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

Optimal. Leaf size=23 \[ -x-\frac{2 i \log (-a-b x+i)}{b} \]

[Out]

-x - ((2*I)*Log[I - a - b*x])/b

________________________________________________________________________________________

Rubi [A]  time = 0.0125188, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5093, 43} \[ -x-\frac{2 i \log (-a-b x+i)}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^((-2*I)*ArcTan[a + b*x]),x]

[Out]

-x - ((2*I)*Log[I - a - b*x])/b

Rule 5093

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]

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+b x)} \, dx &=\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]  time = 0.0110591, size = 32, normalized size = 1.39 \[ -\frac{i \log \left ((a+b x)^2+1\right )}{b}+\frac{2 \tan ^{-1}(a+b x)}{b}-x \]

Warning: Unable to verify antiderivative.

[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]  time = 0.042, size = 40, normalized size = 1.7 \begin{align*} -x-{\frac{i\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) }{b}}+2\,{\frac{\arctan \left ( bx+a \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.986961, size = 26, normalized size = 1.13 \begin{align*} -x - \frac{2 i \, \log \left (i \, b x + i \, a + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 2.09417, size = 50, normalized size = 2.17 \begin{align*} -\frac{b x + 2 i \, \log \left (\frac{b x + a - i}{b}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0.378319, size = 15, normalized size = 0.65 \begin{align*} - x - \frac{2 i \log{\left (a + b x - i \right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Giac [A]  time = 1.10408, size = 51, normalized size = 2.22 \begin{align*} \frac{{\left (b i x + a i + 1\right )} i}{b} + \frac{2 \, i \log \left (\frac{1}{\sqrt{{\left (b x + a\right )}^{2} + 1}{\left | b \right |}}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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