Integrand size = 21, antiderivative size = 121 \[ \int e^{2 (a+i b x)} \tan ^3(d+b x) \, dx=\frac {e^{2 (a-i d)+2 i (d+b x)}}{2 b}+\frac {2 e^{2 a-2 i d}}{b \left (1+e^{2 i (d+b x)}\right )^2}-\frac {6 e^{2 a-2 i d}}{b \left (1+e^{2 i (d+b x)}\right )}-\frac {3 e^{2 a-2 i d} \log \left (1+e^{2 i (d+b x)}\right )}{b} \] Output:
1/2*exp(2*a-2*I*d+2*I*(b*x+d))/b+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)))-3*exp(2*a-2*I*d)*ln(1+exp(2*I*(b* x+d)))/b
Time = 0.28 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.99 \[ \int e^{2 (a+i b x)} \tan ^3(d+b x) \, dx=\frac {e^{2 a} \left (e^{2 i b x}-6 i \arctan \left (\frac {\left (-1+e^{2 i b x}\right ) \tan (d)}{1+e^{2 i b x}}\right ) \cos (2 d)-3 \cos (2 d) \log \left (1+e^{4 i b x}+2 e^{2 i b x} \cos (2 d)\right )+\frac {4 (\cos (d)-i \sin (d))^4}{\left (\left (1+e^{2 i b x}\right ) \cos (d)+i \left (-1+e^{2 i b x}\right ) \sin (d)\right )^2}-\frac {12 (\cos (d)-i \sin (d))^3}{\left (1+e^{2 i b x}\right ) \cos (d)+i \left (-1+e^{2 i b x}\right ) \sin (d)}-6 \arctan \left (\frac {\left (-1+e^{2 i b x}\right ) \tan (d)}{1+e^{2 i b x}}\right ) \sin (2 d)+3 i \log \left (1+e^{4 i b x}+2 e^{2 i b x} \cos (2 d)\right ) \sin (2 d)\right )}{2 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Tan[d + b*x]^3,x]
Output:
(E^(2*a)*(E^((2*I)*b*x) - (6*I)*ArcTan[((-1 + E^((2*I)*b*x))*Tan[d])/(1 + E^((2*I)*b*x))]*Cos[2*d] - 3*Cos[2*d]*Log[1 + E^((4*I)*b*x) + 2*E^((2*I)*b *x)*Cos[2*d]] + (4*(Cos[d] - I*Sin[d])^4)/((1 + E^((2*I)*b*x))*Cos[d] + I* (-1 + E^((2*I)*b*x))*Sin[d])^2 - (12*(Cos[d] - I*Sin[d])^3)/((1 + E^((2*I) *b*x))*Cos[d] + I*(-1 + E^((2*I)*b*x))*Sin[d]) - 6*ArcTan[((-1 + E^((2*I)* b*x))*Tan[d])/(1 + E^((2*I)*b*x))]*Sin[2*d] + (3*I)*Log[1 + E^((4*I)*b*x) + 2*E^((2*I)*b*x)*Cos[2*d]]*Sin[2*d]))/(2*b)
Time = 0.45 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4942, 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.
\(\displaystyle \int e^{2 (a+i b x)} \tan ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 4942 |
\(\displaystyle -i \int \left (-e^{2 (a+i b x)}+\frac {6 e^{2 (a+i b x)}}{1+e^{2 i (d+b x)}}-\frac {12 e^{2 (a+i b x)}}{\left (1+e^{2 i (d+b x)}\right )^2}+\frac {8 e^{2 (a+i b x)}}{\left (1+e^{2 i (d+b x)}\right )^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -i \left (-\frac {6 i e^{2 a-2 i d}}{b \left (1+e^{2 i (b x+d)}\right )}+\frac {2 i e^{2 a-2 i d}}{b \left (1+e^{2 i (b x+d)}\right )^2}-\frac {3 i e^{2 a-2 i d} \log \left (1+e^{2 i (b x+d)}\right )}{b}+\frac {i e^{2 (a+i b x)}}{2 b}\right )\) |
Input:
Int[E^(2*(a + I*b*x))*Tan[d + b*x]^3,x]
Output:
(-I)*(((I/2)*E^(2*(a + I*b*x)))/b + ((2*I)*E^(2*a - (2*I)*d))/(b*(1 + E^(( 2*I)*(d + b*x)))^2) - ((6*I)*E^(2*a - (2*I)*d))/(b*(1 + E^((2*I)*(d + b*x) ))) - ((3*I)*E^(2*a - (2*I)*d)*Log[1 + E^((2*I)*(d + b*x))])/b)
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tan[(d_.) + (e_.)*(x_)]^(n_.), x_Symb ol] :> Simp[I^n Int[ExpandIntegrand[F^(c*(a + b*x))*((1 - E^(2*I*(d + e*x )))^n/(1 + E^(2*I*(d + e*x)))^n), x], x], x] /; FreeQ[{F, a, b, c, d, e}, x ] && IntegerQ[n]
Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {{\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{2 b}+\frac {4 \,{\mathrm e}^{4 i b x} {\mathrm e}^{2 i d} {\mathrm e}^{2 a}+2 \,{\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{b \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{2}}-\frac {3 \,{\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \ln \left (1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}\) | \(90\) |
Input:
int(exp(2*a+2*I*b*x)*tan(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
1/2*exp(2*a)*exp(2*I*b*x)/b+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)*exp(2*I*b*x))-3*exp(2*a)/b*exp(-2*I*d)*ln(1+exp( 2*I*(b*x+d)))
Time = 0.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.03 \[ \int e^{2 (a+i b x)} \tan ^3(d+b x) \, dx=-\frac {6 \, {\left (e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + 2 \, e^{\left (2 i \, b x + 2 \, a\right )} + e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} - 2 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + 11 \, e^{\left (2 i \, b x + 2 \, a\right )} + 8 \, e^{\left (2 \, a - 2 i \, d\right )}}{2 \, {\left (b e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, b e^{\left (2 i \, b x + 2 i \, d\right )} + b\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*tan(b*x+d)^3,x, algorithm="fricas")
Output:
-1/2*(6*(e^(4*I*b*x + 2*a + 2*I*d) + 2*e^(2*I*b*x + 2*a) + e^(2*a - 2*I*d) )*log(e^(2*I*b*x + 2*I*d) + 1) - e^(6*I*b*x + 2*a + 4*I*d) - 2*e^(4*I*b*x + 2*a + 2*I*d) + 11*e^(2*I*b*x + 2*a) + 8*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)
Time = 0.23 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int e^{2 (a+i b x)} \tan ^3(d+b x) \, dx=\frac {- 6 e^{2 a} e^{2 i d} e^{2 i b x} - 4 e^{2 a}}{b e^{6 i d} e^{4 i b x} + 2 b e^{4 i d} e^{2 i b x} + b e^{2 i d}} + \begin {cases} \frac {e^{2 a} e^{2 i b x}}{2 b} & \text {for}\: b \neq 0 \\i x e^{2 a} & \text {otherwise} \end {cases} - \frac {3 e^{2 a} e^{- 2 i d} \log {\left (e^{2 i b x} + e^{- 2 i d} \right )}}{b} \] Input:
integrate(exp(2*a+2*I*b*x)*tan(b*x+d)**3,x)
Output:
(-6*exp(2*a)*exp(2*I*d)*exp(2*I*b*x) - 4*exp(2*a))/(b*exp(6*I*d)*exp(4*I*b *x) + 2*b*exp(4*I*d)*exp(2*I*b*x) + b*exp(2*I*d)) + Piecewise((exp(2*a)*ex p(2*I*b*x)/(2*b), Ne(b, 0)), (I*x*exp(2*a), True)) - 3*exp(2*a)*exp(-2*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. 535 vs. \(2 (92) = 184\).
Time = 0.06 (sec) , antiderivative size = 535, normalized size of antiderivative = 4.42 \[ \int e^{2 (a+i b x)} \tan ^3(d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(2*a+2*I*b*x)*tan(b*x+d)^3,x, algorithm="maxima")
Output:
-(6*(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)) + I*cos(6*b*x + 6*d)*e^(2*a) + 2*I*cos(4*b*x + 4*d)*e^(2*a) - 11*I*cos(2*b*x + 2*d)*e^(2*a) + 3*(-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(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^(2*a)*sin(6*b*x + 6*d) - 2*e^(2*a)*sin(4*b*x + 4*d) + 11*e^(2*a)*sin(2*b*x + 2*d) - 8*I*e^(2*a))/(-2*I*b*cos(4*b*x + 6*d) - 4* I*b*cos(2*b*x + 4*d) - 2*I*b*cos(2*d) + 2*b*sin(4*b*x + 6*d) + 4*b*sin(2*b *x + 4*d) + 2*b*sin(2*d))
Time = 1.11 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.24 \[ \int e^{2 (a+i b x)} \tan ^3(d+b x) \, dx=-\frac {6 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + 12 \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) + 6 \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right ) - e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} - 2 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} + 11 \, e^{\left (2 i \, b x + 2 \, a\right )} + 8 \, e^{\left (2 \, a - 2 i \, d\right )}}{2 \, b {\left (e^{\left (4 i \, b x + 4 i \, d\right )} + 2 \, e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*tan(b*x+d)^3,x, algorithm="giac")
Output:
-1/2*(6*e^(4*I*b*x + 2*a + 2*I*d)*log(e^(2*I*b*x + 2*I*d) + 1) + 12*e^(2*I *b*x + 2*a)*log(e^(2*I*b*x + 2*I*d) + 1) + 6*e^(2*a - 2*I*d)*log(e^(2*I*b* x + 2*I*d) + 1) - e^(6*I*b*x + 2*a + 4*I*d) - 2*e^(4*I*b*x + 2*a + 2*I*d) + 11*e^(2*I*b*x + 2*a) + 8*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. \[ \int e^{2 (a+i b x)} \tan ^3(d+b x) \, dx=\int {\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}}\,{\mathrm {tan}\left (d+b\,x\right )}^3 \,d x \] Input:
int(exp(2*a + b*x*2i)*tan(d + b*x)^3,x)
Output:
int(exp(2*a + b*x*2i)*tan(d + b*x)^3, x)
\[ \int e^{2 (a+i b x)} \tan ^3(d+b x) \, dx=e^{2 a} \left (\int e^{2 b i x} \tan \left (b x +d \right )^{3}d x \right ) \] Input:
int(exp(2*a+2*I*b*x)*tan(b*x+d)^3,x)
Output:
e**(2*a)*int(e**(2*b*i*x)*tan(b*x + d)**3,x)