Integrand size = 21, antiderivative size = 127 \[ \int e^{2 (a+i b x)} \cot ^3(d+b x) \, dx=-\frac {e^{2 (a-i d)+2 i (d+b x)}}{2 b}+\frac {2 e^{2 a-2 i d}}{b \left (1-e^{2 i (d+b x)}\right )^2}-\frac {6 e^{2 a-2 i d}}{b \left (1-e^{2 i (d+b x)}\right )}-\frac {3 e^{2 a-2 i d} \log \left (1-e^{2 i (d+b x)}\right )}{b} \] 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 = 243, normalized size of antiderivative = 1.91 \[ \int e^{2 (a+i b x)} \cot ^3(d+b x) \, dx=\frac {e^{2 a} \left (-e^{2 i b x}-6 i \arctan \left (\frac {\left (1+e^{2 i b x}\right ) \tan (d)}{-1+e^{2 i b x}}\right ) \cos (2 d)-3 \cos (2 d) \log \left (1+e^{4 i b x}-2 e^{2 i b x} \cos (2 d)\right )+\frac {4 (\cos (d)-i \sin (d))^4}{\left (\left (-1+e^{2 i b x}\right ) \cos (d)+i \left (1+e^{2 i b x}\right ) \sin (d)\right )^2}+\frac {12 (\cos (d)-i \sin (d))^3}{\left (-1+e^{2 i b x}\right ) \cos (d)+i \left (1+e^{2 i b x}\right ) \sin (d)}-6 \arctan \left (\frac {\left (1+e^{2 i b x}\right ) \tan (d)}{-1+e^{2 i b x}}\right ) \sin (2 d)+3 i \log \left (1+e^{4 i b x}-2 e^{2 i b x} \cos (2 d)\right ) \sin (2 d)\right )}{2 b} \] Input:
Integrate[E^(2*(a + I*b*x))*Cot[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 = 0.49 (sec) , antiderivative size = 130, 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 = {4943, 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.
\(\displaystyle \int e^{2 (a+i b x)} \cot ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 4943 |
\(\displaystyle i \int \left (-e^{2 (a+i b x)}+\frac {6 e^{2 (a+i b x)}}{1-e^{2 i (d+b x)}}-\frac {12 e^{2 (a+i b x)}}{\left (1-e^{2 i (d+b x)}\right )^2}+\frac {8 e^{2 (a+i b x)}}{\left (1-e^{2 i (d+b x)}\right )^3}\right )dx\) |
\(\Big \downarrow \) 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))*Cot[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[Cot[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), 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 = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {{\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{2 b}+\frac {4 \,{\mathrm e}^{4 i b x} {\mathrm e}^{2 i d} {\mathrm e}^{2 a}-2 \,{\mathrm e}^{2 a} {\mathrm e}^{2 i b x}}{b \left (-1+{\mathrm e}^{2 i \left (b x +d \right )}\right )^{2}}-\frac {3 \,{\mathrm e}^{2 a} {\mathrm e}^{-2 i d} \ln \left (-1+{\mathrm e}^{2 i \left (b x +d \right )}\right )}{b}\) | \(91\) |
Input:
int(exp(2*a+2*I*b*x)*cot(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)*e xp(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+e xp(2*I*(b*x+d)))
Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.97 \[ \int e^{2 (a+i b x)} \cot ^3(d+b x) \, dx=-\frac {6 \, {\left (e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - 2 \, e^{\left (2 i \, b x + 2 \, a\right )} + e^{\left (2 \, a - 2 i \, d\right )}\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) + e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} - 2 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - 11 \, e^{\left (2 i \, b x + 2 \, a\right )} + 8 \, e^{\left (2 \, a - 2 i \, d\right )}}{2 \, {\left (b e^{\left (4 i \, b x + 4 i \, d\right )} - 2 \, b e^{\left (2 i \, b x + 2 i \, d\right )} + b\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*cot(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 = 134, normalized size of antiderivative = 1.06 \[ \int e^{2 (a+i b x)} \cot ^3(d+b x) \, dx=\frac {6 e^{2 a} e^{2 i d} e^{2 i b x} - 4 e^{2 a}}{b e^{6 i d} e^{4 i b x} - 2 b e^{4 i d} e^{2 i b x} + b e^{2 i d}} + \begin {cases} - \frac {e^{2 a} e^{2 i b x}}{2 b} & \text {for}\: b \neq 0 \\- i x e^{2 a} & \text {otherwise} \end {cases} - \frac {3 e^{2 a} e^{- 2 i d} \log {\left (e^{2 i b x} - e^{- 2 i d} \right )}}{b} \] Input:
integrate(exp(2*a+2*I*b*x)*cot(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. 869 vs. \(2 (94) = 188\).
Time = 0.07 (sec) , antiderivative size = 869, normalized size of antiderivative = 6.84 \[ \int e^{2 (a+i b x)} \cot ^3(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(2*a+2*I*b*x)*cot(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(s in(b*x) + sin(d), cos(b*x) - cos(d)) + 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*(co s(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(sin(b*x) - sin(d), cos(b*x) + cos(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)*s in(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(b*x)^2 + 2*cos(b*x)*cos(d) + cos(d)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(d) + sin(d)^2) + 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)*s in(2*d))*sin(2*b*x + 4*d))*log(cos(b*x)^2 - 2*cos(b*x)*cos(d) + cos(d)^...
Time = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17 \[ \int e^{2 (a+i b x)} \cot ^3(d+b x) \, dx=-\frac {6 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) - 12 \, e^{\left (2 i \, b x + 2 \, a\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) + 6 \, e^{\left (2 \, a - 2 i \, d\right )} \log \left (e^{\left (2 i \, b x + 2 i \, d\right )} - 1\right ) + e^{\left (6 i \, b x + 2 \, a + 4 i \, d\right )} - 2 \, e^{\left (4 i \, b x + 2 \, a + 2 i \, d\right )} - 11 \, e^{\left (2 i \, b x + 2 \, a\right )} + 8 \, e^{\left (2 \, a - 2 i \, d\right )}}{2 \, b {\left (e^{\left (4 i \, b x + 4 i \, d\right )} - 2 \, e^{\left (2 i \, b x + 2 i \, d\right )} + 1\right )}} \] Input:
integrate(exp(2*a+2*I*b*x)*cot(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. \[ \int e^{2 (a+i b x)} \cot ^3(d+b x) \, dx=\int {\mathrm {cot}\left (d+b\,x\right )}^3\,{\mathrm {e}}^{2\,a+b\,x\,2{}\mathrm {i}} \,d x \] Input:
int(cot(d + b*x)^3*exp(2*a + b*x*2i),x)
Output:
int(cot(d + b*x)^3*exp(2*a + b*x*2i), x)
\[ \int e^{2 (a+i b x)} \cot ^3(d+b x) \, dx=e^{2 a} \left (\int e^{2 b i x} \cot \left (b x +d \right )^{3}d x \right ) \] Input:
int(exp(2*a+2*I*b*x)*cot(b*x+d)^3,x)
Output:
e**(2*a)*int(e**(2*b*i*x)*cot(b*x + d)**3,x)