Integrand size = 27, antiderivative size = 118 \[ \int e^{a+i b x} \csc ^3(d+b x) \sec ^2(d+b x) \, dx=\frac {2 e^{a-i d}}{b \left (1-e^{2 i (d+b x)}\right )^2}-\frac {4 e^{a-i d}}{b \left (1-e^{2 i (d+b x)}\right )}-\frac {2 e^{a-i d}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {2 e^{a-i d} \text {arctanh}\left (e^{2 i (d+b x)}\right )}{b} \] Output:
2*exp(a-I*d)/b/(1-exp(2*I*(b*x+d)))^2-4*exp(a-I*d)/b/(1-exp(2*I*(b*x+d)))- 2*exp(a-I*d)/b/(1+exp(2*I*(b*x+d)))-2*exp(a-I*d)*arctanh(exp(2*I*(b*x+d))) /b
Time = 0.36 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82 \[ \int e^{a+i b x} \csc ^3(d+b x) \sec ^2(d+b x) \, dx=\frac {e^{a-i d} \left (\frac {2}{\left (-1+e^{2 i (d+b x)}\right )^2}+\frac {4}{-1+e^{2 i (d+b x)}}-\frac {2}{1+e^{2 i (d+b x)}}+\log \left (1-e^{2 i (d+b x)}\right )-\log \left (1+e^{2 i (d+b x)}\right )\right )}{b} \] Input:
Integrate[E^(a + I*b*x)*Csc[d + b*x]^3*Sec[d + b*x]^2,x]
Output:
(E^(a - I*d)*(2/(-1 + E^((2*I)*(d + b*x)))^2 + 4/(-1 + E^((2*I)*(d + b*x)) ) - 2/(1 + E^((2*I)*(d + b*x))) + Log[1 - E^((2*I)*(d + b*x))] - Log[1 + E ^((2*I)*(d + b*x))]))/b
Time = 0.56 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.49, 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.
\(\displaystyle \int e^{a+i b x} \csc ^3(b x+d) \sec ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 4974 |
\(\displaystyle \int \left (\frac {8 i e^{a+6 i b x+5 i d}}{\left (-1+e^{2 i (b x+d)}\right )^2}+\frac {4 i e^{a+6 i b x+5 i d}}{\left (1+e^{2 i (b x+d)}\right )^2}-\frac {8 i e^{a+6 i b x+5 i d}}{\left (-1+e^{2 i (b x+d)}\right )^3}-\frac {12 i e^{a+6 i b x+5 i d}}{-1+e^{4 i (b x+d)}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {6 e^{a-i d} \text {arctanh}\left (e^{2 i (b x+d)}\right )}{b}-\frac {4 e^{a-i d}}{b \left (1-e^{2 i (b x+d)}\right )}-\frac {2 e^{a-i d}}{b \left (1+e^{2 i (b x+d)}\right )}+\frac {2 e^{a-i d}}{b \left (1-e^{2 i (b x+d)}\right )^2}+\frac {4 e^{a-i d} \log \left (1-e^{2 i (b x+d)}\right )}{b}-\frac {4 e^{a-i d} \log \left (1+e^{2 i (b x+d)}\right )}{b}\) |
Input:
Int[E^(a + I*b*x)*Csc[d + b*x]^3*Sec[d + b*x]^2,x]
Output:
(2*E^(a - I*d))/(b*(1 - E^((2*I)*(d + b*x)))^2) - (4*E^(a - I*d))/(b*(1 - E^((2*I)*(d + b*x)))) - (2*E^(a - I*d))/(b*(1 + E^((2*I)*(d + b*x)))) + (6 *E^(a - I*d)*ArcTanh[E^((2*I)*(d + b*x))])/b + (4*E^(a - I*d)*Log[1 - E^(( 2*I)*(d + b*x))])/b - (4*E^(a - I*d)*Log[1 + E^((2*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 = 5.07 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{6 i b x} {\mathrm e}^{5 i d} {\mathrm e}^{a}-2 \,{\mathrm e}^{4 i b x} {\mathrm e}^{3 i d} {\mathrm e}^{a}+2 \,{\mathrm e}^{2 i b x} {\mathrm e}^{i d} {\mathrm e}^{a}}{\left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right ) \left (-1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{2} b}-\frac {{\mathrm e}^{-i d} {\mathrm e}^{a} \ln \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}+\frac {{\mathrm e}^{a} {\mathrm e}^{-i d} \ln \left (-1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}\) | \(125\) |
Input:
int(exp(a+I*b*x)*csc(b*x+d)^3*sec(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
2/(1+exp(2*I*(b*x+d)))/(-1+exp(2*I*(b*x+d)))^2/b*(2*exp(6*I*b*x)*exp(5*I*d )*exp(a)-exp(4*I*b*x)*exp(3*I*d)*exp(a)+exp(2*I*b*x)*exp(I*d)*exp(a))-exp( -I*d)*exp(a)/b*ln(1+exp(2*I*(b*x+d)))+exp(a)*exp(-I*d)/b*ln(-1+exp(2*I*(b* x+d)))
Time = 0.09 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.55 \[ \int e^{a+i b x} \csc ^3(d+b x) \sec ^2(d+b x) \, dx=-\frac {{\left (e^{\left (6 i \, b x + a + 5 i \, d\right )} - e^{\left (4 i \, b x + a + 3 i \, d\right )} - e^{\left (2 i \, b x + a + i \, d\right )} + e^{\left (a - i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - {\left (e^{\left (6 i \, b x + a + 5 i \, d\right )} - e^{\left (4 i \, b x + a + 3 i \, d\right )} - e^{\left (2 i \, b x + a + i \, d\right )} + e^{\left (a - i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) - 2 \, e^{\left (4 i \, b x + a + 3 i \, d\right )} - 6 \, e^{\left (2 i \, b x + a + i \, d\right )} + 4 \, e^{\left (a - i \, d\right )}}{b e^{\left (6 i \, b x + 6 i \, d\right )} - b e^{\left (4 i \, b x + 4 i \, d\right )} - b e^{\left (2 i \, b x + 2 i \, d\right )} + b} \] Input:
integrate(exp(a+I*b*x)*csc(b*x+d)^3*sec(b*x+d)^2,x, algorithm="fricas")
Output:
-((e^(6*I*b*x + a + 5*I*d) - e^(4*I*b*x + a + 3*I*d) - e^(2*I*b*x + a + I* d) + e^(a - I*d))*log(e^(2*I*b*x + 2*I*d) + 1) - (e^(6*I*b*x + a + 5*I*d) - e^(4*I*b*x + a + 3*I*d) - e^(2*I*b*x + a + I*d) + e^(a - I*d))*log(e^(2* I*b*x + 2*I*d) - 1) - 2*e^(4*I*b*x + a + 3*I*d) - 6*e^(2*I*b*x + a + I*d) + 4*e^(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. \[ \int e^{a+i b x} \csc ^3(d+b x) \sec ^2(d+b x) \, dx=\text {Timed out} \] Input:
integrate(exp(a+I*b*x)*csc(b*x+d)**3*sec(b*x+d)**2,x)
Output:
Timed out
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1252 vs. \(2 (94) = 188\).
Time = 0.10 (sec) , antiderivative size = 1252, normalized size of antiderivative = 10.61 \[ \int e^{a+i b x} \csc ^3(d+b x) \sec ^2(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(a+I*b*x)*csc(b*x+d)^3*sec(b*x+d)^2,x, algorithm="maxima")
Output:
-(2*(cos(d)^2*e^a + e^a*sin(d)^2 + (cos(d)*e^a - I*e^a*sin(d))*cos(6*b*x + 7*d) - (cos(d)*e^a - I*e^a*sin(d))*cos(4*b*x + 5*d) - (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(6*b*x + 7*d) + (-I*cos(d)*e^a - e^a*sin(d))*sin(4*b*x + 5*d) + (-I*cos(d)*e^a - 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) ) - 2*(cos(d)^2*e^a + e^a*sin(d)^2 + (cos(d)*e^a - I*e^a*sin(d))*cos(6*b*x + 7*d) - (cos(d)*e^a - I*e^a*sin(d))*cos(4*b*x + 5*d) - (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(6*b*x + 7*d ) - (I*cos(d)*e^a + e^a*sin(d))*sin(4*b*x + 5*d) - (I*cos(d)*e^a + e^a*sin (d))*sin(2*b*x + 3*d))*arctan2(sin(b*x) + sin(d), cos(b*x) - cos(d)) - 2*( cos(d)^2*e^a + e^a*sin(d)^2 + (cos(d)*e^a - I*e^a*sin(d))*cos(6*b*x + 7*d) - (cos(d)*e^a - I*e^a*sin(d))*cos(4*b*x + 5*d) - (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(6*b*x + 7*d) - (I* cos(d)*e^a + e^a*sin(d))*sin(4*b*x + 5*d) - (I*cos(d)*e^a + e^a*sin(d))*si n(2*b*x + 3*d))*arctan2(sin(b*x) - sin(d), cos(b*x) + cos(d)) + 4*I*cos(4* b*x + 4*d)*e^a + 12*I*cos(2*b*x + 2*d)*e^a - (I*cos(d)^2*e^a + I*e^a*sin(d )^2 + (I*cos(d)*e^a + e^a*sin(d))*cos(6*b*x + 7*d) + (-I*cos(d)*e^a - e^a* sin(d))*cos(4*b*x + 5*d) + (-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(6*b*x + 7*d) + (cos(d)*e^a - I*e^a*sin(d) )*sin(4*b*x + 5*d) + (cos(d)*e^a - I*e^a*sin(d))*sin(2*b*x + 3*d))*log(...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (94) = 188\).
Time = 0.14 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.07 \[ \int e^{a+i b x} \csc ^3(d+b x) \sec ^2(d+b x) \, dx=-\frac {e^{\left (6 i \, b x + a + 6 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - e^{\left (4 i \, b x + a + 4 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - e^{\left (2 i \, b x + a + 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + e^{a} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - e^{\left (6 i \, b x + a + 6 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) + e^{\left (4 i \, b x + a + 4 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) + e^{\left (2 i \, b x + a + 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) - e^{a} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) - 2 \, e^{\left (4 i \, b x + a + 4 i \, d\right )} - 6 \, e^{\left (2 i \, b x + a + 2 i \, d\right )} + 4 \, e^{a}}{b {\left (e^{\left (6 i \, b x + 7 i \, d\right )} - e^{\left (4 i \, b x + 5 i \, d\right )} - e^{\left (2 i \, b x + 3 i \, d\right )} + e^{\left (i \, d\right )}\right )}} \] Input:
integrate(exp(a+I*b*x)*csc(b*x+d)^3*sec(b*x+d)^2,x, algorithm="giac")
Output:
-(e^(6*I*b*x + a + 6*I*d)*log(e^(2*I*b*x + 2*I*d) + 1) - e^(4*I*b*x + a + 4*I*d)*log(e^(2*I*b*x + 2*I*d) + 1) - e^(2*I*b*x + a + 2*I*d)*log(e^(2*I*b *x + 2*I*d) + 1) + e^a*log(e^(2*I*b*x + 2*I*d) + 1) - e^(6*I*b*x + a + 6*I *d)*log(e^(2*I*b*x + 2*I*d) - 1) + e^(4*I*b*x + a + 4*I*d)*log(e^(2*I*b*x + 2*I*d) - 1) + e^(2*I*b*x + a + 2*I*d)*log(e^(2*I*b*x + 2*I*d) - 1) - e^a *log(e^(2*I*b*x + 2*I*d) - 1) - 2*e^(4*I*b*x + a + 4*I*d) - 6*e^(2*I*b*x + a + 2*I*d) + 4*e^a)/(b*(e^(6*I*b*x + 7*I*d) - e^(4*I*b*x + 5*I*d) - e^(2* I*b*x + 3*I*d) + e^(I*d)))
Timed out. \[ \int e^{a+i b x} \csc ^3(d+b x) \sec ^2(d+b x) \, dx=\int \frac {{\mathrm {e}}^{a+b\,x\,1{}\mathrm {i}}}{{\cos \left (d+b\,x\right )}^2\,{\sin \left (d+b\,x\right )}^3} \,d x \] Input:
int(exp(a + b*x*1i)/(cos(d + b*x)^2*sin(d + b*x)^3),x)
Output:
int(exp(a + b*x*1i)/(cos(d + b*x)^2*sin(d + b*x)^3), x)
\[ \int e^{a+i b x} \csc ^3(d+b x) \sec ^2(d+b x) \, dx=e^{a} \left (\int e^{b i x} \csc \left (b x +d \right )^{3} \sec \left (b x +d \right )^{2}d x \right ) \] Input:
int(exp(a+I*b*x)*csc(b*x+d)^3*sec(b*x+d)^2,x)
Output:
e**a*int(e**(b*i*x)*csc(b*x + d)**3*sec(b*x + d)**2,x)