\(\int e^{a+i b x} \sec (d+b x) \tan ^2(d+b x) \, dx\) [37]

Optimal result

Integrand size = 25, antiderivative size = 93 \[ \int e^{a+i b x} \sec (d+b x) \tan ^2(d+b x) \, dx=-\frac {2 i e^{a-i d}}{b \left (1+e^{2 i (d+b x)}\right )^2}+\frac {4 i e^{a-i d}}{b \left (1+e^{2 i (d+b x)}\right )}+\frac {i e^{a-i d} \log \left (1+e^{2 i (d+b x)}\right )}{b} \] Output:

-2*I*exp(a-I*d)/b/(1+exp(2*I*(b*x+d)))^2+4*I*exp(a-I*d)/b/(1+exp(2*I*(b*x+ 
d)))+I*exp(a-I*d)*ln(1+exp(2*I*(b*x+d)))/b
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int e^{a+i b x} \sec (d+b x) \tan ^2(d+b x) \, dx=\frac {i e^{a-i d} \left (2+4 e^{2 i (d+b x)}+\left (1+e^{2 i (d+b x)}\right )^2 \log \left (1+e^{2 i (d+b x)}\right )\right )}{b \left (1+e^{2 i (d+b x)}\right )^2} \] Input:

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

Output:

(I*E^(a - I*d)*(2 + 4*E^((2*I)*(d + b*x)) + (1 + E^((2*I)*(d + b*x)))^2*Lo 
g[1 + E^((2*I)*(d + b*x))]))/(b*(1 + E^((2*I)*(d + b*x)))^2)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 93, 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.

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

\(\Big \downarrow \) 4974

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 i e^{a-i d}}{b \left (1+e^{2 i (b x+d)}\right )}-\frac {2 i e^{a-i d}}{b \left (1+e^{2 i (b x+d)}\right )^2}+\frac {i e^{a-i d} \log \left (1+e^{2 i (b x+d)}\right )}{b}\)

Input:

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

Output:

((-2*I)*E^(a - I*d))/(b*(1 + E^((2*I)*(d + b*x)))^2) + ((4*I)*E^(a - I*d)) 
/(b*(1 + E^((2*I)*(d + b*x)))) + (I*E^(a - I*d)*Log[1 + E^((2*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 = 59, normalized size of antiderivative = 0.63

method	result	size
risch	\(-\frac {2 i {\mathrm e}^{4 i b x} {\mathrm e}^{3 i d} {\mathrm e}^{a}}{\left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{2} b}+\frac {i {\mathrm e}^{a} {\mathrm e}^{-i d} \ln \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}\)	\(59\)

Input:

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

Output:

-2*I/(1+exp(2*I*(b*x+d)))^2/b*exp(4*I*b*x)*exp(3*I*d)*exp(a)+I*exp(-I*d)*e 
xp(a)/b*ln(1+exp(2*I*(b*x+d)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02 \[ \int e^{a+i b x} \sec (d+b x) \tan ^2(d+b x) \, dx=\frac {{\left (i \, e^{\left (4 i \, b x + a + 3 i \, d\right )} + 2 i \, e^{\left (2 i \, b x + a + i \, d\right )} + i \, e^{\left (a - i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + 4 i \, e^{\left (2 i \, b x + a + i \, d\right )} + 2 i \, e^{\left (a - i \, d\right )}}{b e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, b e^{\left (2 i \, b x + 2 i \, d\right )} + b} \] Input:

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

Output:

((I*e^(4*I*b*x + a + 3*I*d) + 2*I*e^(2*I*b*x + a + I*d) + I*e^(a - I*d))*l 
og(e^(2*I*b*x + 2*I*d) + 1) + 4*I*e^(2*I*b*x + a + I*d) + 2*I*e^(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]

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

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

Output:

exp(a)*Integral(exp(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. 386 vs. \(2 (72) = 144\).

Time = 0.05 (sec) , antiderivative size = 386, normalized size of antiderivative = 4.15 \[ \int e^{a+i b x} \sec (d+b x) \tan ^2(d+b x) \, dx=-\frac {2 \, {\left (-i \, \cos \left (d\right )^{2} e^{a} - i \, e^{a} \sin \left (d\right )^{2} + {\left (-i \, \cos \left (d\right ) e^{a} - e^{a} \sin \left (d\right )\right )} \cos \left (4 \, b x + 5 \, d\right ) + 2 \, {\left (-i \, \cos \left (d\right ) e^{a} - e^{a} \sin \left (d\right )\right )} \cos \left (2 \, b x + 3 \, d\right ) + {\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \sin \left (4 \, b x + 5 \, d\right ) + 2 \, {\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \sin \left (2 \, b x + 3 \, 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 ) - 8 \, \cos \left (2 \, b x + 2 \, d\right ) e^{a} - {\left (\cos \left (d\right )^{2} e^{a} + e^{a} \sin \left (d\right )^{2} + {\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \cos \left (4 \, b x + 5 \, d\right ) + 2 \, {\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \cos \left (2 \, b x + 3 \, d\right ) + {\left (i \, \cos \left (d\right ) e^{a} + e^{a} \sin \left (d\right )\right )} \sin \left (4 \, b x + 5 \, d\right ) - 2 \, {\left (-i \, \cos \left (d\right ) e^{a} - e^{a} \sin \left (d\right )\right )} \sin \left (2 \, b x + 3 \, 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 ) - 8 i \, e^{a} \sin \left (2 \, b x + 2 \, d\right ) - 4 \, e^{a}}{-2 i \, b \cos \left (4 \, b x + 5 \, d\right ) - 4 i \, b \cos \left (2 \, b x + 3 \, d\right ) - 2 i \, b \cos \left (d\right ) + 2 \, b \sin \left (4 \, b x + 5 \, d\right ) + 4 \, b \sin \left (2 \, b x + 3 \, d\right ) + 2 \, b \sin \left (d\right )} \] Input:

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

Output:

-(2*(-I*cos(d)^2*e^a - I*e^a*sin(d)^2 + (-I*cos(d)*e^a - e^a*sin(d))*cos(4 
*b*x + 5*d) + 2*(-I*cos(d)*e^a - e^a*sin(d))*cos(2*b*x + 3*d) + (cos(d)*e^ 
a - I*e^a*sin(d))*sin(4*b*x + 5*d) + 2*(cos(d)*e^a - I*e^a*sin(d))*sin(2*b 
*x + 3*d))*arctan2(sin(2*b*x) - sin(2*d), cos(2*b*x) + cos(2*d)) - 8*cos(2 
*b*x + 2*d)*e^a - (cos(d)^2*e^a + e^a*sin(d)^2 + (cos(d)*e^a - I*e^a*sin(d 
))*cos(4*b*x + 5*d) + 2*(cos(d)*e^a - I*e^a*sin(d))*cos(2*b*x + 3*d) + (I* 
cos(d)*e^a + e^a*sin(d))*sin(4*b*x + 5*d) - 2*(-I*cos(d)*e^a - e^a*sin(d)) 
*sin(2*b*x + 3*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) - 8*I*e^a*sin(2*b*x + 
2*d) - 4*e^a)/(-2*I*b*cos(4*b*x + 5*d) - 4*I*b*cos(2*b*x + 3*d) - 2*I*b*co 
s(d) + 2*b*sin(4*b*x + 5*d) + 4*b*sin(2*b*x + 3*d) + 2*b*sin(d))
 

Giac [A] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.26 \[ \int e^{a+i b x} \sec (d+b x) \tan ^2(d+b x) \, dx=\frac {i \, e^{\left (4 i \, b x + a + 3 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + 2 i \, e^{\left (2 i \, b x + a + i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + i \, e^{\left (a - i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + 4 i \, e^{\left (2 i \, b x + a + i \, d\right )} + 2 i \, e^{\left (a - i \, d\right )}}{b {\left (e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:

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

Output:

(I*e^(4*I*b*x + a + 3*I*d)*log(e^(2*I*b*x + 2*I*d) + 1) + 2*I*e^(2*I*b*x + 
 a + I*d)*log(e^(2*I*b*x + 2*I*d) + 1) + I*e^(a - I*d)*log(e^(2*I*b*x + 2* 
I*d) + 1) + 4*I*e^(2*I*b*x + a + I*d) + 2*I*e^(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. \[ \int e^{a+i b x} \sec (d+b x) \tan ^2(d+b x) \, dx=\int \frac {{\mathrm {e}}^{a+b\,x\,1{}\mathrm {i}}\,{\mathrm {tan}\left (d+b\,x\right )}^2}{\cos \left (d+b\,x\right )} \,d x \] Input:

int((exp(a + b*x*1i)*tan(d + b*x)^2)/cos(d + b*x),x)
 

Output:

int((exp(a + b*x*1i)*tan(d + b*x)^2)/cos(d + b*x), x)
 

Reduce [F]

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

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

Output:

(e**a*( - 6*e**(b*i*x)*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**4 + 12* 
e**(b*i*x)*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**2 - 6*e**(b*i*x)*co 
s(b*x + d)*sin(b*x + d) - 4*e**(b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**4*i 
+ 8*e**(b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**2*i - 4*e**(b*i*x)*cos(b*x + 
 d)*i + e**(b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**4*i - 4*e**(b*i*x)*si 
n(b*x + d)**2*tan((b*x + d)/2)**3 - 6*e**(b*i*x)*sin(b*x + d)**2*tan((b*x 
+ d)/2)**2*i + 12*e**(b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2) - 87*e**(b*i 
*x)*sin(b*x + d)**2*i - 8*e**(b*i*x)*sin(b*x + d)*tan((b*x + d)/2)**4 + 16 
*e**(b*i*x)*sin(b*x + d)*tan((b*x + d)/2)**2 - 8*e**(b*i*x)*sin(b*x + d) - 
 4*e**(b*i*x)*tan((b*x + d)/2)**4*i + 4*e**(b*i*x)*tan((b*x + d)/2)**3 + 1 
2*e**(b*i*x)*tan((b*x + d)/2)**2*i - 12*e**(b*i*x)*tan((b*x + d)/2) + 84*e 
**(b*i*x)*i + 100*int(e**(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 
- 200*int(e**(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 + 100*int(e* 
*(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*b - 100*int(e**(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 + 200*int(e**(b*i*x)/(tan((b*x + d)/2)**6 - 3*tan((b*x + d)/2)**4 + 3*ta 
n((b*x + d)/2)**2 - 1),x)*tan((b*x + d)/2)**2*b - 100*int(e**(b*i*x)/(t...