\(\int \frac {e^{-2 i \arctan (a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [334]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 54 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {2 i (1-i a x)}{3 a \left (c+a^2 c x^2\right )^{3/2}}+\frac {x}{3 c \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5182, 667, 197} \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {x}{3 c \sqrt {a^2 c x^2+c}}+\frac {2 i (1-i a x)}{3 a \left (a^2 c x^2+c\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 667

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)*((a + c*x^2)^(p + 1)/(c*(p
 + 1))), x] - Dist[e^2*((p + 2)/(c*(p + 1))), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 5182

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[c^(I*(n/2)), Int[(c + d*x^2)^(p
- I*(n/2))*(1 - I*a*x)^(I*n), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0]
) && IGtQ[I*(n/2), 0]

Rubi steps \begin{align*} \text {integral}& = c \int \frac {(1-i a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx \\ & = \frac {2 i (1-i a x)}{3 a \left (c+a^2 c x^2\right )^{3/2}}+\frac {1}{3} \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx \\ & = \frac {2 i (1-i a x)}{3 a \left (c+a^2 c x^2\right )^{3/2}}+\frac {x}{3 c \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-i a x} (2+i a x) \sqrt {1+a^2 x^2}}{3 a c \sqrt {1+i a x} (-i+a x) \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04

method result size
gosper \(-\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (-a x +2 i\right ) \left (a^{2} x^{2}+1\right )}{3 a \left (i a x +1\right )^{2} \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\) \(56\)
default \(-\frac {x}{c \sqrt {a^{2} c \,x^{2}+c}}-\frac {2 i \left (\frac {i}{3 a c \left (x -\frac {i}{a}\right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}+\frac {i \left (2 \left (x -\frac {i}{a}\right ) a^{2} c +2 i a c \right )}{3 a \,c^{2} \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}\right )}{a}\) \(137\)

[In]

int(1/(1+I*a*x)^2*(a^2*x^2+1)/(a^2*c*x^2+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {a^{2} c x^{2} + c} {\left (a x - 2 i\right )}}{3 \, {\left (a^{3} c^{2} x^{2} - 2 i \, a^{2} c^{2} x - a c^{2}\right )}} \]

[In]

integrate(1/(1+I*a*x)^2*(a^2*x^2+1)/(a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {e^{-2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=- \int \frac {a^{2} x^{2}}{a^{4} c x^{4} \sqrt {a^{2} c x^{2} + c} - 2 i a^{3} c x^{3} \sqrt {a^{2} c x^{2} + c} - 2 i a c x \sqrt {a^{2} c x^{2} + c} - c \sqrt {a^{2} c x^{2} + c}}\, dx - \int \frac {1}{a^{4} c x^{4} \sqrt {a^{2} c x^{2} + c} - 2 i a^{3} c x^{3} \sqrt {a^{2} c x^{2} + c} - 2 i a c x \sqrt {a^{2} c x^{2} + c} - c \sqrt {a^{2} c x^{2} + c}}\, dx \]

[In]

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

[Out]

-Integral(a**2*x**2/(a**4*c*x**4*sqrt(a**2*c*x**2 + c) - 2*I*a**3*c*x**3*sqrt(a**2*c*x**2 + c) - 2*I*a*c*x*sqr
t(a**2*c*x**2 + c) - c*sqrt(a**2*c*x**2 + c)), x) - Integral(1/(a**4*c*x**4*sqrt(a**2*c*x**2 + c) - 2*I*a**3*c
*x**3*sqrt(a**2*c*x**2 + c) - 2*I*a*c*x*sqrt(a**2*c*x**2 + c) - c*sqrt(a**2*c*x**2 + c)), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {x}{3 \, \sqrt {a^{2} c x^{2} + c} c} + \frac {2 i}{3 i \, \sqrt {a^{2} c x^{2} + c} a^{2} c x + 3 \, \sqrt {a^{2} c x^{2} + c} a c} \]

[In]

integrate(1/(1+I*a*x)^2*(a^2*x^2+1)/(a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {a^{2} c} {\left (3 \, \sqrt {a^{2} c} x - 3 \, \sqrt {a^{2} c x^{2} + c} - i \, \sqrt {c}\right )}}{3 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c} - i \, \sqrt {c}\right )}^{3} a^{2} c} \]

[In]

integrate(1/(1+I*a*x)^2*(a^2*x^2+1)/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.59 \[ \int \frac {e^{-2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {a^3\,x^3+3\,a\,x+2{}\mathrm {i}}{3\,a\,{\left (c\,\left (a^2\,x^2+1\right )\right )}^{3/2}} \]

[In]

int((a^2*x^2 + 1)/((c + a^2*c*x^2)^(3/2)*(a*x*1i + 1)^2),x)

[Out]

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