3.46 \(\int e^{-2 i \tan ^{-1}(a x)} x \, dx\)

Optimal. Leaf size=30 \[ \frac{2 \log (-a x+i)}{a^2}-\frac{2 i x}{a}-\frac{x^2}{2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0210554, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5062, 77} \[ \frac{2 \log (-a x+i)}{a^2}-\frac{2 i x}{a}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[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 5062

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]

Rule 77

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]))))

Rubi steps

\begin{align*} \int e^{-2 i \tan ^{-1}(a x)} x \, dx &=\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]  time = 0.0101914, size = 30, normalized size = 1. \[ \frac{2 \log (-a x+i)}{a^2}-\frac{2 i x}{a}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[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]  time = 0.041, size = 38, normalized size = 1.3 \begin{align*} -{\frac{{x}^{2}}{2}}-{\frac{2\,ix}{a}}+{\frac{\ln \left ({a}^{2}{x}^{2}+1 \right ) }{{a}^{2}}}+{\frac{2\,i\arctan \left ( ax \right ) }{{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.08784, size = 38, normalized size = 1.27 \begin{align*} \frac{i \,{\left (i \, a x^{2} - 4 \, x\right )}}{2 \, a} + \frac{2 \, \log \left (i \, a x + 1\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 1.51843, size = 69, normalized size = 2.3 \begin{align*} -\frac{a^{2} x^{2} + 4 i \, a x - 4 \, \log \left (\frac{a x - i}{a}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0.175426, size = 22, normalized size = 0.73 \begin{align*} - \frac{x^{2}}{2} - \frac{2 i x}{a} + \frac{2 \log{\left (a x - i \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Giac [B]  time = 1.11053, size = 78, normalized size = 2.6 \begin{align*} \frac{i{\left (\frac{4 \, i \log \left (\frac{1}{\sqrt{a^{2} x^{2} + 1}{\left | a \right |}}\right )}{a} + \frac{{\left (a i x + 1\right )}^{2}{\left (i - \frac{6 \, i}{a i x + 1}\right )}}{a i^{2}}\right )}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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