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

Optimal. Leaf size=19 \[ -x+\frac{2 i \log (a x+i)}{a} \]

[Out]

-x + ((2*I)*Log[I + a*x])/a

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Rubi [A]  time = 0.0091863, antiderivative size = 19, 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 = 0.0092926, size = 30, normalized size = 1.58 \[ \frac{i \log \left (a^2 x^2+1\right )}{a}+\frac{2 \tan ^{-1}(a x)}{a}-x \]

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 = 0.036, size = 30, normalized size = 1.6 \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+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 = 1.50846, size = 38, normalized size = 2. \begin{align*} -x + \frac{2 \, \arctan \left (a x\right )}{a} + \frac{i \, \log \left (a^{2} x^{2} + 1\right )}{a} \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, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.54591, size = 45, normalized size = 2.37 \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+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 = 0.311358, size = 12, normalized size = 0.63 \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+I*a*x)**2/(a**2*x**2+1),x)

[Out]

-x + 2*I*log(a*x + I)/a

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Giac [A]  time = 1.1037, size = 22, normalized size = 1.16 \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+I*a*x)^2/(a^2*x^2+1),x, algorithm="giac")

[Out]

-x + 2*i*log(a*x + i)/a