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

Optimal. Leaf size=13 \[ \log (x)-2 \log (a x+i) \]

[Out]

Log[x] - 2*Log[I + a*x]

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Rubi [A]  time = 0.0194988, antiderivative size = 13, 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[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 = 0.0058213, size = 13, normalized size = 1. \[ \log (x)-2 \log (a x+i) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - 2*Log[I + a*x]

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Maple [A]  time = 0.041, size = 23, normalized size = 1.8 \begin{align*} 2\,i\arctan \left ( ax \right ) -\ln \left ({a}^{2}{x}^{2}+1 \right ) +\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.50648, size = 28, normalized size = 2.15 \begin{align*} 2 i \, \arctan \left (a x\right ) - \log \left (a^{2} x^{2} + 1\right ) + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.68503, size = 39, normalized size = 3. \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+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 = 0.407838, size = 10, normalized size = 0.77 \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+I*a*x)**2/(a**2*x**2+1)/x,x)

[Out]

log(x) - 2*log(x + I/a)

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Giac [A]  time = 1.13246, size = 20, normalized size = 1.54 \begin{align*} 2 \, i^{2} \log \left (a x + i\right ) + \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*i^2*log(a*x + i) + log(abs(x))