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

Optimal. Leaf size=30 \[ \frac {2 \log (-a x+i)}{a^2}-\frac {2 i x}{a}-\frac {x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 = {5062, 77} \[ \frac {2 \log (-a x+i)}{a^2}-\frac {2 i x}{a}-\frac {x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x/E^((2*I)*ArcTan[a*x]),x]

[Out]

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

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]))))

Rule 5062

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 - I*a*x)^((I*n)/2))/(1 + I*a*x)^((I*n)/2
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.00 \[ \frac {2 \log (-a x+i)}{a^2}-\frac {2 i x}{a}-\frac {x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 29, normalized size = 0.97 \[ -\frac {a^{2} x^{2} + 4 i \, a x - 4 \, \log \left (\frac {a x - i}{a}\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.13, size = 58, normalized size = 1.93 \[ \frac {i {\left (\frac {4 \, i \log \left (\frac {1}{\sqrt {a^{2} x^{2} + 1} {\left | a \right |}}\right )}{a} + \frac {{\left (a i x + 1\right )}^{2} {\left (i - \frac {6 \, i}{a i x + 1}\right )}}{a i^{2}}\right )}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*i*(4*i*log(1/(sqrt(a^2*x^2 + 1)*abs(a)))/a + (a*i*x + 1)^2*(i - 6*i/(a*i*x + 1))/(a*i^2))/a

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maple [A]  time = 0.05, size = 38, normalized size = 1.27 \[ -\frac {x^{2}}{2}-\frac {2 i x}{a}+\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{2}}+\frac {2 i \arctan \left (a x \right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 28, normalized size = 0.93 \[ \frac {i \, {\left (i \, a x^{2} - 4 \, x\right )}}{2 \, a} + \frac {2 \, \log \left (i \, a x + 1\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.42, size = 27, normalized size = 0.90 \[ \frac {2\,\ln \left (x-\frac {1{}\mathrm {i}}{a}\right )}{a^2}-\frac {x^2}{2}-\frac {x\,2{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.11, size = 24, normalized size = 0.80 \[ - \frac {x^{2}}{2} - \frac {2 i x}{a} + \frac {2 \log {\left (i a x + 1 \right )}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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