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

Optimal. Leaf size=14 \[ \log (x)-2 \log (-a x+i) \]

[Out]

Log[x] - 2*Log[I - a*x]

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Rubi [A]  time = 0.0200594, antiderivative size = 14, 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, 72} \[ \log (x)-2 \log (-a x+i) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - 2*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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0069279, size = 14, normalized size = 1. \[ \log (x)-2 \log (-a x+i) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - 2*Log[I - a*x]

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Maple [A]  time = 0.044, size = 23, normalized size = 1.6 \begin{align*} -\ln \left ({a}^{2}{x}^{2}+1 \right ) -2\,i\arctan \left ( ax \right ) +\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,x)

[Out]

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

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Maxima [A]  time = 1.02635, size = 16, normalized size = 1.14 \begin{align*} -2 \, \log \left (i \, a x + 1\right ) + \log \left (x\right ) \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,x, algorithm="maxima")

[Out]

-2*log(I*a*x + 1) + log(x)

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Fricas [A]  time = 1.62398, size = 39, normalized size = 2.79 \begin{align*} \log \left (x\right ) - 2 \, \log \left (\frac{a x - i}{a}\right ) \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,x, algorithm="fricas")

[Out]

log(x) - 2*log((a*x - I)/a)

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Sympy [A]  time = 0.418123, size = 10, normalized size = 0.71 \begin{align*} \log{\left (x \right )} - 2 \log{\left (x - \frac{i}{a} \right )} \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,x)

[Out]

log(x) - 2*log(x - I/a)

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Giac [B]  time = 1.09633, size = 63, normalized size = 4.5 \begin{align*} -a i{\left (\frac{i \log \left (-i + \frac{i}{a i x + 1}\right )}{a} + \frac{i \log \left (\frac{1}{\sqrt{a^{2} x^{2} + 1}{\left | a \right |}}\right )}{a}\right )} \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,x, algorithm="giac")

[Out]

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