3.28 \(\int e^{4 i \tan ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=53 \[ -\frac{8 x}{a^2}-\frac{4}{a^3 (a x+i)}+\frac{12 i \log (a x+i)}{a^3}-\frac{2 i x^2}{a}+\frac{x^3}{3} \]

[Out]

(-8*x)/a^2 - ((2*I)*x^2)/a + x^3/3 - 4/(a^3*(I + a*x)) + ((12*I)*Log[I + a*x])/a^3

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Rubi [A]  time = 0.0399861, antiderivative size = 53, normalized size of antiderivative = 1., 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 = {5062, 88} \[ -\frac{8 x}{a^2}-\frac{4}{a^3 (a x+i)}+\frac{12 i \log (a x+i)}{a^3}-\frac{2 i x^2}{a}+\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x])*x^2,x]

[Out]

(-8*x)/a^2 - ((2*I)*x^2)/a + x^3/3 - 4/(a^3*(I + a*x)) + ((12*I)*Log[I + a*x])/a^3

Rule 5062

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]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int e^{4 i \tan ^{-1}(a x)} x^2 \, dx &=\int \frac{x^2 (1+i a x)^2}{(1-i a x)^2} \, dx\\ &=\int \left (-\frac{8}{a^2}-\frac{4 i x}{a}+x^2+\frac{4}{a^2 (i+a x)^2}+\frac{12 i}{a^2 (i+a x)}\right ) \, dx\\ &=-\frac{8 x}{a^2}-\frac{2 i x^2}{a}+\frac{x^3}{3}-\frac{4}{a^3 (i+a x)}+\frac{12 i \log (i+a x)}{a^3}\\ \end{align*}

Mathematica [A]  time = 0.0308238, size = 53, normalized size = 1. \[ -\frac{8 x}{a^2}-\frac{4}{a^3 (a x+i)}+\frac{12 i \log (a x+i)}{a^3}-\frac{2 i x^2}{a}+\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((4*I)*ArcTan[a*x])*x^2,x]

[Out]

(-8*x)/a^2 - ((2*I)*x^2)/a + x^3/3 - 4/(a^3*(I + a*x)) + ((12*I)*Log[I + a*x])/a^3

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Maple [A]  time = 0.046, size = 60, normalized size = 1.1 \begin{align*}{\frac{{x}^{3}}{3}}-{\frac{2\,i{x}^{2}}{a}}-8\,{\frac{x}{{a}^{2}}}-4\,{\frac{1}{{a}^{3} \left ( ax+i \right ) }}+{\frac{6\,i\ln \left ({a}^{2}{x}^{2}+1 \right ) }{{a}^{3}}}+12\,{\frac{\arctan \left ( ax \right ) }{{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^2*x^2,x)

[Out]

1/3*x^3-2*I*x^2/a-8*x/a^2-4/a^3/(a*x+I)+6*I/a^3*ln(a^2*x^2+1)+12/a^3*arctan(a*x)

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Maxima [A]  time = 1.49246, size = 92, normalized size = 1.74 \begin{align*} -\frac{8 \, a x - 8 i}{2 \,{\left (a^{5} x^{2} + a^{3}\right )}} + \frac{a^{2} x^{3} - 6 i \, a x^{2} - 24 \, x}{3 \, a^{2}} + \frac{12 \, \arctan \left (a x\right )}{a^{3}} + \frac{6 i \, \log \left (a^{2} x^{2} + 1\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2*x^2,x, algorithm="maxima")

[Out]

-1/2*(8*a*x - 8*I)/(a^5*x^2 + a^3) + 1/3*(a^2*x^3 - 6*I*a*x^2 - 24*x)/a^2 + 12*arctan(a*x)/a^3 + 6*I*log(a^2*x
^2 + 1)/a^3

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Fricas [A]  time = 1.62085, size = 147, normalized size = 2.77 \begin{align*} \frac{a^{4} x^{4} - 5 i \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 24 i \, a x - 36 \,{\left (-i \, a x + 1\right )} \log \left (\frac{a x + i}{a}\right ) - 12}{3 \,{\left (a^{4} x + i \, a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2*x^2,x, algorithm="fricas")

[Out]

1/3*(a^4*x^4 - 5*I*a^3*x^3 - 18*a^2*x^2 - 24*I*a*x - 36*(-I*a*x + 1)*log((a*x + I)/a) - 12)/(a^4*x + I*a^3)

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Sympy [A]  time = 0.490888, size = 46, normalized size = 0.87 \begin{align*} - \frac{4 a}{a^{5} x + i a^{4}} + \frac{x^{3}}{3} - \frac{2 i x^{2}}{a} - \frac{8 x}{a^{2}} + \frac{12 i \log{\left (a x + i \right )}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**2*x**2,x)

[Out]

-4*a/(a**5*x + I*a**4) + x**3/3 - 2*I*x**2/a - 8*x/a**2 + 12*I*log(a*x + I)/a**3

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Giac [A]  time = 1.11657, size = 72, normalized size = 1.36 \begin{align*} \frac{12 \, i \log \left (a x + i\right )}{a^{3}} - \frac{4}{{\left (a x + i\right )} a^{3}} + \frac{a^{6} x^{3} - 6 \, a^{5} i x^{2} - 24 \, a^{4} x}{3 \, a^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2*x^2,x, algorithm="giac")

[Out]

12*i*log(a*x + i)/a^3 - 4/((a*x + i)*a^3) + 1/3*(a^6*x^3 - 6*a^5*i*x^2 - 24*a^4*x)/a^6