Integrand size = 27, antiderivative size = 93 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \tan (d+b x) \, dx=\frac {2 e^{2 a-2 i d}}{b \left (1+e^{2 i (d+b x)}\right )^2}-\frac {6 e^{2 a-2 i d}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {2 e^{2 a-2 i d} \log \left (1+e^{2 i (d+b x)}\right )}{b} \] Output:
2*exp(2*a-2*I*d)/b/(1+exp(2*I*(b*x+d)))^2-6*exp(2*a-2*I*d)/b/(1+exp(2*I*(b *x+d)))-2*exp(2*a-2*I*d)*ln(1+exp(2*I*(b*x+d)))/b
Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \tan (d+b x) \, dx=-\frac {2 e^{2 a-2 i d} \left (2+3 e^{2 i (d+b x)}+\left (1+e^{2 i (d+b x)}\right )^2 \log \left (1+e^{2 i (d+b x)}\right )\right )}{b \left (1+e^{2 i (d+b x)}\right )^2} \] Input:
Integrate[E^(2*(a + I*b*x))*Sec[d + b*x]^2*Tan[d + b*x],x]
Output:
(-2*E^(2*a - (2*I)*d)*(2 + 3*E^((2*I)*(d + b*x)) + (1 + E^((2*I)*(d + b*x) ))^2*Log[1 + E^((2*I)*(d + b*x))]))/(b*(1 + E^((2*I)*(d + b*x)))^2)
Time = 0.35 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.11, 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.
\(\displaystyle \int e^{2 (a+i b x)} \tan (b x+d) \sec ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 4974 |
\(\displaystyle \int \left (\frac {8 i e^{2 (a+i d)+4 i b x}}{\left (1+e^{2 i (b x+d)}\right )^3}-\frac {4 i e^{2 (a+i d)+4 i b x}}{\left (1+e^{2 i (b x+d)}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 e^{2 a-2 i d}}{b \left (1+e^{2 i (b x+d)}\right )}+\frac {2 e^{2 (a-i d)+4 i (b x+d)}}{b \left (1+e^{2 i (b x+d)}\right )^2}-\frac {2 e^{2 a-2 i d} \log \left (1+e^{2 i (b x+d)}\right )}{b}\) |
Input:
Int[E^(2*(a + I*b*x))*Sec[d + b*x]^2*Tan[d + b*x],x]
Output:
(2*E^(2*(a - I*d) + (4*I)*(d + b*x)))/(b*(1 + E^((2*I)*(d + b*x)))^2) - (2 *E^(2*a - (2*I)*d))/(b*(1 + E^((2*I)*(d + b*x)))) - (2*E^(2*a - (2*I)*d)*L og[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 = 3.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{4 i b x} {\mathrm e}^{2 i d} {\mathrm e}^{2 a}+2 \,{\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{b \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{2}}-\frac {2 \,{\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \ln \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}\) | \(75\) |
Input:
int(exp(2*a+2*I*b*x)*sec(b*x+d)^2*tan(b*x+d),x,method=_RETURNVERBOSE)
Output:
2/b/(1+exp(2*I*(b*x+d)))^2*(2*exp(4*I*b*x)*exp(2*I*d)*exp(2*a)+exp(2*a)*ex p(2*I*b*x))-2*exp(2*a)/b*exp(-2*I*d)*ln(1+exp(2*I*(b*x+d)))
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \tan (d+b x) \, dx=-\frac {2 \, {\left ({\left (e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + 2 \, e^{\left (2 i \, b x + 2 \, a\right )} + e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + 3 \, e^{\left (2 i \, b x + 2 \, a\right )} + 2 \, e^{\left (2 \, a - 2 i \, d\right )}\right )}}{b e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, b e^{\left (2 i \, b x + 2 i \, d\right )} + b} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)^2*tan(b*x+d),x, algorithm="fricas")
Output:
-2*((e^(4*I*b*x + 2*a + 2*I*d) + 2*e^(2*I*b*x + 2*a) + e^(2*a - 2*I*d))*lo g(e^(2*I*b*x + 2*I*d) + 1) + 3*e^(2*I*b*x + 2*a) + 2*e^(2*a - 2*I*d))/(b*e ^(4*I*b*x + 4*I*d) + 2*b*e^(2*I*b*x + 2*I*d) + b)
\[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \tan (d+b x) \, dx=e^{2 a} \int e^{2 i b x} \tan {\left (b x + d \right )} \sec ^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)**2*tan(b*x+d),x)
Output:
exp(2*a)*Integral(exp(2*I*b*x)*tan(b*x + d)*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. 473 vs. \(2 (78) = 156\).
Time = 0.05 (sec) , antiderivative size = 473, normalized size of antiderivative = 5.09 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \tan (d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)^2*tan(b*x+d),x, algorithm="maxima")
Output:
-(2*(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(4*b*x + 6*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(4*b*x + 6*d ) + 2*(I*cos(2*d)*e^(2*a) + e^(2*a)*sin(2*d))*sin(2*b*x + 4*d))*arctan2(si n(2*b*x) - sin(2*d), cos(2*b*x) + cos(2*d)) - 6*I*cos(2*b*x + 2*d)*e^(2*a) - (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(4*b*x + 6*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(4*b*x + 6*d) - 2*(cos(2*d)*e^(2*a) - I*e^(2*a)*sin(2*d))*sin(2*b*x + 4*d))*log(co s(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) + 6*e^(2*a)*sin(2*b*x + 2*d) - 4*I*e^(2*a))/(-I *b*cos(4*b*x + 6*d) - 2*I*b*cos(2*b*x + 4*d) - I*b*cos(2*d) + b*sin(4*b*x + 6*d) + 2*b*sin(2*b*x + 4*d) + b*sin(2*d))
Time = 0.17 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.29 \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \tan (d+b x) \, dx=-\frac {2 \, {\left (e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + 2 \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + 3 \, e^{\left (2 i \, b x + 2 \, a\right )} + 2 \, e^{\left (2 \, a - 2 i \, d\right )}\right )}}{b {\left (e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*sec(b*x+d)^2*tan(b*x+d),x, algorithm="giac")
Output:
-2*(e^(4*I*b*x + 2*a + 2*I*d)*log(e^(2*I*b*x + 2*I*d) + 1) + 2*e^(2*I*b*x + 2*a)*log(e^(2*I*b*x + 2*I*d) + 1) + e^(2*a - 2*I*d)*log(e^(2*I*b*x + 2*I *d) + 1) + 3*e^(2*I*b*x + 2*a) + 2*e^(2*a - 2*I*d))/(b*(e^(4*I*b*x + 4*I*d ) + 2*e^(2*I*b*x + 2*I*d) + 1))
Timed out. \[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \tan (d+b x) \, dx=\int \frac {{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\,\mathrm {tan}\left (d+b\,x\right )}{{\cos \left (d+b\,x\right )}^2} \,d x \] Input:
int((exp(2*a + b*x*2i)*tan(d + b*x))/cos(d + b*x)^2,x)
Output:
int((exp(2*a + b*x*2i)*tan(d + b*x))/cos(d + b*x)^2, x)
\[ \int e^{2 (a+i b x)} \sec ^2(d+b x) \tan (d+b x) \, dx=\text {too large to display} \] Input:
int(exp(2*a+2*I*b*x)*sec(b*x+d)^2*tan(b*x+d),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*i - 84*e**(2*b*i*x)*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**2*i + 42*e* *(2*b*i*x)*cos(b*x + d)*sin(b*x + d)*i - 392*e**(2*b*i*x)*cos(b*x + d)*tan ((b*x + d)/2)**4 + 784*e**(2*b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**2 - 392 *e**(2*b*i*x)*cos(b*x + d) + 80*e**(2*b*i*x)*sin(b*x + d)**2*tan((b*x + d) /2)**4 - 128*e**(2*b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**3*i + 96*e**(2 *b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**2 + 640*e**(2*b*i*x)*sin(b*x + d )**2*tan((b*x + d)/2)*i + 556*e**(2*b*i*x)*sin(b*x + d)**2 - 112*e**(2*b*i *x)*sin(b*x + d)*tan((b*x + d)/2)**4*i + 224*e**(2*b*i*x)*sin(b*x + d)*tan ((b*x + d)/2)**2*i - 112*e**(2*b*i*x)*sin(b*x + d)*i - 437*e**(2*b*i*x)*ta n((b*x + d)/2)**4 + 128*e**(2*b*i*x)*tan((b*x + d)/2)**3*i + 618*e**(2*b*i *x)*tan((b*x + d)/2)**2 - 640*e**(2*b*i*x)*tan((b*x + d)/2)*i - 913*e**(2* b*i*x) + 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*i - 6000*int(e**(2*b*i*x)/(tan((b*x + d)/2)**6 - 3*tan((b*x + d)/2)**4 + 3*t an((b*x + d)/2)**2 - 1),x)*sin(b*x + d)**2*tan((b*x + d)/2)**2*b*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*b*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)*tan((b* x + d)/2)**4*b*i + 6000*int(e**(2*b*i*x)/(tan((b*x + d)/2)**6 - 3*tan((...