\(\int e^{-2 i \arctan (a x)} x^3 \, dx\) [44]

Optimal result

Integrand size = 14, antiderivative size = 49 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=\frac {2 i x}{a^3}+\frac {x^2}{a^2}-\frac {2 i x^3}{3 a}-\frac {x^4}{4}-\frac {2 \log (i-a x)}{a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, 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} \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=-\frac {2 \log (-a x+i)}{a^4}+\frac {2 i x}{a^3}+\frac {x^2}{a^2}-\frac {2 i x^3}{3 a}-\frac {x^4}{4} \]

[In]

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

[Out]

((2*I)*x)/a^3 + x^2/a^2 - (((2*I)/3)*x^3)/a - x^4/4 - (2*Log[I - a*x])/a^4

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^3 (1-i a x)}{1+i a x} \, dx \\ & = \int \left (\frac {2 i}{a^3}+\frac {2 x}{a^2}-\frac {2 i x^2}{a}-x^3-\frac {2}{a^3 (-i+a x)}\right ) \, dx \\ & = \frac {2 i x}{a^3}+\frac {x^2}{a^2}-\frac {2 i x^3}{3 a}-\frac {x^4}{4}-\frac {2 \log (i-a x)}{a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=\frac {2 i x}{a^3}+\frac {x^2}{a^2}-\frac {2 i x^3}{3 a}-\frac {x^4}{4}-\frac {2 \log (i-a x)}{a^4} \]

[In]

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

[Out]

((2*I)*x)/a^3 + x^2/a^2 - (((2*I)/3)*x^3)/a - x^4/4 - (2*Log[I - a*x])/a^4

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {\frac {1}{4} a^{3} x^{4}+\frac {2}{3} i a^{2} x^{3}-a \,x^{2}-2 i x}{a^{3}}-\frac {2 \ln \left (-a x +i\right )}{a^{4}}\)	\(48\)
risch	\(-\frac {x^{4}}{4}-\frac {2 i x^{3}}{3 a}+\frac {x^{2}}{a^{2}}+\frac {2 i x}{a^{3}}-\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{4}}-\frac {2 i \arctan \left (a x \right )}{a^{4}}\)	\(55\)
parallelrisch	\(-\frac {-3 a^{5} x^{5}-5 i a^{4} x^{4}+4 a^{3} x^{3}+24 i+12 i a^{2} x^{2}-24 \ln \left (a x -i\right ) x a +24 i \ln \left (a x -i\right )}{12 a^{4} \left (-a x +i\right )}\)	\(73\)
meijerg	\(-\frac {-\frac {i x a \left (-3 a^{4} x^{4}-5 i a^{3} x^{3}+10 a^{2} x^{2}+30 i a x +60\right )}{12 \left (i a x +1\right )}+5 \ln \left (i a x +1\right )}{a^{4}}+\frac {-\frac {i a x \left (2 a^{2} x^{2}+6 i a x +12\right )}{4 \left (i a x +1\right )}+3 \ln \left (i a x +1\right )}{a^{4}}\)	\(108\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=-\frac {3 \, a^{4} x^{4} + 8 i \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 24 i \, a x + 24 \, \log \left (\frac {a x - i}{a}\right )}{12 \, a^{4}} \]

[In]

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

[Out]

-1/12*(3*a^4*x^4 + 8*I*a^3*x^3 - 12*a^2*x^2 - 24*I*a*x + 24*log((a*x - I)/a))/a^4

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=- \frac {x^{4}}{4} - \frac {2 i x^{3}}{3 a} + \frac {x^{2}}{a^{2}} + \frac {2 i x}{a^{3}} - \frac {2 \log {\left (a x - i \right )}}{a^{4}} \]

[In]

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

[Out]

-x**4/4 - 2*I*x**3/(3*a) + x**2/a**2 + 2*I*x/a**3 - 2*log(a*x - I)/a**4

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=-\frac {i \, {\left (-3 i \, a^{3} x^{4} + 8 \, a^{2} x^{3} + 12 i \, a x^{2} - 24 \, x\right )}}{12 \, a^{3}} - \frac {2 \, \log \left (i \, a x + 1\right )}{a^{4}} \]

[In]

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

[Out]

-1/12*I*(-3*I*a^3*x^4 + 8*a^2*x^3 + 12*I*a*x^2 - 24*x)/a^3 - 2*log(I*a*x + 1)/a^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.39 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=\frac {{\left (i \, a x + 1\right )}^{4} {\left (\frac {20}{i \, a x + 1} - \frac {54}{{\left (i \, a x + 1\right )}^{2}} + \frac {84}{{\left (i \, a x + 1\right )}^{3}} - 3\right )}}{12 \, a^{4}} + \frac {2 \, \log \left (\frac {1}{\sqrt {a^{2} x^{2} + 1} {\left | a \right |}}\right )}{a^{4}} \]

[In]

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

[Out]

1/12*(I*a*x + 1)^4*(20/(I*a*x + 1) - 54/(I*a*x + 1)^2 + 84/(I*a*x + 1)^3 - 3)/a^4 + 2*log(1/(sqrt(a^2*x^2 + 1)
*abs(a)))/a^4

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=\frac {x^2}{a^2}-\frac {x^4}{4}-\frac {2\,\ln \left (x-\frac {1{}\mathrm {i}}{a}\right )}{a^4}+\frac {x\,2{}\mathrm {i}}{a^3}-\frac {x^3\,2{}\mathrm {i}}{3\,a} \]

[In]

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

[Out]

(x*2i)/a^3 - (2*log(x - 1i/a))/a^4 - x^4/4 - (x^3*2i)/(3*a) + x^2/a^2