3.327 \(\int \frac {e^{-4 i \tan ^{-1}(a x)}}{(1+a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=67 \[ \frac {i (1-i a x)^{3/2}}{15 a (1+i a x)^{3/2}}+\frac {i (1-i a x)^{3/2}}{5 a (1+i a x)^{5/2}} \]

[Out]

1/5*I*(1-I*a*x)^(3/2)/a/(1+I*a*x)^(5/2)+1/15*I*(1-I*a*x)^(3/2)/a/(1+I*a*x)^(3/2)

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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} \[ \frac {i (1-i a x)^{3/2}}{15 a (1+i a x)^{3/2}}+\frac {i (1-i a x)^{3/2}}{5 a (1+i a x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((4*I)*ArcTan[a*x])*(1 + a^2*x^2)^(3/2)),x]

[Out]

((I/5)*(1 - I*a*x)^(3/2))/(a*(1 + I*a*x)^(5/2)) + ((I/15)*(1 - I*a*x)^(3/2))/(a*(1 + I*a*x)^(3/2))

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^{-4 i \tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx &=\int \frac {\sqrt {1-i a x}}{(1+i a x)^{7/2}} \, dx\\ &=\frac {i (1-i a x)^{3/2}}{5 a (1+i a x)^{5/2}}+\frac {1}{5} \int \frac {\sqrt {1-i a x}}{(1+i a x)^{5/2}} \, dx\\ &=\frac {i (1-i a x)^{3/2}}{5 a (1+i a x)^{5/2}}+\frac {i (1-i a x)^{3/2}}{15 a (1+i a x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 47, normalized size = 0.70 \[ \frac {(1-i a x)^{3/2} (a x-4 i)}{15 a \sqrt {1+i a x} (a x-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^((4*I)*ArcTan[a*x])*(1 + a^2*x^2)^(3/2)),x]

[Out]

((1 - I*a*x)^(3/2)*(-4*I + a*x))/(15*a*Sqrt[1 + I*a*x]*(-I + a*x)^2)

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fricas [A]  time = 0.46, size = 76, normalized size = 1.13 \[ -\frac {a^{3} x^{3} - 3 i \, a^{2} x^{2} - 3 \, a x + {\left (a^{2} x^{2} - 3 i \, a x + 4\right )} \sqrt {a^{2} x^{2} + 1} + i}{15 \, a^{4} x^{3} - 45 i \, a^{3} x^{2} - 45 \, a^{2} x + 15 i \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^4*(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-(a^3*x^3 - 3*I*a^2*x^2 - 3*a*x + (a^2*x^2 - 3*I*a*x + 4)*sqrt(a^2*x^2 + 1) + I)/(15*a^4*x^3 - 45*I*a^3*x^2 -
45*a^2*x + 15*I*a)

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giac [B]  time = 0.20, size = 111, normalized size = 1.66 \[ -\frac {2 \, {\left (5 \, a^{3} i {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )} - 15 \, a i {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{3} + 4 \, a^{4} - 25 \, a^{2} {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{2} + 15 \, {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{4}\right )}}{15 \, {\left (a i - \sqrt {a^{2} + \frac {1}{x^{2}}} + \frac {1}{x}\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^4*(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

-2/15*(5*a^3*i*(sqrt(a^2 + 1/x^2) - 1/x) - 15*a*i*(sqrt(a^2 + 1/x^2) - 1/x)^3 + 4*a^4 - 25*a^2*(sqrt(a^2 + 1/x
^2) - 1/x)^2 + 15*(sqrt(a^2 + 1/x^2) - 1/x)^4)/(a*i - sqrt(a^2 + 1/x^2) + 1/x)^5

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maple [A]  time = 0.18, size = 92, normalized size = 1.37 \[ \frac {\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{5 a \left (x -\frac {i}{a}\right )^{4}}-\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{15 \left (x -\frac {i}{a}\right )^{3}}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+I*a*x)^4*(a^2*x^2+1)^(1/2),x)

[Out]

1/a^4*(1/5*I/a/(x-I/a)^4*((x-I/a)^2*a^2+2*I*a*(x-I/a))^(3/2)-1/15/(x-I/a)^3*((x-I/a)^2*a^2+2*I*a*(x-I/a))^(3/2
))

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maxima [B]  time = 0.32, size = 100, normalized size = 1.49 \[ \frac {2 i \, \sqrt {a^{2} x^{2} + 1}}{-5 i \, a^{4} x^{3} - 15 \, a^{3} x^{2} + 15 i \, a^{2} x + 5 \, a} + \frac {i \, \sqrt {a^{2} x^{2} + 1}}{15 \, a^{3} x^{2} - 30 i \, a^{2} x - 15 \, a} - \frac {i \, \sqrt {a^{2} x^{2} + 1}}{15 i \, a^{2} x + 15 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^4*(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

2*I*sqrt(a^2*x^2 + 1)/(-5*I*a^4*x^3 - 15*a^3*x^2 + 15*I*a^2*x + 5*a) + I*sqrt(a^2*x^2 + 1)/(15*a^3*x^2 - 30*I*
a^2*x - 15*a) - I*sqrt(a^2*x^2 + 1)/(15*I*a^2*x + 15*a)

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mupad [B]  time = 0.52, size = 40, normalized size = 0.60 \[ \frac {\sqrt {a^2\,x^2+1}\,\left (a^2\,x^2-a\,x\,3{}\mathrm {i}+4\right )\,1{}\mathrm {i}}{15\,a\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*x^2 + 1)^(1/2)/(a*x*1i + 1)^4,x)

[Out]

((a^2*x^2 + 1)^(1/2)*(a^2*x^2 - a*x*3i + 4)*1i)/(15*a*(a*x*1i + 1)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} x^{2} + 1}}{\left (a x - i\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)**4*(a**2*x**2+1)**(1/2),x)

[Out]

Integral(sqrt(a**2*x**2 + 1)/(a*x - I)**4, x)

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