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

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 = 0.0320956, 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[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 = 0.0175488, size = 18, normalized size = 0.95 \[ \frac{i}{2 a (a x+i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[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 = 0.047, size = 15, normalized size = 0.8 \begin{align*}{\frac{{\frac{i}{2}}}{a \left ( ax+i \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^3/(a^2*x^2+1)^3,x)

[Out]

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

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Maxima [B]  time = 1.48044, size = 47, normalized size = 2.47 \begin{align*} \frac{4 i \, a^{2} x^{2} + 8 \, a x - 4 i}{8 \,{\left (a^{5} x^{4} + 2 \, a^{3} x^{2} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^3/(a^2*x^2+1)^3,x, algorithm="maxima")

[Out]

1/8*(4*I*a^2*x^2 + 8*a*x - 4*I)/(a^5*x^4 + 2*a^3*x^2 + a)

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Fricas [A]  time = 1.78899, size = 45, normalized size = 2.37 \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+I*a*x)^3/(a^2*x^2+1)^3,x, algorithm="fricas")

[Out]

I/(2*a^3*x^2 + 4*I*a^2*x - 2*a)

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Sympy [A]  time = 0.589506, size = 26, normalized size = 1.37 \begin{align*} \frac{i a^{6}}{2 a^{9} x^{2} + 4 i a^{8} x - 2 a^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**3/(a**2*x**2+1)**3,x)

[Out]

I*a**6/(2*a**9*x**2 + 4*I*a**8*x - 2*a**7)

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Giac [A]  time = 1.09756, size = 18, normalized size = 0.95 \begin{align*} \frac{i}{2 \,{\left (a x + i\right )}^{2} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^3/(a^2*x^2+1)^3,x, algorithm="giac")

[Out]

1/2*i/((a*x + i)^2*a)