\(\int e^{-2 i \arctan (a x)} \, dx\) [47]

Optimal result

Integrand size = 10, antiderivative size = 20 \[ \int e^{-2 i \arctan (a x)} \, dx=-x-\frac {2 i \log (i-a x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5169, 45} \[ \int e^{-2 i \arctan (a x)} \, dx=-x-\frac {2 i \log (-a x+i)}{a} \]

[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*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int e^{-2 i \arctan (a x)} \, dx=-x+\frac {2 \arctan (a x)}{a}-\frac {i \log \left (1+a^2 x^2\right )}{a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
default	\(-x -\frac {2 i \ln \left (-a x +i\right )}{a}\)	\(19\)
risch	\(-x -\frac {i \ln \left (a^{2} x^{2}+1\right )}{a}+\frac {2 \arctan \left (a x \right )}{a}\)	\(30\)
parallelrisch	\(\frac {2 i \ln \left (a x -i\right ) x a +a^{2} x^{2}+1+2 \ln \left (a x -i\right )}{a \left (-a x +i\right )}\)	\(44\)
meijerg	\(\frac {i \left (\frac {i a x \left (3 i a x +6\right )}{3 i a x +3}-2 \ln \left (i a x +1\right )\right )}{a}+\frac {x}{i a x +1}\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int e^{-2 i \arctan (a x)} \, dx=-\frac {a x + 2 i \, \log \left (\frac {a x - i}{a}\right )}{a} \]

[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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int e^{-2 i \arctan (a x)} \, dx=- x - \frac {2 i \log {\left (a x - i \right )}}{a} \]

[In]

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

[Out]

-x - 2*I*log(a*x - I)/a

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int e^{-2 i \arctan (a x)} \, dx=-x - \frac {2 i \, \log \left (i \, a x + 1\right )}{a} \]

[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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (16) = 32\).

Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.25 \[ \int e^{-2 i \arctan (a x)} \, dx=a^{2} {\left (\frac {i \, {\left (i \, a x + 1\right )}}{a^{3}} + \frac {2 i \, \log \left (\frac {1}{\sqrt {a^{2} x^{2} + 1} {\left | a \right |}}\right )}{a^{3}} - \frac {i}{{\left (i \, a x + 1\right )} a^{3}}\right )} + \frac {i}{{\left (i \, a x + 1\right )} a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int e^{-2 i \arctan (a x)} \, dx=-x-\frac {\ln \left (x-\frac {1{}\mathrm {i}}{a}\right )\,2{}\mathrm {i}}{a} \]

[In]

int((a^2*x^2 + 1)/(a*x*1i + 1)^2,x)

[Out]

- x - (log(x - 1i/a)*2i)/a