\(\int \frac {e^{2 i \arctan (a x)} x^2}{(c+a^2 c x^2)^3} \, dx\) [377]

Optimal result

Integrand size = 26, antiderivative size = 38 \[ \int \frac {e^{2 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {i+2 a x}{6 a^3 c^3 (1-i a x)^3 (1+i a x)} \]

[Out]

1/6*(-I-2*a*x)/a^3/c^3/(1-I*a*x)^3/(1+I*a*x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {5190, 82} \[ \int \frac {e^{2 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {2 a x+i}{6 a^3 c^3 (1-i a x)^3 (1+i a x)} \]

[In]

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

[Out]

-1/6*(I + 2*a*x)/(a^3*c^3*(1 - I*a*x)^3*(1 + I*a*x))

Rule 82

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x
)^(n + 1)*(e + f*x)^(p + 1)*((2*a*d*f*(n + p + 3) - b*(d*e*(n + 2) + c*f*(p + 2)) + b*d*f*(n + p + 2)*x)/(d^2*
f^2*(n + p + 2)*(n + p + 3))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && NeQ[n + p + 3,
0] && EqQ[d*f*(n + p + 2)*(a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1)))) - b*(d*e*(n + 1)
+ c*f*(p + 1))*(a*d*f*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2))), 0]

Rule 5190

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {x^2}{(1-i a x)^4 (1+i a x)^2} \, dx}{c^3} \\ & = -\frac {i+2 a x}{6 a^3 c^3 (1-i a x)^3 (1+i a x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {i+2 a x}{6 a^3 c^3 (-i+a x) (i+a x)^3} \]

[In]

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

[Out]

(I + 2*a*x)/(6*a^3*c^3*(-I + a*x)*(I + a*x)^3)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\frac {x}{3 a^{2}}+\frac {i}{6 a^{3}}}{c^{3} \left (a x +i\right )^{3} \left (a x -i\right )}\)	\(34\)
risch	\(\frac {\frac {x}{3 a^{2}}+\frac {i}{6 a^{3}}}{c^{3} \left (a x +i\right )^{3} \left (a x -i\right )}\)	\(34\)
parallelrisch	\(\frac {i x^{6} a^{3}+3 i x^{4} a +2 x^{3}}{6 c^{3} \left (a^{2} x^{2}+1\right )^{3}}\)	\(39\)
norman	\(\frac {\frac {x^{3}}{3 c}+\frac {i a \,x^{4}}{2 c}+\frac {i a^{3} x^{6}}{6 c}}{\left (a^{2} x^{2}+1\right )^{3} c^{2}}\)	\(47\)
gosper	\(\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (2 a x +i\right ) \left (i a x +1\right )^{2}}{6 a^{3} \left (a^{2} x^{2}+1\right )^{4} c^{3}}\)	\(49\)

[In]

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

[Out]

1/c^3*(1/3*x/a^2+1/6*I/a^3)/(I+a*x)^3/(a*x-I)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int \frac {e^{2 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {2 \, a x + i}{6 \, {\left (a^{7} c^{3} x^{4} + 2 i \, a^{6} c^{3} x^{3} + 2 i \, a^{4} c^{3} x - a^{3} c^{3}\right )}} \]

[In]

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

[Out]

1/6*(2*a*x + I)/(a^7*c^3*x^4 + 2*I*a^6*c^3*x^3 + 2*I*a^4*c^3*x - a^3*c^3)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \frac {e^{2 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^3} \, dx=- \frac {- 2 a x - i}{6 a^{7} c^{3} x^{4} + 12 i a^{6} c^{3} x^{3} + 12 i a^{4} c^{3} x - 6 a^{3} c^{3}} \]

[In]

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

[Out]

-(-2*a*x - I)/(6*a**7*c**3*x**4 + 12*I*a**6*c**3*x**3 + 12*I*a**4*c**3*x - 6*a**3*c**3)

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. 62 vs. \(2 (30) = 60\).

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.63 \[ \int \frac {e^{2 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {2 \, a^{3} x^{3} - 3 i \, a^{2} x^{2} - i}{6 \, {\left (a^{9} c^{3} x^{6} + 3 \, a^{7} c^{3} x^{4} + 3 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )}} \]

[In]

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

[Out]

1/6*(2*a^3*x^3 - 3*I*a^2*x^2 - I)/(a^9*c^3*x^6 + 3*a^7*c^3*x^4 + 3*a^5*c^3*x^2 + a^3*c^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.18 \[ \int \frac {e^{2 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {1}{16 \, {\left (a x - i\right )} a^{3} c^{3}} + \frac {3 \, a^{2} x^{2} + 12 i \, a x - 5}{48 \, {\left (a x + i\right )}^{3} a^{3} c^{3}} \]

[In]

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

[Out]

-1/16/((a*x - I)*a^3*c^3) + 1/48*(3*a^2*x^2 + 12*I*a*x - 5)/((a*x + I)^3*a^3*c^3)

Mupad [B] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {e^{2 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {\frac {x}{3\,a^6\,c^3}+\frac {1{}\mathrm {i}}{6\,a^7\,c^3}}{\frac {x\,2{}\mathrm {i}}{a^3}-\frac {1}{a^4}+x^4+\frac {x^3\,2{}\mathrm {i}}{a}} \]

[In]

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

[Out]

(1i/(6*a^7*c^3) + x/(3*a^6*c^3))/((x*2i)/a^3 - 1/a^4 + x^4 + (x^3*2i)/a)