3.50 \(\int \frac {e^{-2 i \tan ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=37 \[ -2 a^2 \log (x)+2 a^2 \log (-a x+i)+\frac {2 i a}{x}-\frac {1}{2 x^2} \]

[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 = 0.03, antiderivative size = 37, 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 = {5062, 77} \[ -2 a^2 \log (x)+2 a^2 \log (-a x+i)+\frac {2 i a}{x}-\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

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 5062

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 = 37, normalized size = 1.00 \[ -2 a^2 \log (x)+2 a^2 \log (-a x+i)+\frac {2 i a}{x}-\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.40, size = 39, normalized size = 1.05 \[ -\frac {4 \, a^{2} x^{2} \log \relax (x) - 4 \, a^{2} x^{2} \log \left (\frac {a x - i}{a}\right ) - 4 i \, a x + 1}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.13, size = 63, normalized size = 1.70 \[ 2 \, a^{2} i^{2} \log \left (\frac {i^{2}}{a i x + 1} + 1\right ) + \frac {\frac {6 \, a^{2} i^{2}}{a i x + 1} + 5 \, a^{2}}{2 \, {\left (i - \frac {i}{a i x + 1}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 45, normalized size = 1.22 \[ -\frac {1}{2 x^{2}}+\frac {2 i a}{x}-2 a^{2} \ln \relax (x )+2 i a^{2} \arctan \left (a x \right )+a^{2} \ln \left (a^{2} x^{2}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 50, normalized size = 1.35 \[ 2 \, a^{2} \log \left (i \, a x + 1\right ) - 2 \, a^{2} \log \relax (x) - \frac {4 \, a^{2} x^{2} - 3 i \, a x + 1}{2 i \, a x^{3} + 2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.07, size = 26, normalized size = 0.70 \[ a^2\,\mathrm {atan}\left (2\,a\,x-\mathrm {i}\right )\,4{}\mathrm {i}+\frac {-\frac {1}{2}+a\,x\,2{}\mathrm {i}}{x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.18, size = 42, normalized size = 1.14 \[ - 2 a^{2} \left (\log {\left (4 a^{3} x \right )} - \log {\left (4 a^{3} x - 4 i a^{2} \right )}\right ) - \frac {- 4 i a x + 1}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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