Integrand size = 25, antiderivative size = 103 \[ \int e^{a+i b x} \sec ^2(d+b x) \tan (d+b x) \, dx=-\frac {2 e^{a-i d+i (d+b x)}}{b \left (1+e^{2 i (d+b x)}\right )^2}+\frac {3 e^{a-i d+i (d+b x)}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {e^{a-i d} \arctan \left (e^{i (d+b x)}\right )}{b} \] Output:
-2*exp(a-I*d+I*(b*x+d))/b/(1+exp(2*I*(b*x+d)))^2+3*exp(a-I*d+I*(b*x+d))/b/ (1+exp(2*I*(b*x+d)))-exp(a-I*d)*arctan(exp(I*(b*x+d)))/b
Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80 \[ \int e^{a+i b x} \sec ^2(d+b x) \tan (d+b x) \, dx=\frac {e^{a-i d} \left (e^{i (d+b x)}+3 e^{3 i (d+b x)}-\left (1+e^{2 i (d+b x)}\right )^2 \arctan \left (e^{i (d+b x)}\right )\right )}{b \left (1+e^{2 i (d+b x)}\right )^2} \] Input:
Integrate[E^(a + I*b*x)*Sec[d + b*x]^2*Tan[d + b*x],x]
Output:
(E^(a - I*d)*(E^(I*(d + b*x)) + 3*E^((3*I)*(d + b*x)) - (1 + E^((2*I)*(d + b*x)))^2*ArcTan[E^(I*(d + b*x))]))/(b*(1 + E^((2*I)*(d + b*x)))^2)
Time = 0.37 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.
\(\displaystyle \int e^{a+i b x} \tan (b x+d) \sec ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 4974 |
\(\displaystyle \int \left (\frac {8 i e^{a+3 i b x+2 i d}}{\left (1+e^{2 i (b x+d)}\right )^3}-\frac {4 i e^{a+3 i b x+2 i d}}{\left (1+e^{2 i (b x+d)}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{a-i d} \arctan \left (e^{i (b x+d)}\right )}{b}+\frac {3 e^{a+i (b x+d)-i d}}{b \left (1+e^{2 i (b x+d)}\right )}-\frac {2 e^{a+i (b x+d)-i d}}{b \left (1+e^{2 i (b x+d)}\right )^2}\) |
Input:
Int[E^(a + I*b*x)*Sec[d + b*x]^2*Tan[d + b*x],x]
Output:
(-2*E^(a - I*d + I*(d + b*x)))/(b*(1 + E^((2*I)*(d + b*x)))^2) + (3*E^(a - I*d + I*(d + b*x)))/(b*(1 + E^((2*I)*(d + b*x)))) - (E^(a - I*d)*ArcTan[E ^(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 = 3.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.64
method | result | size |
risch | \(\frac {3 \,{\mathrm e}^{a} {\mathrm e}^{3 i b x} {\mathrm e}^{2 i d}+{\mathrm e}^{a} {\mathrm e}^{i b x}}{b \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{2}}-\frac {{\mathrm e}^{a} {\mathrm e}^{-i d} \arctan \left ({\mathrm e}^{i \left (b x +d \right )}\right )}{b}\) | \(66\) |
Input:
int(exp(a+I*b*x)*sec(b*x+d)^2*tan(b*x+d),x,method=_RETURNVERBOSE)
Output:
1/b/(1+exp(2*I*(b*x+d)))^2*(3*exp(a)*exp(3*I*b*x)*exp(2*I*d)+exp(a)*exp(I* b*x))-exp(a)/b*exp(-I*d)*arctan(exp(I*(b*x+d)))
Time = 0.08 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.39 \[ \int e^{a+i b x} \sec ^2(d+b x) \tan (d+b x) \, dx=\frac {{\left (-i \, e^{\left (4 i \, b x + a + 3 i \, d\right )} - 2 i \, e^{\left (2 i \, b x + a + i \, d\right )} - i \, e^{\left (a - i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) + {\left (i \, e^{\left (4 i \, b x + a + 3 i \, d\right )} + 2 i \, e^{\left (2 i \, b x + a + i \, d\right )} + i \, e^{\left (a - i \, d\right )}\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + 6 \, e^{\left (3 i \, b x + a + 2 i \, d\right )} + 2 \, e^{\left (i \, b x + a\right )}}{2 \, {\left (b e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, b e^{\left (2 i \, b x + 2 i \, d\right )} + b\right )}} \] Input:
integrate(exp(a+I*b*x)*sec(b*x+d)^2*tan(b*x+d),x, algorithm="fricas")
Output:
1/2*((-I*e^(4*I*b*x + a + 3*I*d) - 2*I*e^(2*I*b*x + a + I*d) - I*e^(a - I* d))*log(e^(I*b*x + I*d) + I) + (I*e^(4*I*b*x + a + 3*I*d) + 2*I*e^(2*I*b*x + a + I*d) + I*e^(a - I*d))*log(e^(I*b*x + I*d) - I) + 6*e^(3*I*b*x + a + 2*I*d) + 2*e^(I*b*x + a))/(b*e^(4*I*b*x + 4*I*d) + 2*b*e^(2*I*b*x + 2*I*d ) + b)
\[ \int e^{a+i b x} \sec ^2(d+b x) \tan (d+b x) \, dx=e^{a} \int e^{i b x} \tan {\left (b x + d \right )} \sec ^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(a+I*b*x)*sec(b*x+d)**2*tan(b*x+d),x)
Output:
exp(a)*Integral(exp(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. 589 vs. \(2 (72) = 144\).
Time = 0.16 (sec) , antiderivative size = 589, normalized size of antiderivative = 5.72 \[ \int e^{a+i b x} \sec ^2(d+b x) \tan (d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(a+I*b*x)*sec(b*x+d)^2*tan(b*x+d),x, algorithm="maxima")
Output:
-(2*((I*cos(d)*e^a + e^a*sin(d))*cos(4*b*x + 4*d) + 2*(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( 4*b*x + 4*d) - 2*(cos(d)*e^a - I*e^a*sin(d))*sin(2*b*x + 2*d) + 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)) + 12*I*cos(3*b*x + 2*d)* e^a + 4*I*cos(b*x)*e^a - ((cos(d)*e^a - I*e^a*sin(d))*cos(4*b*x + 4*d) + 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(4*b*x + 4*d) - 2*(-I*cos(d)*e^a - e^a*sin(d))*sin(2*b*x + 2*d) - I*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)) - 12*e^a*sin(3*b*x + 2*d) - 4*e^a*sin(b*x)) /(-4*I*b*cos(4*b*x + 4*d) - 8*I*b*cos(2*b*x + 2*d) + 4*b*sin(4*b*x + 4*d) + 8*b*sin(2*b*x + 2*d) - 4*I*b)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (72) = 144\).
Time = 0.20 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.82 \[ \int e^{a+i b x} \sec ^2(d+b x) \tan (d+b x) \, dx=\frac {-i \, e^{\left (4 i \, b x + a + 3 i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) - 2 i \, e^{\left (2 i \, b x + a + i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) - i \, e^{\left (a - i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) + i \, e^{\left (4 i \, b x + a + 3 i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + 2 i \, e^{\left (2 i \, b x + a + i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + i \, e^{\left (a - i \, d\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + 6 \, e^{\left (3 i \, b x + a + 2 i \, d\right )} + 2 \, e^{\left (i \, b x + a\right )}}{2 \, b {\left (e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:
integrate(exp(a+I*b*x)*sec(b*x+d)^2*tan(b*x+d),x, algorithm="giac")
Output:
1/2*(-I*e^(4*I*b*x + a + 3*I*d)*log(e^(I*b*x + I*d) + I) - 2*I*e^(2*I*b*x + a + I*d)*log(e^(I*b*x + I*d) + I) - I*e^(a - I*d)*log(e^(I*b*x + I*d) + I) + I*e^(4*I*b*x + a + 3*I*d)*log(e^(I*b*x + I*d) - I) + 2*I*e^(2*I*b*x + a + I*d)*log(e^(I*b*x + I*d) - I) + I*e^(a - I*d)*log(e^(I*b*x + I*d) - I ) + 6*e^(3*I*b*x + a + 2*I*d) + 2*e^(I*b*x + a))/(b*(e^(4*I*b*x + 4*I*d) + 2*e^(2*I*b*x + 2*I*d) + 1))
Timed out. \[ \int e^{a+i b x} \sec ^2(d+b x) \tan (d+b x) \, dx=\int \frac {{\mathrm {e}}^{a+b\,x\,1{}\mathrm {i}}\,\mathrm {tan}\left (d+b\,x\right )}{{\cos \left (d+b\,x\right )}^2} \,d x \] Input:
int((exp(a + b*x*1i)*tan(d + b*x))/cos(d + b*x)^2,x)
Output:
int((exp(a + b*x*1i)*tan(d + b*x))/cos(d + b*x)^2, x)
\[ \int e^{a+i b x} \sec ^2(d+b x) \tan (d+b x) \, dx=\text {too large to display} \] Input:
int(exp(a+I*b*x)*sec(b*x+d)^2*tan(b*x+d),x)
Output:
(e**a*(3*e**(b*i*x)*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**4*i - 6*e* *(b*i*x)*cos(b*x + d)*sin(b*x + d)*tan((b*x + d)/2)**2*i + 3*e**(b*i*x)*co s(b*x + d)*sin(b*x + d)*i - 7*e**(b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**4 + 14*e**(b*i*x)*cos(b*x + d)*tan((b*x + d)/2)**2 - 7*e**(b*i*x)*cos(b*x + d) - 2*e**(b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**4 - 8*e**(b*i*x)*sin(b *x + d)**2*tan((b*x + d)/2)**3*i + 12*e**(b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)**2 + 24*e**(b*i*x)*sin(b*x + d)**2*tan((b*x + d)/2)*i + 24*e**(b*i *x)*sin(b*x + d)**2 - e**(b*i*x)*sin(b*x + d)*tan((b*x + d)/2)**4*i + 2*e* *(b*i*x)*sin(b*x + d)*tan((b*x + d)/2)**2*i - e**(b*i*x)*sin(b*x + d)*i - 7*e**(b*i*x)*tan((b*x + d)/2)**4 + 8*e**(b*i*x)*tan((b*x + d)/2)**3*i + 6* e**(b*i*x)*tan((b*x + d)/2)**2 - 24*e**(b*i*x)*tan((b*x + d)/2)*i - 33*e** (b*i*x) + 50*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*i - 1 00*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*i + 50*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*b*i - 50*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)*tan((b*x + d)/2)**4*b *i + 100*int(e**(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)*tan((b*x + d)/2)**2*b*i - 50*int(e**(b*i*x)/...