Integrand size = 25, antiderivative size = 99 \[ \int e^{a+i b x} \sin (d+b x) \tan ^2(d+b x) \, dx=\frac {e^{a-i d+2 i (d+b x)}}{4 b}-\frac {2 e^{a-i d}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {1}{2} i e^{a-i d} x-\frac {e^{a-i d} \log \left (1+e^{2 i (d+b x)}\right )}{b} \] Output:
1/4*exp(a-I*d+2*I*(b*x+d))/b-2*exp(a-I*d)/b/(1+exp(2*I*(b*x+d)))-1/2*I*exp (a-I*d)*x-exp(a-I*d)*ln(1+exp(2*I*(b*x+d)))/b
Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int e^{a+i b x} \sin (d+b x) \tan ^2(d+b x) \, dx=\frac {e^{a-i d} \left (e^{2 i (d+b x)}-\frac {8}{1+e^{2 i (d+b x)}}-2 i (d+b x)-4 \log \left (1+e^{2 i (d+b x)}\right )\right )}{4 b} \] Input:
Integrate[E^(a + I*b*x)*Sin[d + b*x]*Tan[d + b*x]^2,x]
Output:
(E^(a - I*d)*(E^((2*I)*(d + b*x)) - 8/(1 + E^((2*I)*(d + b*x))) - (2*I)*(d + b*x) - 4*Log[1 + E^((2*I)*(d + b*x))]))/(4*b)
Time = 0.33 (sec) , antiderivative size = 99, 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.
\(\displaystyle \int e^{a+i b x} \sin (b x+d) \tan ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 4974 |
\(\displaystyle \int \left (\frac {1}{2} i e^{a+2 i (b x+d)-i d}+\frac {6 i e^{a-i d}}{1+e^{2 i (b x+d)}}-\frac {4 i e^{a-i d}}{\left (1+e^{2 i (b x+d)}\right )^2}-\frac {5}{2} i e^{a-i d}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{a+2 i (b x+d)-i d}}{4 b}-\frac {2 e^{a-i d}}{b \left (1+e^{2 i (b x+d)}\right )}-\frac {e^{a-i d} \log \left (1+e^{2 i (b x+d)}\right )}{b}-\frac {1}{2} i x e^{a-i d}\) |
Input:
Int[E^(a + I*b*x)*Sin[d + b*x]*Tan[d + b*x]^2,x]
Output:
E^(a - I*d + (2*I)*(d + b*x))/(4*b) - (2*E^(a - I*d))/(b*(1 + E^((2*I)*(d + b*x)))) - (I/2)*E^(a - I*d)*x - (E^(a - I*d)*Log[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]
\[\int {\mathrm e}^{i b x +a} \sin \left (b x +d \right ) \tan \left (b x +d \right )^{2}d x\]
Input:
int(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d)^2,x)
Output:
int(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d)^2,x)
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.91 \[ \int e^{a+i b x} \sin (d+b x) \tan ^2(d+b x) \, dx=\frac {{\left (-2 i \, b x + 1\right )} e^{\left (2 i \, b x + a + i \, d\right )} - 2 \, {\left (i \, b x + 4\right )} e^{\left (a - i \, d\right )} - 4 \, {\left (e^{\left (2 i \, b x + a + i \, d\right )} + e^{\left (a - i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + e^{\left (4 i \, b x + a + 3 i \, d\right )}}{4 \, {\left (b e^{\left (2 i \, b x + 2 i \, d\right )} + b\right )}} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d)^2,x, algorithm="fricas")
Output:
1/4*((-2*I*b*x + 1)*e^(2*I*b*x + a + I*d) - 2*(I*b*x + 4)*e^(a - I*d) - 4* (e^(2*I*b*x + a + I*d) + e^(a - I*d))*log(e^(2*I*b*x + 2*I*d) + 1) + e^(4* I*b*x + a + 3*I*d))/(b*e^(2*I*b*x + 2*I*d) + b)
Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.29 \[ \int e^{a+i b x} \sin (d+b x) \tan ^2(d+b x) \, dx=- \frac {i x e^{a} e^{- i d}}{2} + \begin {cases} \frac {e^{a} e^{i d} e^{2 i b x}}{4 b} & \text {for}\: b \neq 0 \\x \left (\frac {\left (i e^{a} e^{2 i d} - i e^{a}\right ) e^{- i d}}{2} + \frac {i e^{a} e^{- i d}}{2}\right ) & \text {otherwise} \end {cases} - \frac {2 e^{a}}{b e^{3 i d} e^{2 i b x} + b e^{i d}} - \frac {e^{a} e^{- i d} \log {\left (e^{2 i b x} + e^{- 2 i d} \right )}}{b} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d)**2,x)
Output:
-I*x*exp(a)*exp(-I*d)/2 + Piecewise((exp(a)*exp(I*d)*exp(2*I*b*x)/(4*b), N e(b, 0)), (x*((I*exp(a)*exp(2*I*d) - I*exp(a))*exp(-I*d)/2 + I*exp(a)*exp( -I*d)/2), True)) - 2*exp(a)/(b*exp(3*I*d)*exp(2*I*b*x) + b*exp(I*d)) - exp (a)*exp(-I*d)*log(exp(2*I*b*x) + exp(-2*I*d))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (72) = 144\).
Time = 0.06 (sec) , antiderivative size = 315, normalized size of antiderivative = 3.18 \[ \int e^{a+i b x} \sin (d+b x) \tan ^2(d+b x) \, dx=-\frac {2 \, b x e^{a} + 4 \, {\left (\cos \left (d\right )^{2} e^{a} + e^{a} \sin \left (d\right )^{2} + {\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \cos \left (2 \, b x + 3 \, d\right ) + {\left (i \, \cos \left (d\right ) e^{a} + e^{a} \sin \left (d\right )\right )} \sin \left (2 \, b x + 3 \, d\right )\right )} \arctan \left (\sin \left (2 \, b x\right ) - \sin \left (2 \, d\right ), \cos \left (2 \, b x\right ) + \cos \left (2 \, d\right )\right ) + {\left (2 \, b x e^{a} + i \, e^{a}\right )} \cos \left (2 \, b x + 2 \, d\right ) + i \, \cos \left (4 \, b x + 4 \, d\right ) e^{a} + 2 \, {\left (-i \, \cos \left (d\right )^{2} e^{a} - i \, e^{a} \sin \left (d\right )^{2} + {\left (-i \, \cos \left (d\right ) e^{a} - e^{a} \sin \left (d\right )\right )} \cos \left (2 \, b x + 3 \, d\right ) + {\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \sin \left (2 \, b x + 3 \, d\right )\right )} \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, d\right ) + \cos \left (2 \, d\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, d\right ) + \sin \left (2 \, d\right )^{2}\right ) - e^{a} \sin \left (4 \, b x + 4 \, d\right ) - {\left (-2 i \, b x e^{a} + e^{a}\right )} \sin \left (2 \, b x + 2 \, d\right ) - 8 i \, e^{a}}{-4 i \, b \cos \left (2 \, b x + 3 \, d\right ) - 4 i \, b \cos \left (d\right ) + 4 \, b \sin \left (2 \, b x + 3 \, d\right ) + 4 \, b \sin \left (d\right )} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d)^2,x, algorithm="maxima")
Output:
-(2*b*x*e^a + 4*(cos(d)^2*e^a + e^a*sin(d)^2 + (cos(d)*e^a - I*e^a*sin(d)) *cos(2*b*x + 3*d) + (I*cos(d)*e^a + e^a*sin(d))*sin(2*b*x + 3*d))*arctan2( sin(2*b*x) - sin(2*d), cos(2*b*x) + cos(2*d)) + (2*b*x*e^a + I*e^a)*cos(2* b*x + 2*d) + I*cos(4*b*x + 4*d)*e^a + 2*(-I*cos(d)^2*e^a - I*e^a*sin(d)^2 + (-I*cos(d)*e^a - e^a*sin(d))*cos(2*b*x + 3*d) + (cos(d)*e^a - I*e^a*sin( d))*sin(2*b*x + 3*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) - e^a*sin(4*b*x + 4 *d) - (-2*I*b*x*e^a + e^a)*sin(2*b*x + 2*d) - 8*I*e^a)/(-4*I*b*cos(2*b*x + 3*d) - 4*I*b*cos(d) + 4*b*sin(2*b*x + 3*d) + 4*b*sin(d))
Time = 0.68 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.32 \[ \int e^{a+i b x} \sin (d+b x) \tan ^2(d+b x) \, dx=-\frac {2 \, {\left (i \, b x + i \, d\right )} e^{\left (2 i \, b x + a + i \, d\right )} + 2 \, {\left (i \, b x + i \, d\right )} e^{\left (a - i \, d\right )} + 4 \, e^{\left (2 i \, b x + a + i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + 4 \, e^{\left (a - i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - e^{\left (4 i \, b x + a + 3 i \, d\right )} - e^{\left (2 i \, b x + a + i \, d\right )} + 8 \, e^{\left (a - i \, d\right )}}{4 \, b {\left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d)^2,x, algorithm="giac")
Output:
-1/4*(2*(I*b*x + I*d)*e^(2*I*b*x + a + I*d) + 2*(I*b*x + I*d)*e^(a - I*d) + 4*e^(2*I*b*x + a + I*d)*log(e^(2*I*b*x + 2*I*d) + 1) + 4*e^(a - I*d)*log (e^(2*I*b*x + 2*I*d) + 1) - e^(4*I*b*x + a + 3*I*d) - e^(2*I*b*x + a + I*d ) + 8*e^(a - I*d))/(b*(e^(2*I*b*x + 2*I*d) + 1))
Time = 17.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.01 \[ \int e^{a+i b x} \sin (d+b x) \tan ^2(d+b x) \, dx=\frac {{\mathrm {e}}^{a+d\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}}{4\,b}-\frac {x\,{\mathrm {e}}^{a-d\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {2\,{\mathrm {e}}^{3\,a-d\,3{}\mathrm {i}}}{b\,\left ({\mathrm {e}}^{2\,a-d\,2{}\mathrm {i}}+{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{a-d\,1{}\mathrm {i}}\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\right )}{b} \] Input:
int(exp(a + b*x*1i)*sin(d + b*x)*tan(d + b*x)^2,x)
Output:
exp(a + d*1i + b*x*2i)/(4*b) - (x*exp(a - d*1i)*1i)/2 - (2*exp(3*a - d*3i) )/(b*(exp(2*a - d*2i) + exp(2*a + b*x*2i))) - (exp(a - d*1i)*log(exp(2*a)* exp(b*x*2i) + exp(2*a)*exp(-d*2i)))/b
\[ \int e^{a+i b x} \sin (d+b x) \tan ^2(d+b x) \, dx=e^{a} \left (\int e^{b i x} \sin \left (b x +d \right ) \tan \left (b x +d \right )^{2}d x \right ) \] Input:
int(exp(a+I*b*x)*sin(b*x+d)*tan(b*x+d)^2,x)
Output:
e**a*int(e**(b*i*x)*sin(b*x + d)*tan(b*x + d)**2,x)