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

Optimal. Leaf size=40 \[ \frac{2 (1+i a) \log (-a-b x+i)}{b^2}-\frac{2 i x}{b}-\frac{x^2}{2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0317712, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5095, 77} \[ \frac{2 (1+i a) \log (-a-b x+i)}{b^2}-\frac{2 i x}{b}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x/E^((2*I)*ArcTan[a + b*x]),x]

[Out]

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

Rule 5095

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(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, d, e, m, n}, x]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int e^{-2 i \tan ^{-1}(a+b x)} x \, dx &=\int \frac{x (1-i a-i b x)}{1+i a+i b x} \, dx\\ &=\int \left (-\frac{2 i}{b}-x+\frac{2 (1+i a)}{b (-i+a+b x)}\right ) \, dx\\ &=-\frac{2 i x}{b}-\frac{x^2}{2}+\frac{2 (1+i a) \log (i-a-b x)}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.0203792, size = 40, normalized size = 1. \[ \frac{2 (1+i a) \log (-a-b x+i)}{b^2}-\frac{2 i x}{b}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.043, size = 85, normalized size = 2.1 \begin{align*} -{\frac{{x}^{2}}{2}}-{\frac{2\,ix}{b}}-2\,{\frac{\arctan \left ( bx+a \right ) a}{{b}^{2}}}+{\frac{i\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) a}{{b}^{2}}}+{\frac{2\,i\arctan \left ( bx+a \right ) }{{b}^{2}}}+{\frac{\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.02399, size = 49, normalized size = 1.22 \begin{align*} \frac{i \,{\left (i \, b x^{2} - 4 \, x\right )}}{2 \, b} - \frac{2 \,{\left (-i \, a - 1\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*(I*b*x^2 - 4*x)/b - 2*(-I*a - 1)*log(I*b*x + I*a + 1)/b^2

________________________________________________________________________________________

Fricas [A]  time = 2.19443, size = 89, normalized size = 2.22 \begin{align*} -\frac{b^{2} x^{2} + 4 i \, b x + 4 \,{\left (-i \, a - 1\right )} \log \left (\frac{b x + a - i}{b}\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*x^2 + 4*I*b*x + 4*(-I*a - 1)*log((b*x + a - I)/b))/b^2

________________________________________________________________________________________

Sympy [B]  time = 1.02915, size = 148, normalized size = 3.7 \begin{align*} - \frac{x^{2}}{2} - \frac{x \left (2 i a^{2} + 4 a - 2 i\right )}{a^{2} b - 2 i a b - b} + \frac{2 \left (i a^{5} + 5 a^{4} - 10 i a^{3} - 10 a^{2} + 5 i a + 1\right ) \log{\left (a^{5} - 5 i a^{4} - 10 a^{3} + 10 i a^{2} + 5 a + x \left (a^{4} b - 4 i a^{3} b - 6 a^{2} b + 4 i a b + b\right ) - i \right )}}{b^{2} \left (a^{4} - 4 i a^{3} - 6 a^{2} + 4 i a + 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2/2 - x*(2*I*a**2 + 4*a - 2*I)/(a**2*b - 2*I*a*b - b) + 2*(I*a**5 + 5*a**4 - 10*I*a**3 - 10*a**2 + 5*I*a +
1)*log(a**5 - 5*I*a**4 - 10*a**3 + 10*I*a**2 + 5*a + x*(a**4*b - 4*I*a**3*b - 6*a**2*b + 4*I*a*b + b) - I)/(b
**2*(a**4 - 4*I*a**3 - 6*a**2 + 4*I*a + 1))

________________________________________________________________________________________

Giac [B]  time = 1.11572, size = 108, normalized size = 2.7 \begin{align*} -\frac{i{\left (\frac{4 \,{\left (a - i\right )} \log \left (\frac{1}{\sqrt{{\left (b x + a\right )}^{2} + 1}{\left | b \right |}}\right )}{b} - \frac{{\left (b i x + a i + 1\right )}^{2}{\left (i - \frac{2 \,{\left (a b i + 3 \, b\right )} i}{{\left (b i x + a i + 1\right )} b}\right )}}{b i^{2}}\right )}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*i*(4*(a - i)*log(1/(sqrt((b*x + a)^2 + 1)*abs(b)))/b - (b*i*x + a*i + 1)^2*(i - 2*(a*b*i + 3*b)*i/((b*i*x
+ a*i + 1)*b))/(b*i^2))/b