3.1.14 \(\int e^{2 i \text {ArcTan}(a x)} \, dx\) [14]

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

[Out]

-x+2*I*ln(I+a*x)/a

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Rubi [A]
time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5169, 45} \begin {gather*} -x+\frac {2 i \log (a x+i)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x + ((2*I)*Log[I + a*x])/a

Rule 45

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])

Rule 5169

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]

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*}

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Mathematica [A]
time = 0.01, size = 30, normalized size = 1.58 \begin {gather*} -x+\frac {2 \text {ArcTan}(a x)}{a}+\frac {i \log \left (1+a^2 x^2\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.06, size = 30, normalized size = 1.58

method	result	size
default	\(-x +\frac {i \ln \left (a^{2} x^{2}+1\right )}{a}+\frac {2 \arctan \left (a x \right )}{a}\)	\(30\)
risch	\(-x +\frac {i \ln \left (a^{2} x^{2}+1\right )}{a}+\frac {2 \arctan \left (a x \right )}{a}\)	\(30\)
meijerg	\(\frac {\arctan \left (a x \right )}{a}+\frac {i \ln \left (a^{2} x^{2}+1\right )}{a}-\frac {\frac {2 x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2}}-\frac {2 \left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}}{2 \sqrt {a^{2}}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 28, normalized size = 1.47 \begin {gather*} -x + \frac {2 \, \arctan \left (a x\right )}{a} + \frac {i \, \log \left (a^{2} x^{2} + 1\right )}{a} \end {gather*}

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 = 1.80, size = 21, normalized size = 1.11 \begin {gather*} -\frac {a x - 2 i \, \log \left (\frac {a x + i}{a}\right )}{a} \end {gather*}

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 = 0.05, size = 12, normalized size = 0.63 \begin {gather*} - x + \frac {2 i \log {\left (a x + i \right )}}{a} \end {gather*}

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 = 0.41, size = 15, normalized size = 0.79 \begin {gather*} -x + \frac {2 i \, \log \left (a x + i\right )}{a} \end {gather*}

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

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Mupad [B]
time = 0.04, size = 19, normalized size = 1.00 \begin {gather*} -x+\frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,2{}\mathrm {i}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^2/(a^2*x^2 + 1),x)

[Out]

(log(x + 1i/a)*2i)/a - x

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