3.326 \(\int \frac{e^{-3 i \tan ^{-1}(a x)}}{(1+a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=19 \[ \frac{i}{2 a (1+i a x)^2} \]

[Out]

(I/2)/(a*(1 + I*a*x)^2)

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Rubi [A]  time = 0.0323292, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {5073, 32} \[ \frac{i}{2 a (1+i a x)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((3*I)*ArcTan[a*x])*(1 + a^2*x^2)^(3/2)),x]

[Out]

(I/2)/(a*(1 + I*a*x)^2)

Rule 5073

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - I*a*x)^(p + (I*n
)/2)*(1 + I*a*x)^(p - (I*n)/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[d, a^2*c] && (IntegerQ[p] || GtQ[c,
0])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{e^{-3 i \tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx &=\int \frac{1}{(1+i a x)^3} \, dx\\ &=\frac{i}{2 a (1+i a x)^2}\\ \end{align*}

Mathematica [A]  time = 0.0177464, size = 18, normalized size = 0.95 \[ -\frac{i}{2 a (a x-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^((3*I)*ArcTan[a*x])*(1 + a^2*x^2)^(3/2)),x]

[Out]

(-I/2)/(a*(-I + a*x)^2)

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Maple [A]  time = 0.033, size = 16, normalized size = 0.8 \begin{align*}{\frac{{\frac{i}{2}}}{a \left ( 1+iax \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+I*a*x)^3,x)

[Out]

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

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Maxima [A]  time = 1.00784, size = 18, normalized size = 0.95 \begin{align*} \frac{i}{2 \,{\left (i \, a x + 1\right )}^{2} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^3,x, algorithm="maxima")

[Out]

1/2*I/((I*a*x + 1)^2*a)

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Fricas [A]  time = 1.67099, size = 46, normalized size = 2.42 \begin{align*} -\frac{i}{2 \, a^{3} x^{2} - 4 i \, a^{2} x - 2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^3,x, algorithm="fricas")

[Out]

-I/(2*a^3*x^2 - 4*I*a^2*x - 2*a)

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Sympy [B]  time = 0.395081, size = 27, normalized size = 1.42 \begin{align*} - \frac{i a^{3}}{2 a^{6} x^{2} - 4 i a^{5} x - 2 a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)**3,x)

[Out]

-I*a**3/(2*a**6*x**2 - 4*I*a**5*x - 2*a**4)

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Giac [A]  time = 1.09596, size = 19, normalized size = 1. \begin{align*} \frac{i}{2 \,{\left (a i x + 1\right )}^{2} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^3,x, algorithm="giac")

[Out]

1/2*i/((a*i*x + 1)^2*a)