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

Optimal. Leaf size=27 \[ -2 i a \log (x)+2 i a \log (-a x+i)-\frac{1}{x} \]

[Out]

-x^(-1) - (2*I)*a*Log[x] + (2*I)*a*Log[I - a*x]

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Rubi [A]  time = 0.0255769, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5062, 77} \[ -2 i a \log (x)+2 i a \log (-a x+i)-\frac{1}{x} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((2*I)*ArcTan[a*x])*x^2),x]

[Out]

-x^(-1) - (2*I)*a*Log[x] + (2*I)*a*Log[I - a*x]

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 \frac{e^{-2 i \tan ^{-1}(a x)}}{x^2} \, dx &=\int \frac{1-i a x}{x^2 (1+i a x)} \, dx\\ &=\int \left (\frac{1}{x^2}-\frac{2 i a}{x}+\frac{2 i a^2}{-i+a x}\right ) \, dx\\ &=-\frac{1}{x}-2 i a \log (x)+2 i a \log (i-a x)\\ \end{align*}

Mathematica [A]  time = 0.0092211, size = 27, normalized size = 1. \[ -2 i a \log (x)+2 i a \log (-a x+i)-\frac{1}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^((2*I)*ArcTan[a*x])*x^2),x]

[Out]

-x^(-1) - (2*I)*a*Log[x] + (2*I)*a*Log[I - a*x]

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Maple [A]  time = 0.044, size = 34, normalized size = 1.3 \begin{align*} -2\,a\arctan \left ( ax \right ) +ia\ln \left ({a}^{2}{x}^{2}+1 \right ) -{x}^{-1}-2\,ia\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05493, size = 46, normalized size = 1.7 \begin{align*} 2 i \, a \log \left (i \, a x + 1\right ) - 2 i \, a \log \left (x\right ) - \frac{a x - i}{a x^{2} - i \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.74535, size = 70, normalized size = 2.59 \begin{align*} \frac{-2 i \, a x \log \left (x\right ) + 2 i \, a x \log \left (\frac{a x - i}{a}\right ) - 1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.420873, size = 20, normalized size = 0.74 \begin{align*} - 2 a \left (i \log{\left (x \right )} - i \log{\left (x - \frac{i}{a} \right )}\right ) - \frac{1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*(I*log(x) - I*log(x - I/a)) - 1/x

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Giac [A]  time = 1.09886, size = 50, normalized size = 1.85 \begin{align*} -2 \, a i \log \left (-i + \frac{i}{a i x + 1}\right ) + \frac{a}{i - \frac{i}{a i x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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