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]

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

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Rubi [A]  time = 0.03, antiderivative size = 19, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [A]  time = 0.02, 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]

(-1/2*I)/(a*(-I + a*x)^2)

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fricas [A]  time = 0.37, size = 22, normalized size = 1.16 \[ -\frac {i}{2 \, a^{3} x^{2} - 4 i \, a^{2} x - 2 \, a} \]

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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giac [A]  time = 0.11, size = 14, normalized size = 0.74 \[ \frac {i}{2 \, {\left (a i x + 1\right )}^{2} a} \]

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)

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maple [A]  time = 0.03, size = 16, normalized size = 0.84 \[ \frac {i}{2 a \left (i a x +1\right )^{2}} \]

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 = 0.32, size = 13, normalized size = 0.68 \[ \frac {i}{2 \, {\left (i \, a x + 1\right )}^{2} a} \]

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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mupad [B]  time = 0.05, size = 24, normalized size = 1.26 \[ \frac {1{}\mathrm {i}}{2\,\left (-a^3\,x^2+a^2\,x\,2{}\mathrm {i}+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x*1i + 1)^3,x)

[Out]

1i/(2*(a + a^2*x*2i - a^3*x^2))

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sympy [A]  time = 0.17, size = 22, normalized size = 1.16 \[ - \frac {i}{2 a^{3} x^{2} - 4 i a^{2} x - 2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I/(2*a**3*x**2 - 4*I*a**2*x - 2*a)

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