\(\int \frac {e^{-2 i \arctan (a x)}}{(1+a^2 x^2)^{3/2}} \, dx\) [325]

Optimal result

Integrand size = 24, antiderivative size = 67 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i \sqrt {1-i a x}}{3 a (1+i a x)^{3/2}}+\frac {i \sqrt {1-i a x}}{3 a \sqrt {1+i a x}} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5181, 47, 37} \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i \sqrt {1-i a x}}{3 a \sqrt {1+i a x}}+\frac {i \sqrt {1-i a x}}{3 a (1+i a x)^{3/2}} \]

[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 47

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 5181

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.72 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-i a x} (2+i a x)}{3 a \sqrt {1+i a x} (-i+a x)} \]

[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))

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.69

method	result	size
gosper	\(-\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (-a x +2 i\right )}{3 a \left (i a x +1\right )^{2} \sqrt {a^{2} x^{2}+1}}\)	\(46\)
default	\(-\frac {\frac {i \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{3 a \left (x -\frac {i}{a}\right )^{2}}-\frac {\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{3 \left (x -\frac {i}{a}\right )}}{a^{2}}\)	\(93\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {a^{2} x^{2} - 2 i \, a x + \sqrt {a^{2} x^{2} + 1} {\left (a x - 2 i\right )} - 1}{3 \, {\left (a^{3} x^{2} - 2 i \, a^{2} x - a\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {e^{-2 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=- \int \frac {1}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} - 2 i a x \sqrt {a^{2} x^{2} + 1} - \sqrt {a^{2} x^{2} + 1}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {i \, \sqrt {a^{2} x^{2} + 1}}{3 \, {\left (a^{3} x^{2} - 2 i \, a^{2} x - a\right )}} + \frac {i \, \sqrt {a^{2} x^{2} + 1}}{3 i \, a^{2} x + 3 \, a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (2 \, a^{2} - 3 \, {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{2} + 3 \, a {\left (i \, \sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {i}{x}\right )}\right )}}{3 \, {\left (-i \, a + \sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.46 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {a^2\,x^2+1}\,\left (a\,x-2{}\mathrm {i}\right )}{3\,a\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^2} \]

[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)