\(\int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx\) [16]

Optimal result

Integrand size = 14, antiderivative size = 26 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-\frac {1}{x}+2 i a \log (x)-2 i a \log (i+a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, 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} \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=2 i a \log (x)-2 i a \log (a x+i)-\frac {1}{x} \]

[In]

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

[Out]

-x^(-1) + (2*I)*a*Log[x] - (2*I)*a*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^2 (1-i a x)} \, dx \\ & = \int \left (\frac {1}{x^2}+\frac {2 i a}{x}-\frac {2 i a^2}{i+a x}\right ) \, dx \\ & = -\frac {1}{x}+2 i a \log (x)-2 i a \log (i+a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-\frac {1}{x}+2 i a \log (x)-2 i a \log (i+a x) \]

[In]

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

[Out]

-x^(-1) + (2*I)*a*Log[x] - (2*I)*a*Log[I + a*x]

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {2 i a \ln \left (x \right ) x -2 i a \ln \left (a x +i\right ) x -1}{x}\)	\(26\)
risch	\(-\frac {1}{x}-2 a \arctan \left (a x \right )-i a \ln \left (a^{2} x^{2}+1\right )+2 i a \ln \left (x \right )\)	\(34\)
default	\(-\frac {1}{x}+2 i a \ln \left (x \right )-2 a^{2} \left (\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a}\right )\)	\(43\)
meijerg	\(\frac {a^{2} \left (-\frac {2}{x \sqrt {a^{2}}}-\frac {2 a \arctan \left (a x \right )}{\sqrt {a^{2}}}\right )}{2 \sqrt {a^{2}}}+i a \left (-\ln \left (a^{2} x^{2}+1\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right )-a \arctan \left (a x \right )\)	\(67\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=\frac {2 i \, a x \log \left (x\right ) - 2 i \, a x \log \left (\frac {a x + i}{a}\right ) - 1}{x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=- 2 a \left (- i \log {\left (4 a^{2} x \right )} + i \log {\left (4 a^{2} x + 4 i a \right )}\right ) - \frac {1}{x} \]

[In]

integrate((1+I*a*x)**2/(a**2*x**2+1)/x**2,x)

[Out]

-2*a*(-I*log(4*a**2*x) + I*log(4*a**2*x + 4*I*a)) - 1/x

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-2 \, a \arctan \left (a x\right ) - i \, a \log \left (a^{2} x^{2} + 1\right ) + 2 i \, a \log \left (x\right ) - \frac {1}{x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-2 i \, a \log \left (a x + i\right ) + 2 i \, a \log \left ({\left | x \right |}\right ) - \frac {1}{x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-4\,a\,\mathrm {atan}\left (2\,a\,x+1{}\mathrm {i}\right )-\frac {1}{x} \]

[In]

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

[Out]

- 4*a*atan(2*a*x + 1i) - 1/x