\(\int e^{2 (a+i b x)} \csc ^5(d+b x) \sec ^3(d+b x) \, dx\) [86]

Optimal result

Integrand size = 29, antiderivative size = 225 \[ \int e^{2 (a+i b x)} \csc ^5(d+b x) \sec ^3(d+b x) \, dx=-\frac {4 e^{2 a-2 i d}}{b \left (1-e^{2 i (d+b x)}\right )^4}+\frac {40 e^{2 a-2 i d}}{3 b \left (1-e^{2 i (d+b x)}\right )^3}-\frac {12 e^{2 a-2 i d}}{b \left (1-e^{2 i (d+b x)}\right )^2}-\frac {4 e^{2 a-2 i d}}{b \left (1-e^{2 i (d+b x)}\right )}+\frac {2 e^{2 a-2 i d}}{b \left (1+e^{2 i (d+b x)}\right )^2}-\frac {6 e^{2 a-2 i d}}{b \left (1+e^{2 i (d+b x)}\right )}+\frac {2 e^{2 a-2 i d} \text {arctanh}\left (e^{2 i (d+b x)}\right )}{b} \] Output:

-4*exp(2*a-2*I*d)/b/(1-exp(2*I*(b*x+d)))^4+40/3*exp(2*a-2*I*d)/b/(1-exp(2* 
I*(b*x+d)))^3-12*exp(2*a-2*I*d)/b/(1-exp(2*I*(b*x+d)))^2-4*exp(2*a-2*I*d)/ 
b/(1-exp(2*I*(b*x+d)))+2*exp(2*a-2*I*d)/b/(1+exp(2*I*(b*x+d)))^2-6*exp(2*a 
-2*I*d)/b/(1+exp(2*I*(b*x+d)))+2*exp(2*a-2*I*d)*arctanh(exp(2*I*(b*x+d)))/ 
b
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.67 \[ \int e^{2 (a+i b x)} \csc ^5(d+b x) \sec ^3(d+b x) \, dx=\frac {e^{2 a-2 i d} \left (\frac {-32+58 e^{2 i (d+b x)}+44 e^{4 i (d+b x)}-124 e^{6 i (d+b x)}+12 e^{8 i (d+b x)}-6 e^{10 i (d+b x)}}{\left (-1+e^{2 i (d+b x)}\right )^4 \left (1+e^{2 i (d+b x)}\right )^2}-3 \log \left (1-e^{2 i (d+b x)}\right )+3 \log \left (1+e^{2 i (d+b x)}\right )\right )}{3 b} \] Input:

Integrate[E^(2*(a + I*b*x))*Csc[d + b*x]^5*Sec[d + b*x]^3,x]
 

Output:

(E^(2*a - (2*I)*d)*((-32 + 58*E^((2*I)*(d + b*x)) + 44*E^((4*I)*(d + b*x)) 
 - 124*E^((6*I)*(d + b*x)) + 12*E^((8*I)*(d + b*x)) - 6*E^((10*I)*(d + b*x 
)))/((-1 + E^((2*I)*(d + b*x)))^4*(1 + E^((2*I)*(d + b*x)))^2) - 3*Log[1 - 
 E^((2*I)*(d + b*x))] + 3*Log[1 + E^((2*I)*(d + b*x))]))/(3*b)
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.28, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, 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.

\(\displaystyle \int e^{2 (a+i b x)} \csc ^5(b x+d) \sec ^3(b x+d) \, dx\)

\(\Big \downarrow \) 4974

\(\displaystyle \int \left (-\frac {40 i e^{2 (a+4 i d)+10 i b x}}{\left (-1+e^{2 i (b x+d)}\right )^2}-\frac {20 i e^{2 (a+4 i d)+10 i b x}}{\left (1+e^{2 i (b x+d)}\right )^2}+\frac {48 i e^{2 (a+4 i d)+10 i b x}}{\left (-1+e^{2 i (b x+d)}\right )^3}-\frac {8 i e^{2 (a+4 i d)+10 i b x}}{\left (1+e^{2 i (b x+d)}\right )^3}-\frac {48 i e^{2 (a+4 i d)+10 i b x}}{\left (-1+e^{2 i (b x+d)}\right )^4}+\frac {32 i e^{2 (a+4 i d)+10 i b x}}{\left (-1+e^{2 i (b x+d)}\right )^5}+\frac {60 i e^{2 (a+4 i d)+10 i b x}}{-1+e^{4 i (b x+d)}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {30 e^{2 a-2 i d} \text {arctanh}\left (e^{2 i (b x+d)}\right )}{b}-\frac {4 e^{2 a-2 i d}}{b \left (1-e^{2 i (b x+d)}\right )}-\frac {6 e^{2 a-2 i d}}{b \left (1+e^{2 i (b x+d)}\right )}-\frac {12 e^{2 a-2 i d}}{b \left (1-e^{2 i (b x+d)}\right )^2}+\frac {2 e^{2 a-2 i d}}{b \left (1+e^{2 i (b x+d)}\right )^2}+\frac {40 e^{2 a-2 i d}}{3 b \left (1-e^{2 i (b x+d)}\right )^3}-\frac {4 e^{2 a-2 i d}}{b \left (1-e^{2 i (b x+d)}\right )^4}-\frac {16 e^{2 a-2 i d} \log \left (1-e^{2 i (b x+d)}\right )}{b}+\frac {16 e^{2 a-2 i d} \log \left (1+e^{2 i (b x+d)}\right )}{b}\)

Input:

Int[E^(2*(a + I*b*x))*Csc[d + b*x]^5*Sec[d + b*x]^3,x]
 

Output:

(-4*E^(2*a - (2*I)*d))/(b*(1 - E^((2*I)*(d + b*x)))^4) + (40*E^(2*a - (2*I 
)*d))/(3*b*(1 - E^((2*I)*(d + b*x)))^3) - (12*E^(2*a - (2*I)*d))/(b*(1 - E 
^((2*I)*(d + b*x)))^2) - (4*E^(2*a - (2*I)*d))/(b*(1 - E^((2*I)*(d + b*x)) 
)) + (2*E^(2*a - (2*I)*d))/(b*(1 + E^((2*I)*(d + b*x)))^2) - (6*E^(2*a - ( 
2*I)*d))/(b*(1 + E^((2*I)*(d + b*x)))) - (30*E^(2*a - (2*I)*d)*ArcTanh[E^( 
(2*I)*(d + b*x))])/b - (16*E^(2*a - (2*I)*d)*Log[1 - E^((2*I)*(d + b*x))]) 
/b + (16*E^(2*a - (2*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 = 47.30 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {\frac {32 \,{\mathrm e}^{2 a} {\mathrm e}^{12 i b x} {\mathrm e}^{10 i d}}{3}-\frac {70 \,{\mathrm e}^{10 i b x} {\mathrm e}^{8 i d} {\mathrm e}^{2 a}}{3}-\frac {20 \,{\mathrm e}^{2 a} {\mathrm e}^{8 i b x} {\mathrm e}^{6 i d}}{3}+\frac {4 \,{\mathrm e}^{6 i b x} {\mathrm e}^{4 i d} {\mathrm e}^{2 a}}{3}+4 \,{\mathrm e}^{4 i b x} {\mathrm e}^{2 i d} {\mathrm e}^{2 a}-2 \,{\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{b \left (-1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{4} \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{2}}+\frac {{\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \ln \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}-\frac {{\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \ln \left (-1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}\)	\(182\)

Input:

int(exp(2*a+2*I*b*x)*csc(b*x+d)^5*sec(b*x+d)^3,x,method=_RETURNVERBOSE)
 

Output:

2/3/b/(-1+exp(2*I*(b*x+d)))^4/(1+exp(2*I*(b*x+d)))^2*(16*exp(2*a)*exp(12*I 
*b*x)*exp(10*I*d)-35*exp(10*I*b*x)*exp(8*I*d)*exp(2*a)-10*exp(2*a)*exp(8*I 
*b*x)*exp(6*I*d)+2*exp(6*I*b*x)*exp(4*I*d)*exp(2*a)+6*exp(4*I*b*x)*exp(2*I 
*d)*exp(2*a)-3*exp(2*a)*exp(2*I*b*x))+exp(2*a)/b*exp(-2*I*d)*ln(1+exp(2*I* 
(b*x+d)))-exp(2*a)/b*exp(-2*I*d)*ln(-1+exp(2*I*(b*x+d)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.60 \[ \int e^{2 (a+i b x)} \csc ^5(d+b x) \sec ^3(d+b x) \, dx=\frac {3 \, {\left (e^{\left (12 i \, b x + 2 \, a + 10 i \, d\right )} - 2 \, e^{\left (10 i \, b x + 2 \, a + 8 i \, d\right )} - e^{\left (8 i \, b x + 2 \, a + 6 i \, d\right )} + 4 \, e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} - e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - 2 \, e^{\left (2 i \, b x + 2 \, a\right )} + e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - 3 \, {\left (e^{\left (12 i \, b x + 2 \, a + 10 i \, d\right )} - 2 \, e^{\left (10 i \, b x + 2 \, a + 8 i \, d\right )} - e^{\left (8 i \, b x + 2 \, a + 6 i \, d\right )} + 4 \, e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} - e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - 2 \, e^{\left (2 i \, b x + 2 \, a\right )} + e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) - 6 \, e^{\left (10 i \, b x + 2 \, a + 8 i \, d\right )} + 12 \, e^{\left (8 i \, b x + 2 \, a + 6 i \, d\right )} - 124 \, e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} + 44 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + 58 \, e^{\left (2 i \, b x + 2 \, a\right )} - 32 \, e^{\left (2 \, a - 2 i \, d\right )}}{3 \, {\left (b e^{\left (12 i \, b x + 12 i \, d\right )} - 2 \, b e^{\left (10 i \, b x + 10 i \, d\right )} - b e^{\left (8 i \, b x + 8 i \, d\right )} + 4 \, b e^{\left (6 i \, b x + 6 i \, d\right )} - b e^{\left (4 i \, b x + 4 i \, d\right )} - 2 \, b e^{\left (2 i \, b x + 2 i \, d\right )} + b\right )}} \] Input:

integrate(exp(2*a+2*I*b*x)*csc(b*x+d)^5*sec(b*x+d)^3,x, algorithm="fricas" 
)
 

Output:

1/3*(3*(e^(12*I*b*x + 2*a + 10*I*d) - 2*e^(10*I*b*x + 2*a + 8*I*d) - e^(8* 
I*b*x + 2*a + 6*I*d) + 4*e^(6*I*b*x + 2*a + 4*I*d) - 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^(2*I*b*x + 2*I*d) + 1) 
 - 3*(e^(12*I*b*x + 2*a + 10*I*d) - 2*e^(10*I*b*x + 2*a + 8*I*d) - e^(8*I* 
b*x + 2*a + 6*I*d) + 4*e^(6*I*b*x + 2*a + 4*I*d) - 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^(2*I*b*x + 2*I*d) - 1) - 
 6*e^(10*I*b*x + 2*a + 8*I*d) + 12*e^(8*I*b*x + 2*a + 6*I*d) - 124*e^(6*I* 
b*x + 2*a + 4*I*d) + 44*e^(4*I*b*x + 2*a + 2*I*d) + 58*e^(2*I*b*x + 2*a) - 
 32*e^(2*a - 2*I*d))/(b*e^(12*I*b*x + 12*I*d) - 2*b*e^(10*I*b*x + 10*I*d) 
- b*e^(8*I*b*x + 8*I*d) + 4*b*e^(6*I*b*x + 6*I*d) - b*e^(4*I*b*x + 4*I*d) 
- 2*b*e^(2*I*b*x + 2*I*d) + b)
 

Sympy [F(-1)]

Timed out. \[ \int e^{2 (a+i b x)} \csc ^5(d+b x) \sec ^3(d+b x) \, dx=\text {Timed out} \] Input:

integrate(exp(2*a+2*I*b*x)*csc(b*x+d)**5*sec(b*x+d)**3,x)
 

Output:

Timed out
 

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. 2888 vs. \(2 (180) = 360\).

Time = 0.25 (sec) , antiderivative size = 2888, normalized size of antiderivative = 12.84 \[ \int e^{2 (a+i b x)} \csc ^5(d+b x) \sec ^3(d+b x) \, dx=\text {Too large to display} \] Input:

integrate(exp(2*a+2*I*b*x)*csc(b*x+d)^5*sec(b*x+d)^3,x, algorithm="maxima" 
)
 

Output:

(6*(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(12*b*x + 14*d) - 2*(cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d)) 
*cos(10*b*x + 12*d) - (cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*cos(8*b*x + 
10*d) + 4*(cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*cos(6*b*x + 8*d) - (cos( 
2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*cos(4*b*x + 6*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(12*b*x + 14*d) + 2*(-I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*sin 
(10*b*x + 12*d) + (-I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*sin(8*b*x + 10* 
d) + 4*(I*cos(2*d)*e^(2*a) + e^(2*a)*sin(2*d))*sin(6*b*x + 8*d) + (-I*cos( 
2*d)*e^(2*a) - e^(2*a)*sin(2*d))*sin(4*b*x + 6*d) + 2*(-I*cos(2*d)*e^(2*a) 
 - 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)) - 6*(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(12*b*x + 14*d) - 2*(cos(2*d)*e^(2*a) - 
 I*e^(2*a)*sin(2*d))*cos(10*b*x + 12*d) - (cos(2*d)*e^(2*a) - I*e^(2*a)*si 
n(2*d))*cos(8*b*x + 10*d) + 4*(cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*cos( 
6*b*x + 8*d) - (cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*cos(4*b*x + 6*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(12*b*x + 14*d) - 2*(I*cos(2*d)*e^(2*a) + e 
^(2*a)*sin(2*d))*sin(10*b*x + 12*d) - (I*cos(2*d)*e^(2*a) + e^(2*a)*sin(2* 
d))*sin(8*b*x + 10*d) - 4*(-I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*sin(...
 

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. 501 vs. \(2 (180) = 360\).

Time = 0.20 (sec) , antiderivative size = 501, normalized size of antiderivative = 2.23 \[ \int e^{2 (a+i b x)} \csc ^5(d+b x) \sec ^3(d+b x) \, dx =\text {Too large to display} \] Input:

integrate(exp(2*a+2*I*b*x)*csc(b*x+d)^5*sec(b*x+d)^3,x, algorithm="giac")
 

Output:

1/3*(3*e^(12*I*b*x + 2*a + 10*I*d)*log(e^(2*I*b*x + 2*I*d) + 1) - 6*e^(10* 
I*b*x + 2*a + 8*I*d)*log(e^(2*I*b*x + 2*I*d) + 1) - 3*e^(8*I*b*x + 2*a + 6 
*I*d)*log(e^(2*I*b*x + 2*I*d) + 1) + 12*e^(6*I*b*x + 2*a + 4*I*d)*log(e^(2 
*I*b*x + 2*I*d) + 1) - 3*e^(4*I*b*x + 2*a + 2*I*d)*log(e^(2*I*b*x + 2*I*d) 
 + 1) - 6*e^(2*I*b*x + 2*a)*log(e^(2*I*b*x + 2*I*d) + 1) + 3*e^(2*a - 2*I* 
d)*log(e^(2*I*b*x + 2*I*d) + 1) - 3*e^(12*I*b*x + 2*a + 10*I*d)*log(e^(2*I 
*b*x + 2*I*d) - 1) + 6*e^(10*I*b*x + 2*a + 8*I*d)*log(e^(2*I*b*x + 2*I*d) 
- 1) + 3*e^(8*I*b*x + 2*a + 6*I*d)*log(e^(2*I*b*x + 2*I*d) - 1) - 12*e^(6* 
I*b*x + 2*a + 4*I*d)*log(e^(2*I*b*x + 2*I*d) - 1) + 3*e^(4*I*b*x + 2*a + 2 
*I*d)*log(e^(2*I*b*x + 2*I*d) - 1) + 6*e^(2*I*b*x + 2*a)*log(e^(2*I*b*x + 
2*I*d) - 1) - 3*e^(2*a - 2*I*d)*log(e^(2*I*b*x + 2*I*d) - 1) - 6*e^(10*I*b 
*x + 2*a + 8*I*d) + 12*e^(8*I*b*x + 2*a + 6*I*d) - 124*e^(6*I*b*x + 2*a + 
4*I*d) + 44*e^(4*I*b*x + 2*a + 2*I*d) + 58*e^(2*I*b*x + 2*a) - 32*e^(2*a - 
 2*I*d))/(b*(e^(12*I*b*x + 12*I*d) - 2*e^(10*I*b*x + 10*I*d) - e^(8*I*b*x 
+ 8*I*d) + 4*e^(6*I*b*x + 6*I*d) - e^(4*I*b*x + 4*I*d) - 2*e^(2*I*b*x + 2* 
I*d) + 1))
 

Mupad [F(-1)]

Timed out. \[ \int e^{2 (a+i b x)} \csc ^5(d+b x) \sec ^3(d+b x) \, dx=\int \frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}}{{\cos \left (d+b\,x\right )}^3\,{\sin \left (d+b\,x\right )}^5} \,d x \] Input:

int(exp(2*a + b*x*2i)/(cos(d + b*x)^3*sin(d + b*x)^5),x)
 

Output:

int(exp(2*a + b*x*2i)/(cos(d + b*x)^3*sin(d + b*x)^5), x)
 

Reduce [F]

\[ \int e^{2 (a+i b x)} \csc ^5(d+b x) \sec ^3(d+b x) \, dx=e^{2 a} \left (\int e^{2 b i x} \csc \left (b x +d \right )^{5} \sec \left (b x +d \right )^{3}d x \right ) \] Input:

int(exp(2*a+2*I*b*x)*csc(b*x+d)^5*sec(b*x+d)^3,x)
 

Output:

e**(2*a)*int(e**(2*b*i*x)*csc(b*x + d)**5*sec(b*x + d)**3,x)