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

Optimal result

Integrand size = 14, antiderivative size = 13 \[ \int \frac {e^{2 i \arctan (a x)}}{x} \, dx=\log (x)-2 \log (i+a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, 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[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 = 13, 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[E^((2*I)*ArcTan[a*x])/x,x]

[Out]

Log[x] - 2*Log[I + a*x]

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\ln \left (x \right )-2 \ln \left (a x +i\right )\)	\(13\)
risch	\(\ln \left (-x \right )-\ln \left (a^{2} x^{2}+1\right )+2 i \arctan \left (a x \right )\)	\(25\)
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	\(\ln \left (x \right )+2 a \left (-\frac {\ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {i \arctan \left (a x \right )}{a}\right )\)	\(33\)

[In]

int((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 = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {e^{2 i \arctan (a x)}}{x} \, dx=\log \left (x\right ) - 2 \, \log \left (\frac {a x + i}{a}\right ) \]

[In]

integrate((1+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 = 1.31 \[ \int \frac {e^{2 i \arctan (a x)}}{x} \, dx=\log {\left (3 a x \right )} - 2 \log {\left (3 a x + 3 i \right )} \]

[In]

integrate((1+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 = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.62 \[ \int \frac {e^{2 i \arctan (a x)}}{x} \, dx=2 i \, \arctan \left (a x\right ) - \log \left (a^{2} x^{2} + 1\right ) + \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {e^{2 i \arctan (a x)}}{x} \, dx=-2 \, \log \left (a x + i\right ) + \log \left ({\left | x \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int \frac {e^{2 i \arctan (a x)}}{x} \, dx=\ln \left (x\right )-2\,\ln \left (x+\frac {1{}\mathrm {i}}{a}\right ) \]

[In]

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

[Out]

log(x) - 2*log(x + 1i/a)