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

Optimal. Leaf size=20 \[ -x-\frac{2 i \log (-a x+i)}{a} \]

[Out]

-x - ((2*I)*Log[I - a*x])/a

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Rubi [A]  time = 0.0101968, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5061, 43} \[ -x-\frac{2 i \log (-a x+i)}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^((-2*I)*ArcTan[a*x]),x]

[Out]

-x - ((2*I)*Log[I - a*x])/a

Rule 5061

Int[E^(ArcTan[(a_.)*(x_)]*(n_.)), x_Symbol] :> Int[(1 - I*a*x)^((I*n)/2)/(1 + I*a*x)^((I*n)/2), x] /; FreeQ[{a
, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0088313, size = 30, normalized size = 1.5 \[ -\frac{i \log \left (a^2 x^2+1\right )}{a}+\frac{2 \tan ^{-1}(a x)}{a}-x \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((-2*I)*ArcTan[a*x]),x]

[Out]

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

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Maple [A]  time = 0.041, size = 30, normalized size = 1.5 \begin{align*} -x-{\frac{i\ln \left ({a}^{2}{x}^{2}+1 \right ) }{a}}+2\,{\frac{\arctan \left ( ax \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01925, size = 22, normalized size = 1.1 \begin{align*} -x - \frac{2 i \, \log \left (i \, a x + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*I*log(I*a*x + 1)/a

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Fricas [A]  time = 1.52253, size = 45, normalized size = 2.25 \begin{align*} -\frac{a x + 2 i \, \log \left (\frac{a x - i}{a}\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*x + 2*I*log((a*x - I)/a))/a

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Sympy [A]  time = 0.330293, size = 14, normalized size = 0.7 \begin{align*} - x - \frac{2 i \log{\left (a x - i \right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*I*log(a*x - I)/a

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Giac [B]  time = 1.1094, size = 90, normalized size = 4.5 \begin{align*} a^{2}{\left (\frac{{\left (a i x + 1\right )} i}{a^{3}} + \frac{2 \, i \log \left (\frac{1}{\sqrt{a^{2} x^{2} + 1}{\left | a \right |}}\right )}{a^{3}} - \frac{i}{{\left (a i x + 1\right )} a^{3}}\right )} + \frac{i}{{\left (a i x + 1\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*((a*i*x + 1)*i/a^3 + 2*i*log(1/(sqrt(a^2*x^2 + 1)*abs(a)))/a^3 - i/((a*i*x + 1)*a^3)) + i/((a*i*x + 1)*a)