3.1.11 \(\int e^{2 i \text {ArcTan}(a x)} x^3 \, dx\) [11]

Optimal. Leaf size=48 \[ -\frac {2 i x}{a^3}+\frac {x^2}{a^2}+\frac {2 i x^3}{3 a}-\frac {x^4}{4}-\frac {2 \log (i+a x)}{a^4} \]

[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

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Rubi [A]
time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5170, 78} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^((2*I)*ArcTan[a*x])*x^3,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*} \int e^{2 i \tan ^{-1}(a x)} x^3 \, dx &=\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*}

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Mathematica [A]
time = 0.02, size = 48, normalized size = 1.00 \begin {gather*} -\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 {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [A]
time = 0.10, size = 63, normalized size = 1.31

method	result	size
risch	\(-\frac {x^{4}}{4}+\frac {2 i x^{3}}{3 a}+\frac {x^{2}}{a^{2}}-\frac {2 i x}{a^{3}}-\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{4}}+\frac {2 i \arctan \left (a x \right )}{a^{4}}\)	\(55\)
default	\(\frac {-\frac {1}{4} a^{3} x^{4}+\frac {2}{3} i a^{2} x^{3}+a \,x^{2}-2 i x}{a^{3}}+\frac {-\frac {\ln \left (a^{2} x^{2}+1\right )}{a}+\frac {2 i \arctan \left (a x \right )}{a}}{a^{3}}\)	\(63\)
meijerg	\(\frac {a^{2} x^{2}-\ln \left (a^{2} x^{2}+1\right )}{2 a^{4}}+\frac {i \left (-\frac {2 x \left (a^{2}\right )^{\frac {5}{2}} \left (-5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}\right )}{a^{3} \sqrt {a^{2}}}-\frac {-\frac {x^{2} a^{2} \left (-3 a^{2} x^{2}+6\right )}{6}+\ln \left (a^{2} x^{2}+1\right )}{2 a^{4}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]
time = 0.46, size = 56, normalized size = 1.17 \begin {gather*} -\frac {3 \, a^{3} x^{4} - 8 i \, a^{2} x^{3} - 12 \, a x^{2} + 24 i \, x}{12 \, a^{3}} + \frac {2 i \, \arctan \left (a x\right )}{a^{4}} - \frac {\log \left (a^{2} x^{2} + 1\right )}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.28, size = 46, normalized size = 0.96 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]
time = 0.06, size = 41, normalized size = 0.85 \begin {gather*} - \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [A]
time = 0.42, size = 46, normalized size = 0.96 \begin {gather*} -\frac {3 \, a^{4} x^{4} - 8 i \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 24 i \, a x}{12 \, a^{4}} - \frac {2 \, \log \left (a x + i\right )}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.42, size = 43, normalized size = 0.90 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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