int e^{2 (a+i b x)} \sec (d+b x) \tan ^2(d+b x) \, dx=\text {too large to display} \] Input:

int(exp(2*a+2*I*b*x)*sec(b*x+d)*tan(b*x+d)^2,x)
 

Output:

(e**(2*a)*( - 42*e**(2*b*i*x)*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)** 
4 + 84*e**(2*b*i*x)*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**2 - 42*e** 
(2*b*i*x)*cos(b*x + d)*sin(b*x + d) - 56*e**(2*b*i*x)*cos(b*x + d)*tan((b* 
x + d)/2)**4*i + 112*e**(2*b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**2*i - 56* 
e**(2*b*i*x)*cos(b*x + d)*i + 5*e**(2*b*i*x)*sin(b*x + d)**2*tan((b*x + d) 
/2)**4*i + 8*e**(2*b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**3 + 6*e**(2*b* 
i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**2*i - 40*e**(2*b*i*x)*sin(b*x + d)* 
*2*tan((b*x + d)/2) - 359*e**(2*b*i*x)*sin(b*x + d)**2*i - 56*e**(2*b*i*x) 
*sin(b*x + d)*tan((b*x + d)/2)**4 + 112*e**(2*b*i*x)*sin(b*x + d)*tan((b*x 
 + d)/2)**2 - 56*e**(2*b*i*x)*sin(b*x + d) - 47*e**(2*b*i*x)*tan((b*x + d) 
/2)**4*i - 8*e**(2*b*i*x)*tan((b*x + d)/2)**3 + 78*e**(2*b*i*x)*tan((b*x + 
 d)/2)**2*i + 40*e**(2*b*i*x)*tan((b*x + d)/2) + 317*e**(2*b*i*x)*i + 600* 
int(e**(2*b*i*x)/(tan((b*x + d)/2)**6 - 3*tan((b*x + d)/2)**4 + 3*tan((b*x 
 + d)/2)**2 - 1),x)*sin(b*x + d)**2*tan((b*x + d)/2)**4*b - 1200*int(e**(2 
*b*i*x)/(tan((b*x + d)/2)**6 - 3*tan((b*x + d)/2)**4 + 3*tan((b*x + d)/2)* 
*2 - 1),x)*sin(b*x + d)**2*tan((b*x + d)/2)**2*b + 600*int(e**(2*b*i*x)/(t 
an((b*x + d)/2)**6 - 3*tan((b*x + d)/2)**4 + 3*tan((b*x + d)/2)**2 - 1),x) 
*sin(b*x + d)**2*b - 600*int(e**(2*b*i*x)/(tan((b*x + d)/2)**6 - 3*tan((b* 
x + d)/2)**4 + 3*tan((b*x + d)/2)**2 - 1),x)*tan((b*x + d)/2)**4*b + 1200* 
int(e**(2*b*i*x)/(tan((b*x + d)/2)**6 - 3*tan((b*x + d)/2)**4 + 3*tan((...