3.1.17 \(\int \frac {e^{2 i \text {ArcTan}(a x)}}{x^3} \, dx\) [17]

Optimal. Leaf size=36 \[ -\frac {1}{2 x^2}-\frac {2 i a}{x}-2 a^2 \log (x)+2 a^2 \log (i+a x) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5170, 78} \begin {gather*} -2 a^2 \log (x)+2 a^2 \log (a x+i)-\frac {2 i a}{x}-\frac {1}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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*} \int \frac {e^{2 i \tan ^{-1}(a x)}}{x^3} \, dx &=\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*}

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Mathematica [A]
time = 0.01, size = 36, normalized size = 1.00 \begin {gather*} -\frac {1}{2 x^2}-\frac {2 i a}{x}-2 a^2 \log (x)+2 a^2 \log (i+a x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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]

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Maple [A]
time = 0.11, size = 52, normalized size = 1.44

method	result	size
risch	\(\frac {-2 i a x -\frac {1}{2}}{x^{2}}-2 i a^{2} \arctan \left (a x \right )+a^{2} \ln \left (a^{2} x^{2}+1\right )-2 a^{2} \ln \left (x \right )\)	\(44\)
default	\(-2 a^{3} \left (-\frac {\ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {i \arctan \left (a x \right )}{a}\right )-\frac {1}{2 x^{2}}-\frac {2 i a}{x}-2 a^{2} \ln \left (x \right )\)	\(52\)
meijerg	\(\frac {a^{2} \left (\ln \left (a^{2} x^{2}+1\right )-2 \ln \left (x \right )-\ln \left (a^{2}\right )-\frac {1}{a^{2} x^{2}}\right )}{2}+\frac {i a^{3} \left (-\frac {2}{x \sqrt {a^{2}}}-\frac {2 a \arctan \left (a x \right )}{\sqrt {a^{2}}}\right )}{\sqrt {a^{2}}}-\frac {a^{2} \left (-\ln \left (a^{2} x^{2}+1\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right )}{2}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.47, size = 42, normalized size = 1.17 \begin {gather*} -2 i \, a^{2} \arctan \left (a x\right ) + a^{2} \log \left (a^{2} x^{2} + 1\right ) - 2 \, a^{2} \log \left (x\right ) - \frac {4 i \, a x + 1}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.86, size = 39, normalized size = 1.08 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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

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Sympy [A]
time = 0.12, size = 42, normalized size = 1.17 \begin {gather*} - 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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Giac [A]
time = 0.40, size = 31, normalized size = 0.86 \begin {gather*} 2 \, a^{2} \log \left (a x + i\right ) - 2 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {4 i \, a x + 1}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.08, size = 27, normalized size = 0.75 \begin {gather*} -a^2\,\mathrm {atan}\left (2\,a\,x+1{}\mathrm {i}\right )\,4{}\mathrm {i}-\frac {\frac {1}{2}+a\,x\,2{}\mathrm {i}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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