Integrand size = 19, antiderivative size = 59 \[ \int e^{2 (a+i b x)} \sec (d+b x) \, dx=-\frac {2 i e^{2 (a-i d)+i (d+b x)}}{b}+\frac {2 i e^{2 a-2 i d} \arctan \left (e^{i (d+b x)}\right )}{b} \] Output:
-2*I*exp(2*a-2*I*d+I*(b*x+d))/b+2*I*exp(2*a-2*I*d)*arctan(exp(I*(b*x+d)))/ b
Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75 \[ \int e^{2 (a+i b x)} \sec (d+b x) \, dx=-\frac {2 i e^{2 a-2 i d} \left (e^{i (d+b x)}-\arctan \left (e^{i (d+b x)}\right )\right )}{b} \] Input:
Integrate[E^(2*(a + I*b*x))*Sec[d + b*x],x]
Output:
((-2*I)*E^(2*a - (2*I)*d)*(E^(I*(d + b*x)) - ArcTan[E^(I*(d + b*x))]))/b
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.12, 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.
\(\displaystyle \int e^{2 (a+i b x)} \sec (b x+d) \, dx\) |
\(\Big \downarrow \) 4951 |
\(\displaystyle -\frac {2 i e^{2 (a+i b x)-3 i (b x+d)} \left (e^{2 i (b x+d)}-e^{i (b x+d)} \arctan \left (e^{i (b x+d)}\right )\right )}{b}\) |
Input:
Int[E^(2*(a + I*b*x))*Sec[d + b*x],x]
Output:
((-2*I)*E^(2*(a + I*b*x) - (3*I)*(d + b*x))*(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 = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {2 i {\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \arctan \left ({\mathrm e}^{i \left (b x +d \right )}\right )}{b}-\frac {2 i {\mathrm e}^{2 a} {\mathrm e}^{-i d} {\mathrm e}^{i b x}}{b}\) | \(48\) |
Input:
int(exp(2*a+2*I*b*x)*sec(b*x+d),x,method=_RETURNVERBOSE)
Output:
2*I/b*exp(2*a)*exp(-2*I*d)*arctan(exp(I*(b*x+d)))-2*I/b*exp(2*a)*exp(-I*d) *exp(I*b*x)
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.07 \[ \int e^{2 (a+i b x)} \sec (d+b x) \, dx=-\frac {e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) - e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + 2 i \, e^{\left (i \, b x + 2 \, a - i \, d\right )}}{b} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d),x, algorithm="fricas")
Output:
-(e^(2*a - 2*I*d)*log(e^(I*b*x + I*d) + I) - e^(2*a - 2*I*d)*log(e^(I*b*x + I*d) - I) + 2*I*e^(I*b*x + 2*a - I*d))/b
\[ \int e^{2 (a+i b x)} \sec (d+b x) \, dx=e^{2 a} \int e^{2 i b x} \sec {\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d),x)
Output:
exp(2*a)*Integral(exp(2*I*b*x)*sec(b*x + d), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 358 vs. \(2 (41) = 82\).
Time = 0.16 (sec) , antiderivative size = 358, normalized size of antiderivative = 6.07 \[ \int e^{2 (a+i b x)} \sec (d+b x) \, dx=\frac {2 \, {\left (-i \, \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \arctan \left (\frac {2 \, {\left (\cos \left (b x + 2 \, d\right ) \cos \left (d\right ) + \sin \left (b x + 2 \, d\right ) \sin \left (d\right )\right )}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}, \frac {\cos \left (b x + 2 \, d\right )^{2} - \cos \left (d\right )^{2} + \sin \left (b x + 2 \, d\right )^{2} - \sin \left (d\right )^{2}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}\right ) - 4 i \, \cos \left (b x - d\right ) e^{\left (2 \, a\right )} + {\left (\cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \log \left (\frac {\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} - 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} + 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}\right ) + 4 \, e^{\left (2 \, a\right )} \sin \left (b x - d\right )}{2 \, b} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d),x, algorithm="maxima")
Output:
1/2*(2*(-I*cos(2*d)*e^(2*a) - e^(2*a)*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)*si n(b*x + 2*d) + sin(b*x + 2*d)^2 - 2*cos(b*x + 2*d)*sin(d) + sin(d)^2), (co s(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*I*cos(b*x - d)*e^(2*a) + (cos(2*d)*e^(2*a) - I*e ^(2*a)*sin(2*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*e^(2*a)*sin(b*x - d))/b
Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.14 \[ \int e^{2 (a+i b x)} \sec (d+b x) \, dx=-\frac {e^{\left (2 \, a - 2 i \, d\right )} \log \left (i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) - e^{\left (2 \, a - 2 i \, d\right )} \log \left (-i \, e^{\left (i \, b x + i \, d\right )} - 1\right ) + 2 i \, e^{\left (i \, b x + 2 \, a - i \, d\right )}}{b} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d),x, algorithm="giac")
Output:
-(e^(2*a - 2*I*d)*log(I*e^(I*b*x + I*d) - 1) - e^(2*a - 2*I*d)*log(-I*e^(I *b*x + I*d) - 1) + 2*I*e^(I*b*x + 2*a - I*d))/b
Time = 0.00 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.34 \[ \int e^{2 (a+i b x)} \sec (d+b x) \, dx=-\frac {{\mathrm {e}}^{2\,a-d\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,2{}\mathrm {i}}{b}+\frac {\sqrt {{\mathrm {e}}^{4\,a-d\,4{}\mathrm {i}}}\,\ln \left (2\,{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,3{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}-{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,4{}\mathrm {i}}}\,2{}\mathrm {i}\right )}{b}-\frac {\sqrt {{\mathrm {e}}^{4\,a-d\,4{}\mathrm {i}}}\,\ln \left (2\,{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,3{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-d\,4{}\mathrm {i}}}\,2{}\mathrm {i}\right )}{b} \] Input:
int(exp(2*a + b*x*2i)/cos(d + b*x),x)
Output:
(exp(4*a - d*4i)^(1/2)*log(2*exp(4*a)*exp(-d*3i)*exp(b*x*1i) - exp(2*a)*ex p(-d*2i)*(exp(4*a)*exp(-d*4i))^(1/2)*2i))/b - (exp(2*a - d*1i + b*x*1i)*2i )/b - (exp(4*a - d*4i)^(1/2)*log(2*exp(4*a)*exp(-d*3i)*exp(b*x*1i) + exp(2 *a)*exp(-d*2i)*(exp(4*a)*exp(-d*4i))^(1/2)*2i))/b
\[ \int e^{2 (a+i b x)} \sec (d+b x) \, dx=\frac {e^{2 a} \left (-3 e^{2 b i x} \cos \left (b x +d \right ) \sec \left (b x +d \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} i +3 e^{2 b i x} \cos \left (b x +d \right ) \sec \left (b x +d \right ) i +e^{2 b i x} \cos \left (b x +d \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} i -2 e^{2 b i x} \cos \left (b x +d \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )-5 e^{2 b i x} \cos \left (b x +d \right ) i -12 \cos \left (b x +d \right ) \left (\int \frac {e^{2 b i x}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-1}d x \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} b +12 \cos \left (b x +d \right ) \left (\int \frac {e^{2 b i x}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-1}d x \right ) b -e^{2 b i x} \sin \left (b x +d \right ) \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}+e^{2 b i x} \sin \left (b x +d \right )+e^{2 b i x} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} i -e^{2 b i x} i \right )}{6 \cos \left (b x +d \right ) b \left (\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-1\right )} \] Input:
int(exp(2*a+2*I*b*x)*sec(b*x+d),x)
Output:
(e**(2*a)*( - 3*e**(2*b*i*x)*cos(b*x + d)*sec(b*x + d)*tan((b*x + d)/2)**2 *i + 3*e**(2*b*i*x)*cos(b*x + d)*sec(b*x + d)*i + e**(2*b*i*x)*cos(b*x + d )*tan((b*x + d)/2)**2*i - 2*e**(2*b*i*x)*cos(b*x + d)*tan((b*x + d)/2) - 5 *e**(2*b*i*x)*cos(b*x + d)*i - 12*cos(b*x + d)*int(e**(2*b*i*x)/(tan((b*x + d)/2)**2 - 1),x)*tan((b*x + d)/2)**2*b + 12*cos(b*x + d)*int(e**(2*b*i*x )/(tan((b*x + d)/2)**2 - 1),x)*b - e**(2*b*i*x)*sin(b*x + d)*tan((b*x + d) /2)**2 + e**(2*b*i*x)*sin(b*x + d) + e**(2*b*i*x)*tan((b*x + d)/2)**2*i - e**(2*b*i*x)*i))/(6*cos(b*x + d)*b*(tan((b*x + d)/2)**2 - 1))