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)+ln(x)

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Rubi [A]  time = 0.02, antiderivative size = 16, 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, 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 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]))

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^{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*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ \log (x)+\frac {4 i}{a x+i} \]

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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fricas [A]  time = 0.54, size = 18, normalized size = 1.12 \[ \frac {{\left (a x + i\right )} \log \relax (x) + 4 i}{a x + i} \]

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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giac [A]  time = 0.13, size = 14, normalized size = 0.88 \[ \frac {4 \, i}{a x + i} + \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.05, size = 15, normalized size = 0.94 \[ \frac {4 i}{a x +i}+\ln \relax (x ) \]

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

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maxima [A]  time = 0.42, size = 22, normalized size = 1.38 \[ -\frac {4 \, {\left (-i \, a x - 1\right )}}{a^{2} x^{2} + 1} + \log \relax (x) \]

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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mupad [B]  time = 0.08, size = 14, normalized size = 0.88 \[ \ln \relax (x)+\frac {4{}\mathrm {i}}{a\,x+1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) + 4i/(a*x + 1i)

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sympy [A]  time = 0.21, size = 10, normalized size = 0.62 \[ \log {\relax (x )} + \frac {4 i}{a x + i} \]

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]

log(x) + 4*I/(a*x + I)

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