\(\int \frac {e^{-3 i \arctan (a x)}}{(1+a^2 x^2)^{3/2}} \, dx\) [326]

Optimal result

Integrand size = 24, antiderivative size = 19 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i}{2 a (1+i a x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5181, 32} \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i}{2 a (1+i a x)^2} \]

[In]

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

[Out]

(I/2)/(a*(1 + I*a*x)^2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -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}{(1+i a x)^3} \, dx \\ & = \frac {i}{2 a (1+i a x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {i}{2 a (-i+a x)^2} \]

[In]

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

[Out]

(-1/2*I)/(a*(-I + a*x)^2)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {i}{2 a \left (a x -i\right )^{2}}\)	\(15\)
default	\(\frac {i}{2 a \left (i a x +1\right )^{2}}\)	\(16\)
meijerg	\(\frac {x \left (i a x +2\right )}{2 \left (i a x +1\right )^{2}}\)	\(20\)
gosper	\(\frac {-a x +i}{2 a \left (i a x +1\right )^{3}}\)	\(22\)
parallelrisch	\(-\frac {i a \,x^{2}+2 x}{2 \left (-a x +i\right )^{2}}\)	\(23\)
norman	\(\frac {x -\frac {3}{2} i a \,x^{2}-\frac {1}{2} i a^{3} x^{4}}{\left (a^{2} x^{2}+1\right )^{2}}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {i}{2 \, {\left (a^{3} x^{2} - 2 i \, a^{2} x - a\right )}} \]

[In]

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

[Out]

-1/2*I/(a^3*x^2 - 2*I*a^2*x - a)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=- \frac {i}{2 a^{3} x^{2} - 4 i a^{2} x - 2 a} \]

[In]

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

[Out]

-I/(2*a**3*x**2 - 4*I*a**2*x - 2*a)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i}{2 \, {\left (i \, a x + 1\right )}^{2} a} \]

[In]

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

[Out]

1/2*I/((I*a*x + 1)^2*a)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i}{2 \, {\left (i \, a x + 1\right )}^{2} a} \]

[In]

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

[Out]

1/2*I/((I*a*x + 1)^2*a)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {1{}\mathrm {i}}{2\,\left (-a^3\,x^2+a^2\,x\,2{}\mathrm {i}+a\right )} \]

[In]

int(1/(a*x*1i + 1)^3,x)

[Out]

1i/(2*(a + a^2*x*2i - a^3*x^2))