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]

ln(x)-2*ln(I+a*x)

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \[ \log (x)-2 \log (a x+i) \]

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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fricas [A]  time = 0.39, size = 15, normalized size = 1.15 \[ \log \relax (x) - 2 \, \log \left (\frac {a x + i}{a}\right ) \]

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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giac [A]  time = 0.12, size = 15, normalized size = 1.15 \[ 2 \, i^{2} \log \left (a x + i\right ) + \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.04, size = 23, normalized size = 1.77 \[ \ln \relax (x )+2 i \arctan \left (a x \right )-\ln \left (a^{2} x^{2}+1\right ) \]

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]

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

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maxima [A]  time = 0.45, size = 21, normalized size = 1.62 \[ 2 i \, \arctan \left (a x\right ) - \log \left (a^{2} x^{2} + 1\right ) + \log \relax (x) \]

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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mupad [B]  time = 0.44, size = 14, normalized size = 1.08 \[ \ln \relax (x)-2\,\ln \left (x+\frac {1{}\mathrm {i}}{a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - 2*log(x + 1i/a)

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sympy [A]  time = 0.15, size = 17, normalized size = 1.31 \[ \log {\left (3 a x \right )} - 2 \log {\left (3 a x + 3 i \right )} \]

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(3*a*x) - 2*log(3*a*x + 3*I)

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