Integrand size = 21, antiderivative size = 66 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \, dx=-\frac {2 i e^{2 a-2 i d}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {2 i e^{2 a-2 i d} \log \left (1+e^{2 i (d+b x)}\right )}{b} \] Output:
-2*I*exp(2*a-2*I*d)/b/(1+exp(2*I*(b*x+d)))-2*I*exp(2*a-2*I*d)*ln(1+exp(2*I *(b*x+d)))/b
Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.14 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \, dx=-\frac {2 i e^{2 a-2 i d} \left (-e^{2 i (d+b x)}+\left (1+e^{2 i (d+b x)}\right ) \log \left (1+e^{2 i (d+b x)}\right )\right )}{b \left (1+e^{2 i (d+b x)}\right )} \] Input:
Integrate[E^(2*(a + I*b*x))*Sec[d + b*x]^2,x]
Output:
((-2*I)*E^(2*a - (2*I)*d)*(-E^((2*I)*(d + b*x)) + (1 + E^((2*I)*(d + b*x)) )*Log[1 + E^((2*I)*(d + b*x))]))/(b*(1 + E^((2*I)*(d + b*x))))
Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.73, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, 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.
\(\displaystyle \int e^{2 (a+i b x)} \sec ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 4951 |
\(\displaystyle \frac {2 i e^{2 (a+i b x)+4 i (b x+d)} \left (e^{-4 i (b x+d)} \left (1+e^{2 i (b x+d)}\right )^2-e^{-6 i (b x+d)} \left (1+e^{2 i (b x+d)}\right )^3 \log \left (1+e^{2 i (b x+d)}\right )\right )}{b \left (1+e^{2 i (b x+d)}\right )^3}\) |
Input:
Int[E^(2*(a + I*b*x))*Sec[d + b*x]^2,x]
Output:
((2*I)*E^(2*(a + I*b*x) + (4*I)*(d + b*x))*((1 + E^((2*I)*(d + b*x)))^2/E^ ((4*I)*(d + b*x)) - ((1 + E^((2*I)*(d + b*x)))^3*Log[1 + E^((2*I)*(d + b*x ))])/E^((6*I)*(d + b*x))))/(b*(1 + E^((2*I)*(d + b*x)))^3)
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 = 58, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {2 i {\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{b \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}-\frac {2 i {\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \ln \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}\) | \(58\) |
Input:
int(exp(2*a+2*I*b*x)*sec(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
2*I*exp(2*a)*exp(2*I*b*x)/b/(1+exp(2*I*(b*x+d)))-2*I*exp(2*a)*exp(-2*I*d)* ln(1+exp(2*I*(b*x+d)))/b
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \, dx=-\frac {2 \, {\left ({\left (i \, e^{\left (2 i \, b x + 2 \, a\right )} + i \, e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + i \, e^{\left (2 \, a - 2 i \, d\right )}\right )}}{b e^{\left (2 i \, b x + 2 i \, d\right )} + b} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)^2,x, algorithm="fricas")
Output:
-2*((I*e^(2*I*b*x + 2*a) + I*e^(2*a - 2*I*d))*log(e^(2*I*b*x + 2*I*d) + 1) + I*e^(2*a - 2*I*d))/(b*e^(2*I*b*x + 2*I*d) + b)
\[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \, dx=e^{2 a} \int e^{2 i b x} \sec ^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)**2,x)
Output:
exp(2*a)*Integral(exp(2*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. 293 vs. \(2 (52) = 104\).
Time = 0.04 (sec) , antiderivative size = 293, normalized size of antiderivative = 4.44 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \, dx=\frac {2 \, {\left (-i \, \cos \left (2 \, d\right )^{2} e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (2 \, d\right )^{2} + {\left (-i \, \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \cos \left (2 \, b x + 4 \, d\right ) + {\left (\cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \sin \left (2 \, b x + 4 \, d\right )\right )} \arctan \left (\sin \left (2 \, b x\right ) - \sin \left (2 \, d\right ), \cos \left (2 \, b x\right ) + \cos \left (2 \, d\right )\right ) - {\left (\cos \left (2 \, d\right )^{2} e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )} \sin \left (2 \, d\right )^{2} + {\left (\cos \left (2 \, d\right ) e^{\left (2 \, a\right )} - i \, e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \cos \left (2 \, b x + 4 \, d\right ) + {\left (i \, \cos \left (2 \, d\right ) e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )} \sin \left (2 \, d\right )\right )} \sin \left (2 \, b x + 4 \, d\right )\right )} \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, d\right ) + \cos \left (2 \, d\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, d\right ) + \sin \left (2 \, d\right )^{2}\right ) - 2 \, e^{\left (2 \, a\right )}}{-i \, b \cos \left (2 \, b x + 4 \, d\right ) - i \, b \cos \left (2 \, d\right ) + b \sin \left (2 \, b x + 4 \, d\right ) + b \sin \left (2 \, d\right )} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)^2,x, algorithm="maxima")
Output:
(2*(-I*cos(2*d)^2*e^(2*a) - I*e^(2*a)*sin(2*d)^2 + (-I*cos(2*d)*e^(2*a) - e^(2*a)*sin(2*d))*cos(2*b*x + 4*d) + (cos(2*d)*e^(2*a) - I*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)) - (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(2*b*x + 4*d) + (I*cos(2*d)*e^(2*a) + e^(2*a)*sin(2*d))*sin (2*b*x + 4*d))*log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*d) + cos(2*d)^2 + sin (2*b*x)^2 - 2*sin(2*b*x)*sin(2*d) + sin(2*d)^2) - 2*e^(2*a))/(-I*b*cos(2*b *x + 4*d) - I*b*cos(2*d) + b*sin(2*b*x + 4*d) + b*sin(2*d))
Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.12 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \, dx=-\frac {2 \, {\left (i \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + i \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + i \, e^{\left (2 \, a - 2 i \, d\right )}\right )}}{b {\left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)^2,x, algorithm="giac")
Output:
-2*(I*e^(2*I*b*x + 2*a)*log(e^(2*I*b*x + 2*I*d) + 1) + I*e^(2*a - 2*I*d)*l og(e^(2*I*b*x + 2*I*d) + 1) + I*e^(2*a - 2*I*d))/(b*(e^(2*I*b*x + 2*I*d) + 1))
Time = 0.00 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \, dx=-\frac {{\mathrm {e}}^{2\,a-d\,2{}\mathrm {i}}\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\right )\,2{}\mathrm {i}}{b}-\frac {{\mathrm {e}}^{4\,a-d\,4{}\mathrm {i}}\,2{}\mathrm {i}}{b\,\left ({\mathrm {e}}^{2\,a-d\,2{}\mathrm {i}}+{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\right )} \] Input:
int(exp(2*a + b*x*2i)/cos(d + b*x)^2,x)
Output:
- (exp(2*a - d*2i)*log(exp(2*a)*exp(b*x*2i) + exp(2*a)*exp(-d*2i))*2i)/b - (exp(4*a - d*4i)*2i)/(b*(exp(2*a - d*2i) + exp(2*a + b*x*2i)))
\[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \, dx=\text {too large to display} \] Input:
int(exp(2*a+2*I*b*x)*sec(b*x+d)^2,x)
Output:
(e**(2*a)*( - 42*e**(2*b*i*x)*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)** 4 + 84*e**(2*b*i*x)*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**2 - 42*e** (2*b*i*x)*cos(b*x + d)*sin(b*x + d) - 392*e**(2*b*i*x)*cos(b*x + d)*tan((b *x + d)/2)**4*i + 784*e**(2*b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**2*i - 39 2*e**(2*b*i*x)*cos(b*x + d)*i - 315*e**(2*b*i*x)*sec(b*x + d)**2*sin(b*x + d)**2*tan((b*x + d)/2)**4*i + 630*e**(2*b*i*x)*sec(b*x + d)**2*sin(b*x + d)**2*tan((b*x + d)/2)**2*i - 315*e**(2*b*i*x)*sec(b*x + d)**2*sin(b*x + d )**2*i + 315*e**(2*b*i*x)*sec(b*x + d)**2*tan((b*x + d)/2)**4*i - 630*e**( 2*b*i*x)*sec(b*x + d)**2*tan((b*x + d)/2)**2*i + 315*e**(2*b*i*x)*sec(b*x + d)**2*i + 80*e**(2*b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**4*i + 128*e* *(2*b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**3 + 96*e**(2*b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**2*i - 640*e**(2*b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2) + 556*e**(2*b*i*x)*sin(b*x + d)**2*i + 112*e**(2*b*i*x)*sin(b*x + d )*tan((b*x + d)/2)**4 - 224*e**(2*b*i*x)*sin(b*x + d)*tan((b*x + d)/2)**2 + 112*e**(2*b*i*x)*sin(b*x + d) - 437*e**(2*b*i*x)*tan((b*x + d)/2)**4*i - 128*e**(2*b*i*x)*tan((b*x + d)/2)**3 + 618*e**(2*b*i*x)*tan((b*x + d)/2)* *2*i + 640*e**(2*b*i*x)*tan((b*x + d)/2) - 913*e**(2*b*i*x)*i - 3000*int(e **(2*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 + 6000*int(e**(2*b*i* x)/(tan((b*x + d)/2)**6 - 3*tan((b*x + d)/2)**4 + 3*tan((b*x + d)/2)**2...