\(\int \frac {e^{3 i \arctan (a x)}}{(1+a^2 x^2)^{3/2}} \, dx\) [321]

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 = 0.03 (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[E^((3*I)*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

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

[Out]

(I/2)/(a*(I + a*x)^2)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {i}{2 a \left (a x +i\right )^{2}}\)	\(15\)
risch	\(\frac {i}{2 a \left (a x +i\right )^{2}}\)	\(15\)
norman	\(\frac {x +\frac {3}{2} i a \,x^{2}+\frac {1}{2} i a^{3} x^{4}}{\left (a^{2} x^{2}+1\right )^{2}}\)	\(31\)
gosper	\(-\frac {\left (a x +i\right ) \left (i a x +1\right )^{3}}{2 a \left (a^{2} x^{2}+1\right )^{3}}\)	\(32\)
parallelrisch	\(\frac {i a^{3} x^{4}+3 i a \,x^{2}+2 x}{2 \left (a^{2} x^{2}+1\right )^{2}}\)	\(34\)
meijerg	\(\frac {\frac {x \sqrt {a^{2}}\, \left (3 a^{2} x^{2}+5\right )}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 \sqrt {a^{2}}\, \arctan \left (a x \right )}{2 a}}{4 \sqrt {a^{2}}}+\frac {3 i a \,x^{2} \left (a^{2} x^{2}+2\right )}{4 \left (a^{2} x^{2}+1\right )^{2}}-\frac {3 \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}} \left (-3 a^{2} x^{2}+3\right )}{6 a^{2} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{2 a^{3}}\right )}{4 \sqrt {a^{2}}}-\frac {i a^{3} x^{4}}{4 \left (a^{2} x^{2}+1\right )^{2}}\)	\(154\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (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+I*a*x)^3/(a^2*x^2+1)^3,x, algorithm="fricas")

[Out]

1/2*I/(a^3*x^2 + 2*I*a^2*x - a)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i}{2 a^{3} x^{2} + 4 i a^{2} x - 2 a} \]

[In]

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

[Out]

I/(2*a**3*x**2 + 4*I*a**2*x - 2*a)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (13) = 26\).

Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84 \[ \int \frac {e^{3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {-i \, a^{2} x^{2} - 2 \, a x + i}{2 \, {\left (a^{5} x^{4} + 2 \, a^{3} x^{2} + a\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {e^{3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i}{2 \, {\left (a x + i\right )}^{2} a} \]

[In]

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

[Out]

1/2*I/((a*x + I)^2*a)

Mupad [B] (verification not implemented)

Time = 0.62 (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((a*x*1i + 1)^3/(a^2*x^2 + 1)^3,x)

[Out]

1i/(2*(a^2*x*2i - a + a^3*x^2))