\(\int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx\) [50]

Optimal result

Integrand size = 14, antiderivative size = 37 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-\frac {1}{2 x^2}+\frac {2 i a}{x}-2 a^2 \log (x)+2 a^2 \log (i-a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, 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} \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-2 a^2 \log (x)+2 a^2 \log (-a x+i)+\frac {2 i a}{x}-\frac {1}{2 x^2} \]

[In]

Int[1/(E^((2*I)*ArcTan[a*x])*x^3),x]

[Out]

-1/2*1/x^2 + ((2*I)*a)/x - 2*a^2*Log[x] + 2*a^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^3 (1+i a x)} \, dx \\ & = \int \left (\frac {1}{x^3}-\frac {2 i a}{x^2}-\frac {2 a^2}{x}+\frac {2 a^3}{-i+a x}\right ) \, dx \\ & = -\frac {1}{2 x^2}+\frac {2 i a}{x}-2 a^2 \log (x)+2 a^2 \log (i-a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-\frac {1}{2 x^2}+\frac {2 i a}{x}-2 a^2 \log (x)+2 a^2 \log (i-a x) \]

[In]

Integrate[1/(E^((2*I)*ArcTan[a*x])*x^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {1}{2 x^{2}}+\frac {2 i a}{x}-2 a^{2} \ln \left (x \right )+2 a^{2} \ln \left (-a x +i\right )\)	\(34\)
risch	\(\frac {2 i a x -\frac {1}{2}}{x^{2}}-2 a^{2} \ln \left (-x \right )+2 i a^{2} \arctan \left (a x \right )+a^{2} \ln \left (a^{2} x^{2}+1\right )\)	\(46\)
parallelrisch	\(-\frac {-4 \ln \left (x \right ) x^{3} a^{3}+4 \ln \left (a x -i\right ) x^{3} a^{3}+4 i \ln \left (x \right ) x^{2} a^{2}-4 i \ln \left (a x -i\right ) x^{2} a^{2}+i+4 a^{3} x^{3}+3 a x}{2 \left (-a x +i\right ) x^{2}}\)	\(82\)
meijerg	\(a^{2} \left (-\frac {2 i a x}{2 i a x +2}-\ln \left (i a x +1\right )+1+\ln \left (x \right )+\ln \left (i a \right )\right )-a^{2} \left (-\frac {4 i a x}{4 i a x +4}-3 \ln \left (i a x +1\right )+1+3 \ln \left (x \right )+3 \ln \left (i a \right )+\frac {1}{2 a^{2} x^{2}}-\frac {2 i}{x a}\right )\)	\(98\)

[In]

int(1/(1+I*a*x)^2*(a^2*x^2+1)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-\frac {4 \, a^{2} x^{2} \log \left (x\right ) - 4 \, a^{2} x^{2} \log \left (\frac {a x - i}{a}\right ) - 4 i \, a x + 1}{2 \, x^{2}} \]

[In]

integrate(1/(1+I*a*x)^2*(a^2*x^2+1)/x^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=- 2 a^{2} \left (\log {\left (4 a^{3} x \right )} - \log {\left (4 a^{3} x - 4 i a^{2} \right )}\right ) - \frac {- 4 i a x + 1}{2 x^{2}} \]

[In]

integrate(1/(1+I*a*x)**2*(a**2*x**2+1)/x**3,x)

[Out]

-2*a**2*(log(4*a**3*x) - log(4*a**3*x - 4*I*a**2)) - (-4*I*a*x + 1)/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.35 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=2 \, a^{2} \log \left (i \, a x + 1\right ) - 2 \, a^{2} \log \left (x\right ) - \frac {4 \, a^{2} x^{2} - 3 i \, a x + 1}{2 i \, a x^{3} + 2 \, x^{2}} \]

[In]

integrate(1/(1+I*a*x)^2*(a^2*x^2+1)/x^3,x, algorithm="maxima")

[Out]

2*a^2*log(I*a*x + 1) - 2*a^2*log(x) - (4*a^2*x^2 - 3*I*a*x + 1)/(2*I*a*x^3 + 2*x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-2 \, a^{2} \log \left (\frac {i}{i \, a x + 1} - i\right ) + \frac {5 \, a^{2} - \frac {6 \, a^{2}}{i \, a x + 1}}{2 \, {\left (\frac {i}{i \, a x + 1} - i\right )}^{2}} \]

[In]

integrate(1/(1+I*a*x)^2*(a^2*x^2+1)/x^3,x, algorithm="giac")

[Out]

-2*a^2*log(I/(I*a*x + 1) - I) + 1/2*(5*a^2 - 6*a^2/(I*a*x + 1))/(I/(I*a*x + 1) - I)^2

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=a^2\,\mathrm {atan}\left (2\,a\,x-\mathrm {i}\right )\,4{}\mathrm {i}+\frac {-\frac {1}{2}+a\,x\,2{}\mathrm {i}}{x^2} \]

[In]

int((a^2*x^2 + 1)/(x^3*(a*x*1i + 1)^2),x)

[Out]

a^2*atan(2*a*x - 1i)*4i + (a*x*2i - 1/2)/x^2