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

Optimal. Leaf size=16 \[ \log (x)+\frac{4 i}{a x+i} \]

[Out]

(4*I)/(I + a*x) + Log[x]

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Rubi [A]  time = 0.0215648, antiderivative size = 16, 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} \[ \log (x)+\frac{4 i}{a x+i} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0096094, size = 16, normalized size = 1. \[ \log (x)+\frac{4 i}{a x+i} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*I)/(I + a*x) + Log[x]

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Maple [A]  time = 0.047, size = 15, normalized size = 0.9 \begin{align*}{\frac{4\,i}{ax+i}}+\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,x)

[Out]

4*I/(a*x+I)+ln(x)

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Maxima [A]  time = 1.52583, size = 30, normalized size = 1.88 \begin{align*} -\frac{4 \,{\left (-i \, a x - 1\right )}}{a^{2} x^{2} + 1} + \log \left (x\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,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.60104, size = 49, normalized size = 3.06 \begin{align*} \frac{{\left (a x + i\right )} \log \left (x\right ) + 4 i}{a x + i} \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,x, algorithm="fricas")

[Out]

((a*x + I)*log(x) + 4*I)/(a*x + I)

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Sympy [A]  time = 0.475003, size = 15, normalized size = 0.94 \begin{align*} \frac{4 i a}{a^{2} x + i a} + \log{\left (x \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,x)

[Out]

4*I*a/(a**2*x + I*a) + log(x)

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Giac [A]  time = 1.10788, size = 19, normalized size = 1.19 \begin{align*} \frac{4 \, i}{a x + i} + \log \left ({\left | x \right |}\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,x, algorithm="giac")

[Out]

4*i/(a*x + i) + log(abs(x))