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

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]

((-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])

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Rubi [A]  time = 0.0389113, antiderivative size = 67, normalized size of antiderivative = 1., 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[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 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 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 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]

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}{(1-i a x)^{5/2} \sqrt{1+i a x}} \, dx\\ &=-\frac{i \sqrt{1+i a x}}{3 a (1-i a x)^{3/2}}+\frac{1}{3} \int \frac{1}{(1-i a x)^{3/2} \sqrt{1+i a x}} \, 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*}

Mathematica [A]  time = 0.0125056, size = 48, normalized size = 0.72 \[ \frac{(2-i a x) \sqrt{1+i a x}}{3 a \sqrt{1-i a x} (a x+i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.062, size = 104, normalized size = 1.6 \begin{align*} -{a}^{2} \left ( -{\frac{x}{2\,{a}^{2}} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{1}{2\,{a}^{2}} \left ({\frac{x}{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,x}{3}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) } \right ) -{\frac{{\frac{2\,i}{3}}}{a} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{x}{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,x}{3}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^2/(a^2*x^2+1)^(5/2),x)

[Out]

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

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Maxima [A]  time = 0.967046, size = 61, normalized size = 0.91 \begin{align*} \frac{x}{3 \, \sqrt{a^{2} x^{2} + 1}} + \frac{2 \, x}{3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}} - \frac{2 i}{3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^2/(a^2*x^2+1)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.94631, size = 117, normalized size = 1.75 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^2/(a^2*x^2+1)^(5/2),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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (i a x + 1\right )^{2}}{\left (a^{2} x^{2} + 1\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**2/(a**2*x**2+1)**(5/2),x)

[Out]

Integral((I*a*x + 1)**2/(a**2*x**2 + 1)**(5/2), x)

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Giac [A]  time = 1.14777, size = 89, normalized size = 1.33 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^2/(a^2*x^2+1)^(5/2),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