\(\int \frac {e^{-2 i \arctan (a x)}}{x^4} \, dx\) [51]

Optimal result

Integrand size = 14, antiderivative size = 49 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^4} \, dx=-\frac {1}{3 x^3}+\frac {i a}{x^2}+\frac {2 a^2}{x}+2 i a^3 \log (x)-2 i a^3 \log (i-a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, 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} \[ \int \frac {e^{-2 i \arctan (a x)}}{x^4} \, dx=2 i a^3 \log (x)-2 i a^3 \log (-a x+i)+\frac {2 a^2}{x}+\frac {i a}{x^2}-\frac {1}{3 x^3} \]

[In]

Int[1/(E^((2*I)*ArcTan[a*x])*x^4),x]

[Out]

-1/3*1/x^3 + (I*a)/x^2 + (2*a^2)/x + (2*I)*a^3*Log[x] - (2*I)*a^3*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^4 (1+i a x)} \, dx \\ & = \int \left (\frac {1}{x^4}-\frac {2 i a}{x^3}-\frac {2 a^2}{x^2}+\frac {2 i a^3}{x}-\frac {2 i a^4}{-i+a x}\right ) \, dx \\ & = -\frac {1}{3 x^3}+\frac {i a}{x^2}+\frac {2 a^2}{x}+2 i a^3 \log (x)-2 i a^3 \log (i-a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^4} \, dx=-\frac {1}{3 x^3}+\frac {i a}{x^2}+\frac {2 a^2}{x}+2 i a^3 \log (x)-2 i a^3 \log (i-a x) \]

[In]

Integrate[1/(E^((2*I)*ArcTan[a*x])*x^4),x]

[Out]

-1/3*1/x^3 + (I*a)/x^2 + (2*a^2)/x + (2*I)*a^3*Log[x] - (2*I)*a^3*Log[I - a*x]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {1}{3 x^{3}}+\frac {i a}{x^{2}}+\frac {2 a^{2}}{x}+2 i a^{3} \ln \left (x \right )-2 i a^{3} \ln \left (-a x +i\right )\)	\(44\)
risch	\(\frac {2 a^{2} x^{2}+i a x -\frac {1}{3}}{x^{3}}+2 a^{3} \arctan \left (a x \right )-i a^{3} \ln \left (a^{2} x^{2}+1\right )+2 i a^{3} \ln \left (-x \right )\)	\(56\)
parallelrisch	\(-\frac {6 i \ln \left (x \right ) x^{4} a^{5}-6 i \ln \left (a x -i\right ) x^{4} a^{5}+6 \ln \left (x \right ) x^{3} a^{4}-6 \ln \left (a x -i\right ) x^{3} a^{4}+6 a^{4} x^{3}-3 i a^{3} x^{2}+2 a^{2} x +i a}{3 a \left (-a x +i\right ) x^{3}}\)	\(98\)
meijerg	\(i a^{3} \left (\frac {3 i a x}{3 i a x +3}+2 \ln \left (i a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (i a \right )+\frac {i}{x a}\right )-i a^{3} \left (\frac {5 i a x}{5 i a x +5}+4 \ln \left (i a x +1\right )-1-4 \ln \left (x \right )-4 \ln \left (i a \right )-\frac {i}{3 x^{3} a^{3}}-\frac {1}{a^{2} x^{2}}+\frac {3 i}{x a}\right )\)	\(123\)

[In]

int(1/(1+I*a*x)^2*(a^2*x^2+1)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^4} \, dx=\frac {6 i \, a^{3} x^{3} \log \left (x\right ) - 6 i \, a^{3} x^{3} \log \left (\frac {a x - i}{a}\right ) + 6 \, a^{2} x^{2} + 3 i \, a x - 1}{3 \, x^{3}} \]

[In]

integrate(1/(1+I*a*x)^2*(a^2*x^2+1)/x^4,x, algorithm="fricas")

[Out]

1/3*(6*I*a^3*x^3*log(x) - 6*I*a^3*x^3*log((a*x - I)/a) + 6*a^2*x^2 + 3*I*a*x - 1)/x^3

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^4} \, dx=- 2 a^{3} \left (- i \log {\left (4 a^{4} x \right )} + i \log {\left (4 a^{4} x - 4 i a^{3} \right )}\right ) - \frac {- 6 a^{2} x^{2} - 3 i a x + 1}{3 x^{3}} \]

[In]

integrate(1/(1+I*a*x)**2*(a**2*x**2+1)/x**4,x)

[Out]

-2*a**3*(-I*log(4*a**4*x) + I*log(4*a**4*x - 4*I*a**3)) - (-6*a**2*x**2 - 3*I*a*x + 1)/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^4} \, dx=-2 i \, a^{3} \log \left (i \, a x + 1\right ) + 2 i \, a^{3} \log \left (x\right ) + \frac {6 i \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 2 i \, a x - 1}{3 i \, a x^{4} + 3 \, x^{3}} \]

[In]

integrate(1/(1+I*a*x)^2*(a^2*x^2+1)/x^4,x, algorithm="maxima")

[Out]

-2*I*a^3*log(I*a*x + 1) + 2*I*a^3*log(x) + (6*I*a^3*x^3 + 3*a^2*x^2 + 2*I*a*x - 1)/(3*I*a*x^4 + 3*x^3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^4} \, dx=2 i \, a^{3} \log \left (\frac {i}{i \, a x + 1} - i\right ) - \frac {10 \, a^{3} - \frac {24 \, a^{3}}{i \, a x + 1} + \frac {15 \, a^{3}}{{\left (i \, a x + 1\right )}^{2}}}{3 \, {\left (\frac {i}{i \, a x + 1} - i\right )}^{3}} \]

[In]

integrate(1/(1+I*a*x)^2*(a^2*x^2+1)/x^4,x, algorithm="giac")

[Out]

2*I*a^3*log(I/(I*a*x + 1) - I) - 1/3*(10*a^3 - 24*a^3/(I*a*x + 1) + 15*a^3/(I*a*x + 1)^2)/(I/(I*a*x + 1) - I)^
3

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^4} \, dx=4\,a^3\,\mathrm {atan}\left (2\,a\,x-\mathrm {i}\right )+\frac {2\,a^2\,x^2+a\,x\,1{}\mathrm {i}-\frac {1}{3}}{x^3} \]

[In]

int((a^2*x^2 + 1)/(x^4*(a*x*1i + 1)^2),x)

[Out]

4*a^3*atan(2*a*x - 1i) + (a*x*1i + 2*a^2*x^2 - 1/3)/x^3