Integrand size = 29, antiderivative size = 207 \[ \int e^{2 (a+i b x)} \csc ^2(d+b x) \sec ^3(d+b x) \, dx=\frac {2 i e^{2 (a-i d)+i (d+b x)}}{b \left (1+e^{2 i (d+b x)}\right )^2}+\frac {8 i e^{2 (a-i d)+3 i (d+b x)}}{b \left (1-e^{2 i (d+b x)}\right ) \left (1+e^{2 i (d+b x)}\right )^2}+\frac {3 i e^{2 (a-i d)+i (d+b x)}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {i e^{2 a-2 i d} \arctan \left (e^{i (d+b x)}\right )}{b}-\frac {4 i e^{2 a-2 i d} \text {arctanh}\left (e^{i (d+b x)}\right )}{b} \] Output:
2*I*exp(2*a-2*I*d+I*(b*x+d))/b/(1+exp(2*I*(b*x+d)))^2+8*I*exp(2*a-2*I*d+3* I*(b*x+d))/b/(1-exp(2*I*(b*x+d)))/(1+exp(2*I*(b*x+d)))^2+3*I*exp(2*a-2*I*d +I*(b*x+d))/b/(1+exp(2*I*(b*x+d)))-I*exp(2*a-2*I*d)*arctan(exp(I*(b*x+d))) /b-4*I*exp(2*a-2*I*d)*arctanh(exp(I*(b*x+d)))/b
Time = 0.52 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.72 \[ \int e^{2 (a+i b x)} \csc ^2(d+b x) \sec ^3(d+b x) \, dx=-\frac {i e^{2 a-2 i d} \left (\frac {2 e^{i (d+b x)}}{-1+e^{2 i (d+b x)}}+\frac {2 e^{i (d+b x)}}{\left (1+e^{2 i (d+b x)}\right )^2}-\frac {5 e^{i (d+b x)}}{1+e^{2 i (d+b x)}}+\arctan \left (e^{i (d+b x)}\right )-2 \log \left (1-e^{i (d+b x)}\right )+2 \log \left (1+e^{i (d+b x)}\right )\right )}{b} \] Input:
Integrate[E^(2*(a + I*b*x))*Csc[d + b*x]^2*Sec[d + b*x]^3,x]
Output:
((-I)*E^(2*a - (2*I)*d)*((2*E^(I*(d + b*x)))/(-1 + E^((2*I)*(d + b*x))) + (2*E^(I*(d + b*x)))/(1 + E^((2*I)*(d + b*x)))^2 - (5*E^(I*(d + b*x)))/(1 + E^((2*I)*(d + b*x))) + ArcTan[E^(I*(d + b*x))] - 2*Log[1 - E^(I*(d + b*x) )] + 2*Log[1 + E^(I*(d + b*x))]))/b
Time = 0.62 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.41, 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.
\(\displaystyle \int e^{2 (a+i b x)} \csc ^2(b x+d) \sec ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 4974 |
\(\displaystyle \int \left (-\frac {4 e^{2 a+7 i b x+5 i d}}{\left (-1+e^{2 i (b x+d)}\right )^2}-\frac {8 e^{2 a+7 i b x+5 i d}}{\left (1+e^{2 i (b x+d)}\right )^2}-\frac {8 e^{2 a+7 i b x+5 i d}}{\left (1+e^{2 i (b x+d)}\right )^3}+\frac {12 e^{2 a+7 i b x+5 i d}}{-1+e^{4 i (b x+d)}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i e^{2 a-2 i d} \arctan \left (e^{i (b x+d)}\right )}{b}-\frac {4 i e^{2 a-2 i d} \text {arctanh}\left (e^{i (b x+d)}\right )}{b}+\frac {5 i e^{2 (a-i d)+i (b x+d)}}{b}+\frac {6 i e^{2 (a-i d)+3 i (b x+d)}}{b}+\frac {2 i e^{2 (a-i d)+5 i (b x+d)}}{b \left (1-e^{2 i (b x+d)}\right )}-\frac {5 i e^{2 (a-i d)+3 i (b x+d)}}{b \left (1+e^{2 i (b x+d)}\right )}-\frac {4 i e^{2 (a-i d)+5 i (b x+d)}}{b \left (1+e^{2 i (b x+d)}\right )}-\frac {2 i e^{2 (a-i d)+5 i (b x+d)}}{b \left (1+e^{2 i (b x+d)}\right )^2}\) |
Input:
Int[E^(2*(a + I*b*x))*Csc[d + b*x]^2*Sec[d + b*x]^3,x]
Output:
((5*I)*E^(2*(a - I*d) + I*(d + b*x)))/b + ((6*I)*E^(2*(a - I*d) + (3*I)*(d + b*x)))/b + ((2*I)*E^(2*(a - I*d) + (5*I)*(d + b*x)))/(b*(1 - E^((2*I)*( d + b*x)))) - ((2*I)*E^(2*(a - I*d) + (5*I)*(d + b*x)))/(b*(1 + E^((2*I)*( d + b*x)))^2) - ((5*I)*E^(2*(a - I*d) + (3*I)*(d + b*x)))/(b*(1 + E^((2*I) *(d + b*x)))) - ((4*I)*E^(2*(a - I*d) + (5*I)*(d + b*x)))/(b*(1 + E^((2*I) *(d + b*x)))) - (I*E^(2*a - (2*I)*d)*ArcTan[E^(I*(d + b*x))])/b - ((4*I)*E ^(2*a - (2*I)*d)*ArcTanh[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 = 4.54 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.75
method | result | size |
risch | \(-\frac {i \left (5 \,{\mathrm e}^{7 i b x} {\mathrm e}^{5 i d} {\mathrm e}^{2 a}+2 \,{\mathrm e}^{5 i b x} {\mathrm e}^{3 i d} {\mathrm e}^{2 a}+{\mathrm e}^{3 i b x} {\mathrm e}^{i d} {\mathrm e}^{2 a}\right )}{\left (-1+{\mathrm e}^{2 i \left (b x +d \right )}\right ) \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{2} b}-\frac {4 i {\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \operatorname {arctanh}\left ({\mathrm e}^{i \left (b x +d \right )}\right )}{b}-\frac {i {\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \arctan \left ({\mathrm e}^{i \left (b x +d \right )}\right )}{b}+\frac {5 i {\mathrm e}^{2 a} {\mathrm e}^{i b x} {\mathrm e}^{-i d}}{b}\) | \(156\) |
Input:
int(exp(2*a+2*I*b*x)*csc(b*x+d)^2*sec(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-I/(-1+exp(2*I*(b*x+d)))/(1+exp(2*I*(b*x+d)))^2/b*(5*exp(7*I*b*x)*exp(5*I* d)*exp(2*a)+2*exp(5*I*b*x)*exp(3*I*d)*exp(2*a)+exp(3*I*b*x)*exp(I*d)*exp(2 *a))-4*I*exp(2*a)/b*exp(-2*I*d)*arctanh(exp(I*(b*x+d)))-I*exp(2*a)/b*exp(- 2*I*d)*arctan(exp(I*(b*x+d)))+5*I/b*exp(2*a)*exp(I*b*x)*exp(-I*d)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (150) = 300\).
Time = 0.09 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.60 \[ \int e^{2 (a+i b x)} \csc ^2(d+b x) \sec ^3(d+b x) \, dx=-\frac {4 \, {\left (i \, e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} + i \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - i \, e^{\left (2 i \, b x + 2 \, a\right )} - i \, e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + 1\right ) - {\left (e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} + e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - e^{\left (2 i \, b x + 2 \, a\right )} - e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) + {\left (e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} + e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - e^{\left (2 i \, b x + 2 \, a\right )} - e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + 4 \, {\left (-i \, e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} - i \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + i \, e^{\left (2 i \, b x + 2 \, a\right )} + i \, e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - 1\right ) - 6 i \, e^{\left (5 i \, b x + 2 \, a + 3 i \, d\right )} + 12 i \, e^{\left (3 i \, b x + 2 \, a + i \, d\right )} + 10 i \, e^{\left (i \, b x + 2 \, a - i \, d\right )}}{2 \, {\left (b e^{\left (6 i \, b x + 6 i \, d\right )} + b e^{\left (4 i \, b x + 4 i \, d\right )} - b e^{\left (2 i \, b x + 2 i \, d\right )} - b\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*csc(b*x+d)^2*sec(b*x+d)^3,x, algorithm="fricas" )
Output:
-1/2*(4*(I*e^(6*I*b*x + 2*a + 4*I*d) + I*e^(4*I*b*x + 2*a + 2*I*d) - I*e^( 2*I*b*x + 2*a) - I*e^(2*a - 2*I*d))*log(e^(I*b*x + I*d) + 1) - (e^(6*I*b*x + 2*a + 4*I*d) + e^(4*I*b*x + 2*a + 2*I*d) - e^(2*I*b*x + 2*a) - e^(2*a - 2*I*d))*log(e^(I*b*x + I*d) + I) + (e^(6*I*b*x + 2*a + 4*I*d) + e^(4*I*b* x + 2*a + 2*I*d) - e^(2*I*b*x + 2*a) - e^(2*a - 2*I*d))*log(e^(I*b*x + I*d ) - I) + 4*(-I*e^(6*I*b*x + 2*a + 4*I*d) - I*e^(4*I*b*x + 2*a + 2*I*d) + I *e^(2*I*b*x + 2*a) + I*e^(2*a - 2*I*d))*log(e^(I*b*x + I*d) - 1) - 6*I*e^( 5*I*b*x + 2*a + 3*I*d) + 12*I*e^(3*I*b*x + 2*a + I*d) + 10*I*e^(I*b*x + 2* a - I*d))/(b*e^(6*I*b*x + 6*I*d) + b*e^(4*I*b*x + 4*I*d) - b*e^(2*I*b*x + 2*I*d) - b)
Timed out. \[ \int e^{2 (a+i b x)} \csc ^2(d+b x) \sec ^3(d+b x) \, dx=\text {Timed out} \] Input:
integrate(exp(2*a+2*I*b*x)*csc(b*x+d)**2*sec(b*x+d)**3,x)
Output:
Timed out
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1938 vs. \(2 (150) = 300\).
Time = 0.21 (sec) , antiderivative size = 1938, normalized size of antiderivative = 9.36 \[ \int e^{2 (a+i b x)} \csc ^2(d+b x) \sec ^3(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(2*a+2*I*b*x)*csc(b*x+d)^2*sec(b*x+d)^3,x, algorithm="maxima" )
Output:
(2*((cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*cos(6*b*x + 7*d) + (cos(2*d)*e ^(2*a) - I*e^(2*a)*sin(2*d))*cos(4*b*x + 5*d) - (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(6*b*x + 7*d) - (-I*cos( 2*d)*e^(2*a) - e^(2*a)*sin(2*d))*sin(4*b*x + 5*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))*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*co s(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)) - 8*((-I*cos(2*d )*e^(2*a) - e^(2*a)*sin(2*d))*cos(6*b*x + 7*d) + (-I*cos(2*d)*e^(2*a) - e^ (2*a)*sin(2*d))*cos(4*b*x + 5*d) + (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(6*b*x + 7*d) + (cos(2*d)*e^(2*a) - I* e^(2*a)*sin(2*d))*sin(4*b*x + 5*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))*arctan 2(sin(b*x) + sin(d), cos(b*x) - cos(d)) - 8*((I*cos(2*d)*e^(2*a) + e^(2*a) *sin(2*d))*cos(6*b*x + 7*d) + (I*cos(2*d)*e^(2*a) + e^(2*a)*sin(2*d))*cos( 4*b*x + 5*d) + (-I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*cos(2*b*x + 3*d...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (150) = 300\).
Time = 0.19 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.32 \[ \int e^{2 (a+i b x)} \csc ^2(d+b x) \sec ^3(d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(2*a+2*I*b*x)*csc(b*x+d)^2*sec(b*x+d)^3,x, algorithm="giac")
Output:
1/2*(-4*I*e^(6*I*b*x + 2*a + 4*I*d)*log(e^(I*b*x + I*d) + 1) - 4*I*e^(4*I* b*x + 2*a + 2*I*d)*log(e^(I*b*x + I*d) + 1) + 4*I*e^(2*I*b*x + 2*a)*log(e^ (I*b*x + I*d) + 1) + 4*I*e^(2*a - 2*I*d)*log(e^(I*b*x + I*d) + 1) + 4*I*e^ (6*I*b*x + 2*a + 4*I*d)*log(e^(I*b*x + I*d) - 1) + 4*I*e^(4*I*b*x + 2*a + 2*I*d)*log(e^(I*b*x + I*d) - 1) - 4*I*e^(2*I*b*x + 2*a)*log(e^(I*b*x + I*d ) - 1) - 4*I*e^(2*a - 2*I*d)*log(e^(I*b*x + I*d) - 1) + e^(6*I*b*x + 2*a + 4*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) - 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) - e^(6*I*b*x + 2*a + 4*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) + 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) + 6*I*e^(5*I*b*x + 2*a + 3*I*d) - 12*I*e^(3*I*b*x + 2*a + I*d) - 10*I*e^(I*b*x + 2*a - I*d))/(b*(e^(6*I*b*x + 6*I*d) + e^(4*I*b*x + 4*I*d) - e^(2*I*b*x + 2*I*d) - 1))
Timed out. \[ \int e^{2 (a+i b x)} \csc ^2(d+b x) \sec ^3(d+b x) \, dx=\int \frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}}{{\cos \left (d+b\,x\right )}^3\,{\sin \left (d+b\,x\right )}^2} \,d x \] Input:
int(exp(2*a + b*x*2i)/(cos(d + b*x)^3*sin(d + b*x)^2),x)
Output:
int(exp(2*a + b*x*2i)/(cos(d + b*x)^3*sin(d + b*x)^2), x)
\[ \int e^{2 (a+i b x)} \csc ^2(d+b x) \sec ^3(d+b x) \, dx=e^{2 a} \left (\int e^{2 b i x} \csc \left (b x +d \right )^{2} \sec \left (b x +d \right )^{3}d x \right ) \] Input:
int(exp(2*a+2*I*b*x)*csc(b*x+d)^2*sec(b*x+d)^3,x)
Output:
e**(2*a)*int(e**(2*b*i*x)*csc(b*x + d)**2*sec(b*x + d)**3,x)