\(\int e^{-2 i \arctan (a x)} x^2 \, dx\) [45]

Optimal result

Integrand size = 14, antiderivative size = 40 \[ \int e^{-2 i \arctan (a x)} x^2 \, dx=\frac {2 x}{a^2}-\frac {i x^2}{a}-\frac {x^3}{3}+\frac {2 i \log (i-a x)}{a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5170, 78} \[ \int e^{-2 i \arctan (a x)} x^2 \, dx=\frac {2 i \log (-a x+i)}{a^3}+\frac {2 x}{a^2}-\frac {i x^2}{a}-\frac {x^3}{3} \]

[In]

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

[Out]

(2*x)/a^2 - (I*x^2)/a - x^3/3 + ((2*I)*Log[I - a*x])/a^3

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 5170

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^(I*(n/2))/(1 + I*a*x)^(I*(n/2))
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int e^{-2 i \arctan (a x)} x^2 \, dx=\frac {2 x}{a^2}-\frac {i x^2}{a}-\frac {x^3}{3}+\frac {2 i \log (i-a x)}{a^3} \]

[In]

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

[Out]

(2*x)/a^2 - (I*x^2)/a - x^3/3 + ((2*I)*Log[I - a*x])/a^3

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {\frac {1}{3} a^{2} x^{3}+i a \,x^{2}-2 x}{a^{2}}+\frac {2 i \ln \left (-a x +i\right )}{a^{3}}\)	\(40\)
risch	\(-\frac {x^{3}}{3}-\frac {i x^{2}}{a}+\frac {2 x}{a^{2}}+\frac {i \ln \left (a^{2} x^{2}+1\right )}{a^{3}}-\frac {2 \arctan \left (a x \right )}{a^{3}}\)	\(47\)
parallelrisch	\(-\frac {-a^{4} x^{4}-2 i a^{3} x^{3}+6+6 i \ln \left (a x -i\right ) x a +3 a^{2} x^{2}+6 \ln \left (a x -i\right )}{3 a^{3} \left (-a x +i\right )}\)	\(63\)
meijerg	\(-\frac {i \left (\frac {i x a \left (-5 i a^{3} x^{3}+10 a^{2} x^{2}+30 i a x +60\right )}{15 i a x +15}-4 \ln \left (i a x +1\right )\right )}{a^{3}}+\frac {i \left (\frac {i a x \left (3 i a x +6\right )}{3 i a x +3}-2 \ln \left (i a x +1\right )\right )}{a^{3}}\)	\(95\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int e^{-2 i \arctan (a x)} x^2 \, dx=-\frac {a^{3} x^{3} + 3 i \, a^{2} x^{2} - 6 \, a x - 6 i \, \log \left (\frac {a x - i}{a}\right )}{3 \, a^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int e^{-2 i \arctan (a x)} x^2 \, dx=- \frac {x^{3}}{3} - \frac {i x^{2}}{a} + \frac {2 x}{a^{2}} + \frac {2 i \log {\left (a x - i \right )}}{a^{3}} \]

[In]

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

[Out]

-x**3/3 - I*x**2/a + 2*x/a**2 + 2*I*log(a*x - I)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88 \[ \int e^{-2 i \arctan (a x)} x^2 \, dx=-\frac {a^{2} x^{3} + 3 i \, a x^{2} - 6 \, x}{3 \, a^{2}} + \frac {2 i \, \log \left (i \, a x + 1\right )}{a^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.45 \[ \int e^{-2 i \arctan (a x)} x^2 \, dx=\frac {i \, {\left (i \, a x + 1\right )}^{3} {\left (\frac {6}{i \, a x + 1} - \frac {15}{{\left (i \, a x + 1\right )}^{2}} - 1\right )}}{3 \, a^{3}} - \frac {2 i \, \log \left (\frac {1}{\sqrt {a^{2} x^{2} + 1} {\left | a \right |}}\right )}{a^{3}} \]

[In]

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

[Out]

1/3*I*(I*a*x + 1)^3*(6/(I*a*x + 1) - 15/(I*a*x + 1)^2 - 1)/a^3 - 2*I*log(1/(sqrt(a^2*x^2 + 1)*abs(a)))/a^3

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90 \[ \int e^{-2 i \arctan (a x)} x^2 \, dx=\frac {\ln \left (x-\frac {1{}\mathrm {i}}{a}\right )\,2{}\mathrm {i}}{a^3}+\frac {2\,x}{a^2}-\frac {x^3}{3}-\frac {x^2\,1{}\mathrm {i}}{a} \]

[In]

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

[Out]

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