\(\int e^{2 i \arctan (a x)} x^2 \, dx\) [12]

Optimal result

Integrand size = 14, antiderivative size = 39 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=\frac {2 x}{a^2}+\frac {i x^2}{a}-\frac {x^3}{3}-\frac {2 i \log (i+a x)}{a^3} \]

[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 = 39, 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[E^((2*I)*ArcTan[a*x])*x^2,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 = 39, 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[E^((2*I)*ArcTan[a*x])*x^2,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 = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(-\frac {a^{3} x^{3}-3 i a^{2} x^{2}+6 i \ln \left (a x +i\right )-6 a x}{3 a^{3}}\)	\(37\)
risch	\(\frac {2 x}{a^{2}}-\frac {x^{3}}{3}+\frac {i x^{2}}{a}-\frac {i \ln \left (a^{2} x^{2}+1\right )}{a^{3}}-\frac {2 \arctan \left (a x \right )}{a^{3}}\)	\(47\)
default	\(\frac {2 x -\frac {1}{3} a^{2} x^{3}+i a \,x^{2}}{a^{2}}+\frac {-\frac {i \ln \left (a^{2} x^{2}+1\right )}{a}-\frac {2 \arctan \left (a x \right )}{a}}{a^{2}}\)	\(55\)
meijerg	\(\frac {\frac {2 x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2}}-\frac {2 \left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}}{2 a^{2} \sqrt {a^{2}}}+\frac {i \left (a^{2} x^{2}-\ln \left (a^{2} x^{2}+1\right )\right )}{a^{3}}-\frac {-\frac {2 x \left (a^{2}\right )^{\frac {5}{2}} \left (-5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}}{2 a^{2} \sqrt {a^{2}}}\)	\(110\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=-\frac {a^{3} x^{3} - 3 i \, a^{2} x^{2} - 6 \, a x + 6 i \, \log \left (\frac {a x + i}{a}\right )}{3 \, a^{3}} \]

[In]

integrate((1+I*a*x)^2/(a^2*x^2+1)*x^2,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 = 0.79 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=- \frac {x^{3}}{3} + \frac {i x^{2}}{a} + \frac {2 x}{a^{2}} - \frac {2 i \log {\left (a x + i \right )}}{a^{3}} \]

[In]

integrate((1+I*a*x)**2/(a**2*x**2+1)*x**2,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 = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=-\frac {a^{2} x^{3} - 3 i \, a x^{2} - 6 \, x}{3 \, a^{2}} - \frac {2 \, \arctan \left (a x\right )}{a^{3}} - \frac {i \, \log \left (a^{2} x^{2} + 1\right )}{a^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=-\frac {a^{3} x^{3} - 3 i \, a^{2} x^{2} - 6 \, a x}{3 \, a^{3}} - \frac {2 i \, \log \left (a x + i\right )}{a^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=\frac {2\,x}{a^2}-\frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,2{}\mathrm {i}}{a^3}-\frac {x^3}{3}+\frac {x^2\,1{}\mathrm {i}}{a} \]

[In]

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

[Out]

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