Integrand size = 25, antiderivative size = 100 \[ \int e^{a+i b x} \sin ^2(d+b x) \tan (d+b x) \, dx=\frac {e^{a-i d-i (d+b x)}}{4 b}-\frac {e^{a-i d+i (d+b x)}}{b}+\frac {e^{a-i d+3 i (d+b x)}}{12 b}+\frac {2 e^{a-i d} \arctan \left (e^{i (d+b x)}\right )}{b} \] Output:
1/4*exp(a-I*d-I*(b*x+d))/b-exp(a-I*d+I*(b*x+d))/b+1/12*exp(a-I*d+3*I*(b*x+ d))/b+2*exp(a-I*d)*arctan(exp(I*(b*x+d)))/b
Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.71 \[ \int e^{a+i b x} \sin ^2(d+b x) \tan (d+b x) \, dx=\frac {e^a \left (-12 e^{i b x}+e^{2 i d+3 i b x}+3 e^{-i (2 d+b x)}+24 e^{-i d} \arctan \left (e^{i (d+b x)}\right )\right )}{12 b} \] Input:
Integrate[E^(a + I*b*x)*Sin[d + b*x]^2*Tan[d + b*x],x]
Output:
(E^a*(-12*E^(I*b*x) + E^((2*I)*d + (3*I)*b*x) + 3/E^(I*(2*d + b*x)) + (24* ArcTan[E^(I*(d + b*x))])/E^(I*d)))/(12*b)
Time = 0.34 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12, 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.
\(\displaystyle \int e^{a+i b x} \sin ^2(b x+d) \tan (b x+d) \, dx\) |
\(\Big \downarrow \) 4974 |
\(\displaystyle \int \left (\frac {7}{4} i e^{a-i b x-2 i d}-i e^{a+2 i (b x+d)-i b x-2 i d}+\frac {1}{4} i e^{a+4 i (b x+d)-i b x-2 i d}-\frac {2 i e^{a-i b x-2 i d}}{1+e^{2 i (b x+d)}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 e^{a-i d} \arctan \left (e^{-i b x-i d}\right )}{b}+\frac {e^{a-i b x-2 i d}}{4 b}-\frac {e^{a+2 i (b x+d)-i b x-2 i d}}{b}+\frac {e^{a+4 i (b x+d)-i b x-2 i d}}{12 b}\) |
Input:
Int[E^(a + I*b*x)*Sin[d + b*x]^2*Tan[d + b*x],x]
Output:
E^(a - (2*I)*d - I*b*x)/(4*b) - E^(a - (2*I)*d - I*b*x + (2*I)*(d + b*x))/ b + E^(a - (2*I)*d - I*b*x + (4*I)*(d + b*x))/(12*b) - (2*E^(a - I*d)*ArcT an[E^((-I)*d - I*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]
\[\int {\mathrm e}^{i b x +a} \sin \left (b x +d \right )^{2} \tan \left (b x +d \right )d x\]
Input:
int(exp(a+I*b*x)*sin(b*x+d)^2*tan(b*x+d),x)
Output:
int(exp(a+I*b*x)*sin(b*x+d)^2*tan(b*x+d),x)
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int e^{a+i b x} \sin ^2(d+b x) \tan (d+b x) \, dx=\frac {{\left (12 i \, e^{\left (i \, b x + a\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) - 12 i \, e^{\left (i \, b x + a\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) + e^{\left (4 i \, b x + a + 3 i \, d\right )} - 12 \, e^{\left (2 i \, b x + a + i \, d\right )} + 3 \, e^{\left (a - i \, d\right )}\right )} e^{\left (-i \, b x - i \, d\right )}}{12 \, b} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)^2*tan(b*x+d),x, algorithm="fricas")
Output:
1/12*(12*I*e^(I*b*x + a)*log(e^(I*b*x + I*d) + I) - 12*I*e^(I*b*x + a)*log (e^(I*b*x + I*d) - I) + e^(4*I*b*x + a + 3*I*d) - 12*e^(2*I*b*x + a + I*d) + 3*e^(a - I*d))*e^(-I*b*x - I*d)/b
Time = 0.38 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.53 \[ \int e^{a+i b x} \sin ^2(d+b x) \tan (d+b x) \, dx=\begin {cases} \frac {\left (4 b^{2} e^{a} e^{4 i d} e^{3 i b x} - 48 b^{2} e^{a} e^{2 i d} e^{i b x} + 12 b^{2} e^{a} e^{- i b x}\right ) e^{- 2 i d}}{48 b^{3}} & \text {for}\: b^{3} e^{2 i d} \neq 0 \\\frac {x \left (i e^{a} e^{4 i d} - 4 i e^{a} e^{2 i d} - i e^{a}\right ) e^{- 2 i d}}{4} & \text {otherwise} \end {cases} + \frac {e^{a} e^{- i d} \operatorname {RootSum} {\left (z^{2} + 1, \left ( i \mapsto i \log {\left (i e^{- i d} + e^{i b x} \right )} \right )\right )}}{b} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)**2*tan(b*x+d),x)
Output:
Piecewise(((4*b**2*exp(a)*exp(4*I*d)*exp(3*I*b*x) - 48*b**2*exp(a)*exp(2*I *d)*exp(I*b*x) + 12*b**2*exp(a)*exp(-I*b*x))*exp(-2*I*d)/(48*b**3), Ne(b** 3*exp(2*I*d), 0)), (x*(I*exp(a)*exp(4*I*d) - 4*I*exp(a)*exp(2*I*d) - I*exp (a))*exp(-2*I*d)/4, True)) + exp(a)*exp(-I*d)*RootSum(_z**2 + 1, Lambda(_i , _i*log(_i*exp(-I*d) + exp(I*b*x))))/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (64) = 128\).
Time = 0.16 (sec) , antiderivative size = 381, normalized size of antiderivative = 3.81 \[ \int e^{a+i b x} \sin ^2(d+b x) \tan (d+b x) \, dx=-\frac {12 \, {\left (\cos \left (d\right ) e^{a} - i \, e^{a} \sin \left (d\right )\right )} \arctan \left (\frac {2 \, {\left (\cos \left (b x + 2 \, d\right ) \cos \left (d\right ) + \sin \left (b x + 2 \, d\right ) \sin \left (d\right )\right )}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}, \frac {\cos \left (b x + 2 \, d\right )^{2} - \cos \left (d\right )^{2} + \sin \left (b x + 2 \, d\right )^{2} - \sin \left (d\right )^{2}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}\right ) - \cos \left (3 \, b x + 2 \, d\right ) e^{a} - 3 \, \cos \left (b x + 2 \, d\right ) e^{a} + 12 \, \cos \left (b x\right ) e^{a} - 6 \, {\left (-i \, \cos \left (d\right ) e^{a} - e^{a} \sin \left (d\right )\right )} \log \left (\frac {\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} - 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} + 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}{\cos \left (b x + 2 \, d\right )^{2} + \cos \left (d\right )^{2} + 2 \, \cos \left (d\right ) \sin \left (b x + 2 \, d\right ) + \sin \left (b x + 2 \, d\right )^{2} - 2 \, \cos \left (b x + 2 \, d\right ) \sin \left (d\right ) + \sin \left (d\right )^{2}}\right ) - i \, e^{a} \sin \left (3 \, b x + 2 \, d\right ) + 3 i \, e^{a} \sin \left (b x + 2 \, d\right ) + 12 i \, e^{a} \sin \left (b x\right )}{12 \, b} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)^2*tan(b*x+d),x, algorithm="maxima")
Output:
-1/12*(12*(cos(d)*e^a - I*e^a*sin(d))*arctan2(2*(cos(b*x + 2*d)*cos(d) + s in(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)) - cos(3*b*x + 2*d)*e^a - 3*cos(b*x + 2*d)*e^a + 12*cos(b*x)*e^ a - 6*(-I*cos(d)*e^a - e^a*sin(d))*log((cos(b*x + 2*d)^2 + cos(d)^2 - 2*co s(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)) - I*e^a*sin(3*b*x + 2*d) + 3*I*e ^a*sin(b*x + 2*d) + 12*I*e^a*sin(b*x))/b
Time = 0.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.89 \[ \int e^{a+i b x} \sin ^2(d+b x) \tan (d+b x) \, dx=-\frac {{\left (-12 i \, e^{\left (i \, b x + a\right )} \log \left (e^{\left (i \, b x + i \, d\right )} + i\right ) + 12 i \, e^{\left (i \, b x + a\right )} \log \left (e^{\left (i \, b x + i \, d\right )} - i\right ) - e^{\left (4 i \, b x + a + 3 i \, d\right )} + 12 \, e^{\left (2 i \, b x + a + i \, d\right )} - 3 \, e^{\left (a - i \, d\right )}\right )} e^{\left (-i \, b x - i \, d\right )}}{12 \, b} \] Input:
integrate(exp(a+I*b*x)*sin(b*x+d)^2*tan(b*x+d),x, algorithm="giac")
Output:
-1/12*(-12*I*e^(I*b*x + a)*log(e^(I*b*x + I*d) + I) + 12*I*e^(I*b*x + a)*l og(e^(I*b*x + I*d) - I) - e^(4*I*b*x + a + 3*I*d) + 12*e^(2*I*b*x + a + I* d) - 3*e^(a - I*d))*e^(-I*b*x - I*d)/b
Time = 0.94 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.73 \[ \int e^{a+i b x} \sin ^2(d+b x) \tan (d+b x) \, dx=\frac {{\mathrm {e}}^{a-d\,2{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{4\,b}+\frac {{\mathrm {e}}^{a+d\,2{}\mathrm {i}+b\,x\,3{}\mathrm {i}}}{12\,b}-\frac {{\mathrm {e}}^{a+b\,x\,1{}\mathrm {i}}}{b}+\frac {\sqrt {-{\mathrm {e}}^{2\,a-d\,2{}\mathrm {i}}}\,\ln \left (-{\mathrm {e}}^{3\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,2{}\mathrm {i}-{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {-{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}}\,2{}\mathrm {i}\right )}{b}-\frac {\sqrt {-{\mathrm {e}}^{2\,a-d\,2{}\mathrm {i}}}\,\ln \left (-{\mathrm {e}}^{3\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,2{}\mathrm {i}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}\,\sqrt {-{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d\,2{}\mathrm {i}}}\,2{}\mathrm {i}\right )}{b} \] Input:
int(exp(a + b*x*1i)*sin(d + b*x)^2*tan(d + b*x),x)
Output:
exp(a - d*2i - b*x*1i)/(4*b) + exp(a + d*2i + b*x*3i)/(12*b) - exp(a + b*x *1i)/b + ((-exp(2*a - d*2i))^(1/2)*log(- exp(3*a)*exp(-d*2i)*exp(b*x*1i)*2 i - exp(2*a)*exp(-d*2i)*(-exp(2*a)*exp(-d*2i))^(1/2)*2i))/b - ((-exp(2*a - d*2i))^(1/2)*log(exp(2*a)*exp(-d*2i)*(-exp(2*a)*exp(-d*2i))^(1/2)*2i - ex p(3*a)*exp(-d*2i)*exp(b*x*1i)*2i))/b
\[ \int e^{a+i b x} \sin ^2(d+b x) \tan (d+b x) \, dx=e^{a} \left (\int e^{b i x} \sin \left (b x +d \right )^{2} \tan \left (b x +d \right )d x \right ) \] Input:
int(exp(a+I*b*x)*sin(b*x+d)^2*tan(b*x+d),x)
Output:
e**a*int(e**(b*i*x)*sin(b*x + d)**2*tan(b*x + d),x)