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

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

[Out]

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

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Rubi [A]  time = 0.0268669, antiderivative size = 36, normalized size of antiderivative = 1., 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[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 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]

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]))))

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*}

Mathematica [A]  time = 0.0098901, size = 36, normalized size = 1. \[ -2 a^2 \log (x)+2 a^2 \log (a x+i)-\frac{2 i a}{x}-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[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]

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Maple [A]  time = 0.041, size = 45, normalized size = 1.3 \begin{align*} -2\,i{a}^{2}\arctan \left ( ax \right ) +{a}^{2}\ln \left ({a}^{2}{x}^{2}+1 \right ) -{\frac{1}{2\,{x}^{2}}}-{\frac{2\,ia}{x}}-2\,{a}^{2}\ln \left ( x \right ) \end{align*}

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)

[Out]

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

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Maxima [A]  time = 1.51449, size = 57, normalized size = 1.58 \begin{align*} -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{align*}

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 = 1.59904, size = 97, normalized size = 2.69 \begin{align*} -\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{align*}

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 = 0.459619, size = 29, normalized size = 0.81 \begin{align*} - 2 a^{2} \left (\log{\left (x \right )} - \log{\left (x + \frac{i}{a} \right )}\right ) - \frac{4 i a x + 1}{2 x^{2}} \end{align*}

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

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Giac [A]  time = 1.11688, size = 43, normalized size = 1.19 \begin{align*} 2 \, a^{2} \log \left (a x + i\right ) - 2 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac{4 \, a i x + 1}{2 \, x^{2}} \end{align*}

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*a*i*x + 1)/x^2