\(\int e^{-2 i \arctan (a x)} x \, dx\) [46]

Optimal result

Integrand size = 12, antiderivative size = 30 \[ \int e^{-2 i \arctan (a x)} x \, dx=-\frac {2 i x}{a}-\frac {x^2}{2}+\frac {2 \log (i-a x)}{a^2} \]

[Out]

-2*I*x/a-1/2*x^2+2*ln(I-a*x)/a^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5170, 78} \[ \int e^{-2 i \arctan (a x)} x \, dx=\frac {2 \log (-a x+i)}{a^2}-\frac {2 i x}{a}-\frac {x^2}{2} \]

[In]

Int[x/E^((2*I)*ArcTan[a*x]),x]

[Out]

((-2*I)*x)/a - x^2/2 + (2*Log[I - a*x])/a^2

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 5170

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^(I*(n/2))/(1 + I*a*x)^(I*(n/2))
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x (1-i a x)}{1+i a x} \, dx \\ & = \int \left (-\frac {2 i}{a}-x+\frac {2}{a (-i+a x)}\right ) \, dx \\ & = -\frac {2 i x}{a}-\frac {x^2}{2}+\frac {2 \log (i-a x)}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int e^{-2 i \arctan (a x)} x \, dx=-\frac {2 i x}{a}-\frac {x^2}{2}+\frac {2 \log (i-a x)}{a^2} \]

[In]

Integrate[x/E^((2*I)*ArcTan[a*x]),x]

[Out]

((-2*I)*x)/a - x^2/2 + (2*Log[I - a*x])/a^2

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03

method	result	size
default	\(-\frac {\frac {1}{2} a \,x^{2}+2 i x}{a}+\frac {2 \ln \left (-a x +i\right )}{a^{2}}\)	\(31\)
risch	\(-\frac {x^{2}}{2}-\frac {2 i x}{a}+\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{2}}+\frac {2 i \arctan \left (a x \right )}{a^{2}}\)	\(38\)
parallelrisch	\(\frac {a^{3} x^{3}+3 i a^{2} x^{2}-4 \ln \left (a x -i\right ) x a +4 i \ln \left (a x -i\right )+4 a x}{2 a^{2} \left (-a x +i\right )}\)	\(57\)
meijerg	\(\frac {-\frac {i a x \left (2 a^{2} x^{2}+6 i a x +12\right )}{4 \left (i a x +1\right )}+3 \ln \left (i a x +1\right )}{a^{2}}-\frac {-\frac {i a x}{i a x +1}+\ln \left (i a x +1\right )}{a^{2}}\)	\(74\)

[In]

int(x/(1+I*a*x)^2*(a^2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

-1/a*(1/2*a*x^2+2*I*x)+2*ln(I-a*x)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int e^{-2 i \arctan (a x)} x \, dx=-\frac {a^{2} x^{2} + 4 i \, a x - 4 \, \log \left (\frac {a x - i}{a}\right )}{2 \, a^{2}} \]

[In]

integrate(x/(1+I*a*x)^2*(a^2*x^2+1),x, algorithm="fricas")

[Out]

-1/2*(a^2*x^2 + 4*I*a*x - 4*log((a*x - I)/a))/a^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int e^{-2 i \arctan (a x)} x \, dx=- \frac {x^{2}}{2} - \frac {2 i x}{a} + \frac {2 \log {\left (a x - i \right )}}{a^{2}} \]

[In]

integrate(x/(1+I*a*x)**2*(a**2*x**2+1),x)

[Out]

-x**2/2 - 2*I*x/a + 2*log(a*x - I)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int e^{-2 i \arctan (a x)} x \, dx=\frac {i \, {\left (i \, a x^{2} - 4 \, x\right )}}{2 \, a} + \frac {2 \, \log \left (i \, a x + 1\right )}{a^{2}} \]

[In]

integrate(x/(1+I*a*x)^2*(a^2*x^2+1),x, algorithm="maxima")

[Out]

1/2*I*(I*a*x^2 - 4*x)/a + 2*log(I*a*x + 1)/a^2

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int e^{-2 i \arctan (a x)} x \, dx=-\frac {i \, {\left (\frac {{\left (i \, a x + 1\right )}^{2} {\left (-\frac {6 i}{i \, a x + 1} + i\right )}}{a} - \frac {4 i \, \log \left (\frac {1}{\sqrt {a^{2} x^{2} + 1} {\left | a \right |}}\right )}{a}\right )}}{2 \, a} \]

[In]

integrate(x/(1+I*a*x)^2*(a^2*x^2+1),x, algorithm="giac")

[Out]

-1/2*I*((I*a*x + 1)^2*(-6*I/(I*a*x + 1) + I)/a - 4*I*log(1/(sqrt(a^2*x^2 + 1)*abs(a)))/a)/a

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int e^{-2 i \arctan (a x)} x \, dx=\frac {2\,\ln \left (x-\frac {1{}\mathrm {i}}{a}\right )}{a^2}-\frac {x^2}{2}-\frac {x\,2{}\mathrm {i}}{a} \]

[In]

int((x*(a^2*x^2 + 1))/(a*x*1i + 1)^2,x)

[Out]

(2*log(x - 1i/a))/a^2 - (x*2i)/a - x^2/2