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

Optimal. Leaf size=37 \[ \frac{2 (1-i a) \log (a+b x+i)}{b^2}+\frac{2 i x}{b}-\frac{x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.0300955, antiderivative size = 37, 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[E^((2*I)*ArcTan[a + b*x])*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 = 0.0192281, size = 37, 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[E^((2*I)*ArcTan[a + b*x])*x,x]

[Out]

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

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Maple [B]  time = 0.038, size = 107, normalized size = 2.9 \begin{align*} -{\frac{{x}^{2}}{2}}+{\frac{2\,ix}{b}}-{\frac{i\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) a}{{b}^{2}}}+{\frac{\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) }{{b}^{2}}}-{\frac{2\,i}{{b}^{2}}\arctan \left ({\frac{2\,{b}^{2}x+2\,ab}{2\,b}} \right ) }-2\,{\frac{a}{{b}^{2}}\arctan \left ( 1/2\,{\frac{2\,{b}^{2}x+2\,ab}{b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.46228, size = 89, normalized size = 2.41 \begin{align*} -\frac{b x^{2} - 4 i \, x}{2 \, b} - \frac{{\left (2 \, a + 2 i\right )} \arctan \left (\frac{b^{2} x + a b}{b}\right )}{b^{2}} + \frac{{\left (-i \, a + 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.87842, size = 88, normalized size = 2.38 \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((1+I*(b*x+a))^2/(1+(b*x+a)^2)*x,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

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Sympy [B]  time = 1.0342, size = 148, normalized size = 4. \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((1+I*(b*x+a))**2/(1+(b*x+a)**2)*x,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))

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Giac [A]  time = 1.09022, size = 49, normalized size = 1.32 \begin{align*} -\frac{2 \,{\left (a i - 1\right )} \log \left (b x + a + i\right )}{b^{2}} - \frac{b^{2} x^{2} - 4 \, b i x}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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