Integrand size = 19, antiderivative size = 69 \[ \int e^{a+i b x} \sec ^2(d+b x) \, dx=\frac {2 i e^{a-i d+i (d+b x)}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {2 i e^{a-i d} \arctan \left (e^{i (d+b x)}\right )}{b} \] Output:
2*I*exp(a-I*d+I*(b*x+d))/b/(1+exp(2*I*(b*x+d)))-2*I*exp(a-I*d)*arctan(exp( I*(b*x+d)))/b
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81 \[ \int e^{a+i b x} \sec ^2(d+b x) \, dx=\frac {2 i e^a \left (\frac {e^{i b x}}{1+e^{2 i (d+b x)}}-e^{-i d} \arctan \left (e^{i (d+b x)}\right )\right )}{b} \] Input:
Integrate[E^(a + I*b*x)*Sec[d + b*x]^2,x]
Output:
((2*I)*E^a*(E^(I*b*x)/(1 + E^((2*I)*(d + b*x))) - ArcTan[E^(I*(d + b*x))]/ E^(I*d)))/b
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {4951}
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.
\(\displaystyle \int e^{a+i b x} \sec ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 4951 |
\(\displaystyle \frac {2 i e^{a-2 i (b x+d)+i b x} \left (\frac {e^{2 i (b x+d)}}{1+e^{2 i (b x+d)}}-e^{i (b x+d)} \arctan \left (e^{i (b x+d)}\right )\right )}{b}\) |
Input:
Int[E^(a + I*b*x)*Sec[d + b*x]^2,x]
Output:
((2*I)*E^(a + I*b*x - (2*I)*(d + b*x))*(E^((2*I)*(d + b*x))/(1 + E^((2*I)* (d + b*x))) - E^(I*(d + b*x))*ArcTan[E^(I*(d + b*x))]))/b
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symb ol] :> Simp[2^n*E^(I*n*(d + e*x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hy pergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)), 1 + n/2 - I*b*c*(Log[F]/(2*e )), -E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {2 i {\mathrm e}^{a} {\mathrm e}^{i b x}}{b \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}-\frac {2 i {\mathrm e}^{a} {\mathrm e}^{-i d} \arctan \left ({\mathrm e}^{i \left (b x +d \right )}\right )}{b}\) | \(52\) |
Input:
int(exp(a+I*b*x)*sec(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
2*I*exp(a)*exp(I*b*x)/b/(1+exp(2*I*(b*x+d)))-2*I*exp(a)*exp(-I*d)*arctan(e xp(I*(b*x+d)))/b
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int e^{a+i b x} \sec ^2(d+b x) \, dx=\frac {{\left (e^{\left (2 i \, b x + a + i \, d\right )} + e^{\left (a - i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) - {\left (e^{\left (2 i \, b x + a + i \, d\right )} + e^{\left (a - i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + 2 i \, e^{\left (i \, b x + a\right )}}{b e^{\left (2 i \, b x + 2 i \, d\right )} + b} \] Input:
integrate(exp(a+I*b*x)*sec(b*x+d)^2,x, algorithm="fricas")
Output:
((e^(2*I*b*x + a + I*d) + e^(a - I*d))*log(e^(I*b*x + I*d) + I) - (e^(2*I* b*x + a + I*d) + e^(a - I*d))*log(e^(I*b*x + I*d) - I) + 2*I*e^(I*b*x + a) )/(b*e^(2*I*b*x + 2*I*d) + b)
\[ \int e^{a+i b x} \sec ^2(d+b x) \, dx=e^{a} \int e^{i b x} \sec ^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(a+I*b*x)*sec(b*x+d)**2,x)
Output:
exp(a)*Integral(exp(I*b*x)*sec(b*x + d)**2, x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (47) = 94\).
Time = 0.17 (sec) , antiderivative size = 446, normalized size of antiderivative = 6.46 \[ \int e^{a+i b x} \sec ^2(d+b x) \, dx=\frac {2 \, {\left ({\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \cos \left (2 \, b x + 2 \, d\right ) + \cos \left (d\right ) e^{a} - {\left (-i \, \cos \left (d\right ) e^{a} - e^{a} \sin \left (d\right )\right )} \sin \left (2 \, b x + 2 \, d\right ) - i \, e^{a} \sin \left (d\right )\right )} \arctan \left (\frac {2 \, {\left (\cos \left (b x + 2 \, d\right ) \cos \left (d\right ) + \sin \left (b x + 2 \, d\right ) \sin \left (d\right )\right )}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}, \frac {\cos \left (b x + 2 \, d\right )^{2} - \cos \left (d\right )^{2} + \sin \left (b x + 2 \, d\right )^{2} - \sin \left (d\right )^{2}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}\right ) + 4 \, \cos \left (b x\right ) e^{a} + {\left ({\left (i \, \cos \left (d\right ) e^{a} + e^{a} \sin \left (d\right )\right )} \cos \left (2 \, b x + 2 \, d\right ) + i \, \cos \left (d\right ) e^{a} - {\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \sin \left (2 \, b x + 2 \, d\right ) + e^{a} \sin \left (d\right )\right )} \log \left (\frac {\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} - 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} + 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}\right ) + 4 i \, e^{a} \sin \left (b x\right )}{-2 i \, b \cos \left (2 \, b x + 2 \, d\right ) + 2 \, b \sin \left (2 \, b x + 2 \, d\right ) - 2 i \, b} \] Input:
integrate(exp(a+I*b*x)*sec(b*x+d)^2,x, algorithm="maxima")
Output:
(2*((cos(d)*e^a - I*e^a*sin(d))*cos(2*b*x + 2*d) + cos(d)*e^a - (-I*cos(d) *e^a - e^a*sin(d))*sin(2*b*x + 2*d) - I*e^a*sin(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 + 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)) + 4*cos(b*x)*e^a + ((I*cos(d)*e^a + e^a*sin(d))* cos(2*b*x + 2*d) + I*cos(d)*e^a - (cos(d)*e^a - I*e^a*sin(d))*sin(2*b*x + 2*d) + e^a*sin(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)) + 4*I*e^a*sin(b*x))/(-2*I*b*cos(2*b*x + 2*d) + 2 *b*sin(2*b*x + 2*d) - 2*I*b)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (47) = 94\).
Time = 0.16 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.74 \[ \int e^{a+i b x} \sec ^2(d+b x) \, dx=-\frac {e^{\left (2 i \, b x + a + 2 i \, d\right )} \log \left (i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) + e^{a} \log \left (i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) - e^{\left (2 i \, b x + a + 2 i \, d\right )} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) - e^{a} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} + 1\right ) - 2 i \, e^{\left (i \, b x + a + i \, d\right )}}{b {\left (e^{\left (2 i \, b x + 3 i \, d\right )} + e^{\left (i \, d\right )}\right )}} \] Input:
integrate(exp(a+I*b*x)*sec(b*x+d)^2,x, algorithm="giac")
Output:
-(e^(2*I*b*x + a + 2*I*d)*log(I*e^(I*b*x + I*d) + 1) + e^a*log(I*e^(I*b*x + I*d) + 1) - e^(2*I*b*x + a + 2*I*d)*log(-I*e^(I*b*x + I*d) + 1) - e^a*lo g(-I*e^(I*b*x + I*d) + 1) - 2*I*e^(I*b*x + a + I*d))/(b*(e^(2*I*b*x + 3*I* d) + e^(I*d)))
Time = 0.00 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.38 \[ \int e^{a+i b x} \sec ^2(d+b x) \, dx=\frac {\ln \left (-2\,{\mathrm {e}}^{3\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}-{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}}\,2{}\mathrm {i}\right )\,\sqrt {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}}}{b}-\frac {\ln \left (-2\,{\mathrm {e}}^{3\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}}\,2{}\mathrm {i}\right )\,\sqrt {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}}}{b}+\frac {{\mathrm {e}}^{3\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,2{}\mathrm {i}}{b\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}+b\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}} \] Input:
int(exp(a + b*x*1i)/cos(d + b*x)^2,x)
Output:
(log(- 2*exp(3*a)*exp(-d*2i)*exp(b*x*1i) - exp(2*a)*exp(-d*2i)*(exp(2*a)*e xp(-d*2i))^(1/2)*2i)*(exp(2*a)*exp(-d*2i))^(1/2))/b - (log(exp(2*a)*exp(-d *2i)*(exp(2*a)*exp(-d*2i))^(1/2)*2i - 2*exp(3*a)*exp(-d*2i)*exp(b*x*1i))*( exp(2*a)*exp(-d*2i))^(1/2))/b + (exp(3*a)*exp(-d*2i)*exp(b*x*1i)*2i)/(b*ex p(2*a)*exp(-d*2i) + b*exp(2*a)*exp(b*x*2i))
\[ \int e^{a+i b x} \sec ^2(d+b x) \, dx=\text {too large to display} \] Input:
int(exp(a+I*b*x)*sec(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) - 14*e**(b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**4*i + 28*e**(b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**2*i - 14*e**(b*i*x)*cos(b* x + d)*i - 15*e**(b*i*x)*sec(b*x + d)**2*sin(b*x + d)**2*tan((b*x + d)/2)* *4*i + 30*e**(b*i*x)*sec(b*x + d)**2*sin(b*x + d)**2*tan((b*x + d)/2)**2*i - 15*e**(b*i*x)*sec(b*x + d)**2*sin(b*x + d)**2*i + 15*e**(b*i*x)*sec(b*x + d)**2*tan((b*x + d)/2)**4*i - 30*e**(b*i*x)*sec(b*x + d)**2*tan((b*x + d)/2)**2*i + 15*e**(b*i*x)*sec(b*x + d)**2*i - 4*e**(b*i*x)*sin(b*x + d)** 2*tan((b*x + d)/2)**4*i + 16*e**(b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)** 3 + 24*e**(b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**2*i - 48*e**(b*i*x)*si n(b*x + d)**2*tan((b*x + d)/2) + 48*e**(b*i*x)*sin(b*x + d)**2*i + 2*e**(b *i*x)*sin(b*x + d)*tan((b*x + d)/2)**4 - 4*e**(b*i*x)*sin(b*x + d)*tan((b* x + d)/2)**2 + 2*e**(b*i*x)*sin(b*x + d) - 14*e**(b*i*x)*tan((b*x + d)/2)* *4*i - 16*e**(b*i*x)*tan((b*x + d)/2)**3 + 12*e**(b*i*x)*tan((b*x + d)/2)* *2*i + 48*e**(b*i*x)*tan((b*x + d)/2) - 66*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)*...