\(\int e^{-2 i \arctan (a+b x)} \, dx\) [202]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 12, antiderivative size = 23 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x-\frac {2 i \log (i-a-b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5201, 45} \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x-\frac {2 i \log (-a-b x+i)}{b} \]

[In]

Int[E^((-2*I)*ArcTan[a + b*x]),x]

[Out]

-x - ((2*I)*Log[I - a - b*x])/b

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 5201

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1-i a-i b x}{1+i a+i b x} \, dx \\ & = \int \left (-1-\frac {2 i}{-i+a+b x}\right ) \, dx \\ & = -x-\frac {2 i \log (i-a-b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x+\frac {2 \arctan (a+b x)}{b}-\frac {i \log \left (1+(a+b x)^2\right )}{b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method result size
default \(-x -\frac {2 i \ln \left (-b x -a +i\right )}{b}\) \(22\)
risch \(-x -\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b}+\frac {2 \arctan \left (b x +a \right )}{b}\) \(40\)
parallelrisch \(\frac {2 i \ln \left (b x +a -i\right ) x b +b^{2} x^{2}+2 i \ln \left (b x +a -i\right ) a +1+2 i a -a^{2}+2 \ln \left (b x +a -i\right )}{b \left (-b x -a +i\right )}\) \(70\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-\frac {b x + 2 i \, \log \left (\frac {b x + a - i}{b}\right )}{b} \]

[In]

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

[Out]

-(b*x + 2*I*log((b*x + a - I)/b))/b

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int e^{-2 i \arctan (a+b x)} \, dx=- x - \frac {2 i \log {\left (a + b x - i \right )}}{b} \]

[In]

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

[Out]

-x - 2*I*log(a + b*x - I)/b

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x - \frac {2 i \, \log \left (i \, b x + i \, a + 1\right )}{b} \]

[In]

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

[Out]

-x - 2*I*log(I*b*x + I*a + 1)/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int e^{-2 i \arctan (a+b x)} \, dx=\frac {i \, {\left (i \, b x + i \, a + 1\right )}}{b} + \frac {2 i \, \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x-\frac {\ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,2{}\mathrm {i}}{b} \]

[In]

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

[Out]

- x - (log(x + (a - 1i)/b)*2i)/b