3.27 \(\int e^{4 i \tan ^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=65 \[ -\frac{4 x^2}{a^2}+\frac{12 i x}{a^3}+\frac{4 i}{a^4 (a x+i)}+\frac{16 \log (a x+i)}{a^4}-\frac{4 i x^3}{3 a}+\frac{x^4}{4} \]

[Out]

((12*I)*x)/a^3 - (4*x^2)/a^2 - (((4*I)/3)*x^3)/a + x^4/4 + (4*I)/(a^4*(I + a*x)) + (16*Log[I + a*x])/a^4

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Rubi [A]  time = 0.0445987, antiderivative size = 65, 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} \[ -\frac{4 x^2}{a^2}+\frac{12 i x}{a^3}+\frac{4 i}{a^4 (a x+i)}+\frac{16 \log (a x+i)}{a^4}-\frac{4 i x^3}{3 a}+\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x])*x^3,x]

[Out]

((12*I)*x)/a^3 - (4*x^2)/a^2 - (((4*I)/3)*x^3)/a + x^4/4 + (4*I)/(a^4*(I + a*x)) + (16*Log[I + a*x])/a^4

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^3 \, dx &=\int \frac{x^3 (1+i a x)^2}{(1-i a x)^2} \, dx\\ &=\int \left (\frac{12 i}{a^3}-\frac{8 x}{a^2}-\frac{4 i x^2}{a}+x^3-\frac{4 i}{a^3 (i+a x)^2}+\frac{16}{a^3 (i+a x)}\right ) \, dx\\ &=\frac{12 i x}{a^3}-\frac{4 x^2}{a^2}-\frac{4 i x^3}{3 a}+\frac{x^4}{4}+\frac{4 i}{a^4 (i+a x)}+\frac{16 \log (i+a x)}{a^4}\\ \end{align*}

Mathematica [A]  time = 0.042928, size = 65, normalized size = 1. \[ -\frac{4 x^2}{a^2}+\frac{12 i x}{a^3}+\frac{4 i}{a^4 (a x+i)}+\frac{16 \log (a x+i)}{a^4}-\frac{4 i x^3}{3 a}+\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((4*I)*ArcTan[a*x])*x^3,x]

[Out]

((12*I)*x)/a^3 - (4*x^2)/a^2 - (((4*I)/3)*x^3)/a + x^4/4 + (4*I)/(a^4*(I + a*x)) + (16*Log[I + a*x])/a^4

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Maple [A]  time = 0.046, size = 70, normalized size = 1.1 \begin{align*}{\frac{{x}^{4}}{4}}-{\frac{{\frac{4\,i}{3}}{x}^{3}}{a}}-4\,{\frac{{x}^{2}}{{a}^{2}}}+{\frac{12\,ix}{{a}^{3}}}+{\frac{4\,i}{{a}^{4} \left ( ax+i \right ) }}+8\,{\frac{\ln \left ({a}^{2}{x}^{2}+1 \right ) }{{a}^{4}}}-{\frac{16\,i\arctan \left ( ax \right ) }{{a}^{4}}} \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^3,x)

[Out]

1/4*x^4-4/3*I*x^3/a-4*x^2/a^2+12*I*x/a^3+4*I/a^4/(a*x+I)+8/a^4*ln(a^2*x^2+1)-16*I/a^4*arctan(a*x)

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Maxima [A]  time = 1.53823, size = 104, normalized size = 1.6 \begin{align*} -\frac{4 \,{\left (-i \, a x - 1\right )}}{a^{6} x^{2} + a^{4}} + \frac{3 \, a^{3} x^{4} - 16 i \, a^{2} x^{3} - 48 \, a x^{2} + 144 i \, x}{12 \, a^{3}} - \frac{16 i \, \arctan \left (a x\right )}{a^{4}} + \frac{8 \, \log \left (a^{2} x^{2} + 1\right )}{a^{4}} \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^3,x, algorithm="maxima")

[Out]

-4*(-I*a*x - 1)/(a^6*x^2 + a^4) + 1/12*(3*a^3*x^4 - 16*I*a^2*x^3 - 48*a*x^2 + 144*I*x)/a^3 - 16*I*arctan(a*x)/
a^4 + 8*log(a^2*x^2 + 1)/a^4

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Fricas [A]  time = 1.49774, size = 177, normalized size = 2.72 \begin{align*} \frac{3 \, a^{5} x^{5} - 13 i \, a^{4} x^{4} - 32 \, a^{3} x^{3} + 96 i \, a^{2} x^{2} - 144 \, a x +{\left (192 \, a x + 192 i\right )} \log \left (\frac{a x + i}{a}\right ) + 48 i}{12 \,{\left (a^{5} x + i \, a^{4}\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^3,x, algorithm="fricas")

[Out]

1/12*(3*a^5*x^5 - 13*I*a^4*x^4 - 32*a^3*x^3 + 96*I*a^2*x^2 - 144*a*x + (192*a*x + 192*I)*log((a*x + I)/a) + 48
*I)/(a^5*x + I*a^4)

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Sympy [A]  time = 0.500013, size = 58, normalized size = 0.89 \begin{align*} \frac{4 i a}{a^{6} x + i a^{5}} + \frac{x^{4}}{4} - \frac{4 i x^{3}}{3 a} - \frac{4 x^{2}}{a^{2}} + \frac{12 i x}{a^{3}} + \frac{16 \log{\left (a x + i \right )}}{a^{4}} \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**3,x)

[Out]

4*I*a/(a**6*x + I*a**5) + x**4/4 - 4*I*x**3/(3*a) - 4*x**2/a**2 + 12*I*x/a**3 + 16*log(a*x + I)/a**4

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Giac [A]  time = 1.09623, size = 85, normalized size = 1.31 \begin{align*} \frac{16 \, \log \left (a x + i\right )}{a^{4}} + \frac{4 \, i}{{\left (a x + i\right )} a^{4}} + \frac{3 \, a^{8} x^{4} - 16 \, a^{7} i x^{3} - 48 \, a^{6} x^{2} + 144 \, a^{5} i x}{12 \, a^{8}} \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^3,x, algorithm="giac")

[Out]

16*log(a*x + i)/a^4 + 4*i/((a*x + i)*a^4) + 1/12*(3*a^8*x^4 - 16*a^7*i*x^3 - 48*a^6*x^2 + 144*a^5*i*x)/a^8