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]

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

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Rubi [A]  time = 0.02, antiderivative size = 14, 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[1/(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 = 14, normalized size = 1.00 \[ \log (x)-2 \log (-a x+i) \]

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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fricas [A]  time = 0.41, size = 15, normalized size = 1.07 \[ \log \relax (x) - 2 \, \log \left (\frac {a x - i}{a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 [B]  time = 0.13, size = 47, normalized size = 3.36 \[ -a i {\left (\frac {i \log \left (-i + \frac {i}{a i x + 1}\right )}{a} + \frac {i \log \left (\frac {1}{\sqrt {a^{2} x^{2} + 1} {\left | a \right |}}\right )}{a}\right )} \]

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)

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maple [A]  time = 0.06, size = 23, normalized size = 1.64 \[ \ln \relax (x )-\ln \left (a^{2} x^{2}+1\right )-2 i \arctan \left (a x \right ) \]

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

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maxima [A]  time = 0.32, size = 12, normalized size = 0.86 \[ -2 \, \log \left (i \, a x + 1\right ) + \log \relax (x) \]

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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mupad [B]  time = 0.45, size = 14, normalized size = 1.00 \[ \ln \relax (x)-2\,\ln \left (x-\frac {1{}\mathrm {i}}{a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - 2*log(x - 1i/a)

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sympy [A]  time = 0.15, size = 17, normalized size = 1.21 \[ \log {\left (3 a x \right )} - 2 \log {\left (3 a x - 3 i \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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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