3.1.15 \(\int \frac {e^{2 i \text {ArcTan}(a x)}}{x} \, dx\) [15]

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {5170, 78} \begin {gather*} \log (x)-2 \log (a x+i) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - 2*Log[I + a*x]

Rule 78

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]))))

Rule 5170

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 \begin {gather*} \log (x)-2 \log (i+a x) \end {gather*}

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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (12 ) = 24\).
time = 0.07, size = 33, normalized size = 2.54

method	result	size
risch	\(-\ln \left (a^{2} x^{2}+1\right )+2 i \arctan \left (a x \right )+\ln \left (-x \right )\)	\(25\)
meijerg	\(-\ln \left (a^{2} x^{2}+1\right )+\ln \left (x \right )+\frac {\ln \left (a^{2}\right )}{2}+2 i \arctan \left (a x \right )\)	\(29\)
default	\(2 a \left (-\frac {\ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {i \arctan \left (a x \right )}{a}\right )+\ln \left (x \right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.47, size = 21, normalized size = 1.62 \begin {gather*} 2 i \, \arctan \left (a x\right ) - \log \left (a^{2} x^{2} + 1\right ) + \log \left (x\right ) \end {gather*}

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 = 1.87, size = 15, normalized size = 1.15 \begin {gather*} \log \left (x\right ) - 2 \, \log \left (\frac {a x + i}{a}\right ) \end {gather*}

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 = 0.09, size = 17, normalized size = 1.31 \begin {gather*} \log {\left (3 a x \right )} - 2 \log {\left (3 a x + 3 i \right )} \end {gather*}

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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Giac [A]
time = 0.41, size = 12, normalized size = 0.92 \begin {gather*} -2 \, \log \left (a x + i\right ) + \log \left ({\left | x \right |}\right ) \end {gather*}

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*log(a*x + I) + log(abs(x))

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

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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