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]

-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[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[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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fricas [A]  time = 0.38, 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+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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giac [A]  time = 0.14, size = 13, normalized size = 0.68 \[ \frac {i}{2 \, {\left (a x + i\right )}^{2} a} \]

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)

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maple [A]  time = 0.06, size = 15, normalized size = 0.79 \[ \frac {i}{2 a \left (a x +i\right )^{2}} \]

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/(I+a*x)^2

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maxima [B]  time = 0.43, size = 35, normalized size = 1.84 \[ \frac {4 i \, a^{2} x^{2} + 8 \, a x - 4 i}{8 \, {\left (a^{5} x^{4} + 2 \, a^{3} x^{2} + a\right )}} \]

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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mupad [B]  time = 0.50, 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((a*x*1i + 1)^3/(a^2*x^2 + 1)^3,x)

[Out]

1i/(2*(a^2*x*2i - a + a^3*x^2))

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sympy [A]  time = 0.19, size = 20, normalized size = 1.05 \[ \frac {i}{2 a^{3} x^{2} + 4 i a^{2} x - 2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/(2*a**3*x**2 + 4*I*a**2*x - 2*a)

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