3.33 \(\int \frac{e^{4 i \tan ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=52 \[ -\frac{4 i a^2}{a x+i}-8 a^2 \log (x)+8 a^2 \log (a x+i)-\frac{4 i a}{x}-\frac{1}{2 x^2} \]

[Out]

-1/(2*x^2) - ((4*I)*a)/x - ((4*I)*a^2)/(I + a*x) - 8*a^2*Log[x] + 8*a^2*Log[I + a*x]

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Rubi [A]  time = 0.0353444, antiderivative size = 52, 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, 88} \[ -\frac{4 i a^2}{a x+i}-8 a^2 \log (x)+8 a^2 \log (a x+i)-\frac{4 i a}{x}-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x])/x^3,x]

[Out]

-1/(2*x^2) - ((4*I)*a)/x - ((4*I)*a^2)/(I + a*x) - 8*a^2*Log[x] + 8*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{e^{4 i \tan ^{-1}(a x)}}{x^3} \, dx &=\int \frac{(1+i a x)^2}{x^3 (1-i a x)^2} \, dx\\ &=\int \left (\frac{1}{x^3}+\frac{4 i a}{x^2}-\frac{8 a^2}{x}+\frac{4 i a^3}{(i+a x)^2}+\frac{8 a^3}{i+a x}\right ) \, dx\\ &=-\frac{1}{2 x^2}-\frac{4 i a}{x}-\frac{4 i a^2}{i+a x}-8 a^2 \log (x)+8 a^2 \log (i+a x)\\ \end{align*}

Mathematica [A]  time = 0.0358388, size = 52, normalized size = 1. \[ -\frac{4 i a^2}{a x+i}-8 a^2 \log (x)+8 a^2 \log (a x+i)-\frac{4 i a}{x}-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((4*I)*ArcTan[a*x])/x^3,x]

[Out]

-1/(2*x^2) - ((4*I)*a)/x - ((4*I)*a^2)/(I + a*x) - 8*a^2*Log[x] + 8*a^2*Log[I + a*x]

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Maple [A]  time = 0.048, size = 60, normalized size = 1.2 \begin{align*}{\frac{-4\,i{a}^{2}}{ax+i}}+4\,{a}^{2}\ln \left ({a}^{2}{x}^{2}+1 \right ) -8\,i{a}^{2}\arctan \left ( ax \right ) -{\frac{1}{2\,{x}^{2}}}-{\frac{4\,ia}{x}}-8\,{a}^{2}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^2/x^3,x)

[Out]

-4*I*a^2/(a*x+I)+4*a^2*ln(a^2*x^2+1)-8*I*a^2*arctan(a*x)-1/2/x^2-4*I*a/x-8*a^2*ln(x)

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Maxima [A]  time = 1.53399, size = 93, normalized size = 1.79 \begin{align*} -8 i \, a^{2} \arctan \left (a x\right ) + 4 \, a^{2} \log \left (a^{2} x^{2} + 1\right ) - 8 \, a^{2} \log \left (x\right ) + \frac{-16 i \, a^{3} x^{3} - 9 \, a^{2} x^{2} - 8 i \, a x - 1}{2 \,{\left (a^{2} x^{4} + x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/x^3,x, algorithm="maxima")

[Out]

-8*I*a^2*arctan(a*x) + 4*a^2*log(a^2*x^2 + 1) - 8*a^2*log(x) + 1/2*(-16*I*a^3*x^3 - 9*a^2*x^2 - 8*I*a*x - 1)/(
a^2*x^4 + x^2)

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Fricas [A]  time = 1.71169, size = 171, normalized size = 3.29 \begin{align*} \frac{-16 i \, a^{2} x^{2} + 7 \, a x - 16 \,{\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (x\right ) + 16 \,{\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (\frac{a x + i}{a}\right ) - i}{2 \,{\left (a x^{3} + i \, x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/x^3,x, algorithm="fricas")

[Out]

1/2*(-16*I*a^2*x^2 + 7*a*x - 16*(a^3*x^3 + I*a^2*x^2)*log(x) + 16*(a^3*x^3 + I*a^2*x^2)*log((a*x + I)/a) - I)/
(a*x^3 + I*x^2)

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Sympy [A]  time = 0.685398, size = 51, normalized size = 0.98 \begin{align*} 8 a^{2} \left (- \log{\left (x \right )} + \log{\left (x + \frac{i}{a} \right )}\right ) - \frac{16 i a^{3} x^{2} - 7 a^{2} x + i a}{2 a^{2} x^{3} + 2 i a x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**2/x**3,x)

[Out]

8*a**2*(-log(x) + log(x + I/a)) - (16*I*a**3*x**2 - 7*a**2*x + I*a)/(2*a**2*x**3 + 2*I*a*x**2)

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Giac [A]  time = 1.09499, size = 63, normalized size = 1.21 \begin{align*} 8 \, a^{2} \log \left (a x + i\right ) - 8 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac{16 \, a^{2} i x^{2} - 7 \, a x + i}{2 \,{\left (a x + i\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/x^3,x, algorithm="giac")

[Out]

8*a^2*log(a*x + i) - 8*a^2*log(abs(x)) - 1/2*(16*a^2*i*x^2 - 7*a*x + i)/((a*x + i)*x^2)