\(\int \frac {e^{-2 i \arctan (a x)}}{x} \, dx\) [48]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int \frac {e^{-2 i \arctan (a x)}}{x} \, dx=\log (x)-2 \log (i-a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5170, 78} \[ \int \frac {e^{-2 i \arctan (a x)}}{x} \, dx=\log (x)-2 \log (-a x+i) \]

[In]

Int[1/(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*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 i \arctan (a x)}}{x} \, dx=\log (x)-2 \log (i-a x) \]

[In]

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

[Out]

Log[x] - 2*Log[I - a*x]

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00

method	result	size
default	\(\ln \left (x \right )-2 \ln \left (-a x +i\right )\)	\(14\)
parallelrisch	\(\frac {\ln \left (x \right ) a -2 \ln \left (a x -i\right ) a}{a}\)	\(20\)
risch	\(\ln \left (x \right )-\ln \left (a^{2} x^{2}+1\right )-2 i \arctan \left (a x \right )\)	\(23\)
meijerg	\(\frac {i a x}{i a x +1}-2 \ln \left (i a x +1\right )-\frac {2 i a x}{2 i a x +2}+1+\ln \left (x \right )+\ln \left (i a \right )\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2 i \arctan (a x)}}{x} \, dx=\log \left (x\right ) - 2 \, \log \left (\frac {a x - i}{a}\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-2 i \arctan (a x)}}{x} \, dx=\log {\left (3 a x \right )} - 2 \log {\left (3 a x - 3 i \right )} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-2 i \arctan (a x)}}{x} \, dx=-2 \, \log \left (i \, a x + 1\right ) + \log \left (x\right ) \]

[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)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (12) = 24\).

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 3.14 \[ \int \frac {e^{-2 i \arctan (a x)}}{x} \, dx=i \, a {\left (-\frac {i \, \log \left (\frac {i}{i \, a x + 1} - i\right )}{a} - \frac {i \, \log \left (\frac {1}{\sqrt {a^{2} x^{2} + 1} {\left | a \right |}}\right )}{a}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 i \arctan (a x)}}{x} \, dx=\ln \left (x\right )-2\,\ln \left (x-\frac {1{}\mathrm {i}}{a}\right ) \]

[In]

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

[Out]

log(x) - 2*log(x - 1i/a)