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

Optimal. Leaf size=67 \[ \frac {i \sqrt {1-i a x}}{3 a \sqrt {1+i a x}}+\frac {i \sqrt {1-i a x}}{3 a (1+i a x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5073, 45, 37} \[ \frac {i \sqrt {1-i a x}}{3 a \sqrt {1+i a x}}+\frac {i \sqrt {1-i a x}}{3 a (1+i a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*Sqrt[1 - I*a*x])/(a*(1 + I*a*x)^(3/2)) + ((I/3)*Sqrt[1 - I*a*x])/(a*Sqrt[1 + I*a*x])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 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^{-2 i \tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx &=\int \frac {1}{\sqrt {1-i a x} (1+i a x)^{5/2}} \, dx\\ &=\frac {i \sqrt {1-i a x}}{3 a (1+i a x)^{3/2}}+\frac {1}{3} \int \frac {1}{\sqrt {1-i a x} (1+i a x)^{3/2}} \, dx\\ &=\frac {i \sqrt {1-i a x}}{3 a (1+i a x)^{3/2}}+\frac {i \sqrt {1-i a x}}{3 a \sqrt {1+i a x}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 48, normalized size = 0.72 \[ \frac {\sqrt {1-i a x} (2+i a x)}{3 a \sqrt {1+i a x} (a x-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - I*a*x]*(2 + I*a*x))/(3*a*Sqrt[1 + I*a*x]*(-I + a*x))

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fricas [A]  time = 0.50, size = 51, normalized size = 0.76 \[ \frac {a^{2} x^{2} - 2 i \, a x + \sqrt {a^{2} x^{2} + 1} {\left (a x - 2 i\right )} - 1}{3 \, a^{3} x^{2} - 6 i \, a^{2} x - 3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*x^2 - 2*I*a*x + sqrt(a^2*x^2 + 1)*(a*x - 2*I) - 1)/(3*a^3*x^2 - 6*I*a^2*x - 3*a)

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giac [A]  time = 0.16, size = 66, normalized size = 0.99 \[ \frac {2 \, {\left (3 \, a i {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )} + 2 \, a^{2} - 3 \, {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{2}\right )}}{3 \, {\left (a i - \sqrt {a^{2} + \frac {1}{x^{2}}} + \frac {1}{x}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*a*i*(sqrt(a^2 + 1/x^2) - 1/x) + 2*a^2 - 3*(sqrt(a^2 + 1/x^2) - 1/x)^2)/(a*i - sqrt(a^2 + 1/x^2) + 1/x)^
3

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maple [A]  time = 0.18, size = 93, normalized size = 1.39 \[ -\frac {\frac {i \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{3 a \left (x -\frac {i}{a}\right )^{2}}-\frac {\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{3 \left (x -\frac {i}{a}\right )}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 59, normalized size = 0.88 \[ -\frac {i \, \sqrt {a^{2} x^{2} + 1}}{3 \, a^{3} x^{2} - 6 i \, a^{2} x - 3 \, a} + \frac {i \, \sqrt {a^{2} x^{2} + 1}}{3 i \, a^{2} x + 3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*sqrt(a^2*x^2 + 1)/(3*a^3*x^2 - 6*I*a^2*x - 3*a) + I*sqrt(a^2*x^2 + 1)/(3*I*a^2*x + 3*a)

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mupad [B]  time = 0.06, size = 31, normalized size = 0.46 \[ -\frac {\sqrt {a^2\,x^2+1}\,\left (a\,x-2{}\mathrm {i}\right )}{3\,a\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2*x^2 + 1)^(1/2)*(a*x - 2i))/(3*a*(a*x*1i + 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} - 2 i a x \sqrt {a^{2} x^{2} + 1} - \sqrt {a^{2} x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(a**2*x**2*sqrt(a**2*x**2 + 1) - 2*I*a*x*sqrt(a**2*x**2 + 1) - sqrt(a**2*x**2 + 1)), x)

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