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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 53, 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 = {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 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]))

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]

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*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 1.00 \[ -\frac {4}{a^3 (a x+i)}+\frac {12 i \log (a x+i)}{a^3}-\frac {8 x}{a^2}-\frac {2 i x^2}{a}+\frac {x^3}{3} \]

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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fricas [A]  time = 0.85, size = 62, normalized size = 1.17 \[ \frac {a^{4} x^{4} - 5 i \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 24 i \, a x - 36 \, {\left (-i \, a x + 1\right )} \log \left (\frac {a x + i}{a}\right ) - 12}{3 \, a^{4} x + 3 i \, a^{3}} \]

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]

(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)/(3*a^4*x + 3*I*a^3)

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giac [A]  time = 0.12, size = 53, normalized size = 1.00 \[ \frac {12 \, i \log \left (a x + i\right )}{a^{3}} - \frac {4}{{\left (a x + i\right )} a^{3}} + \frac {a^{6} x^{3} - 6 \, a^{5} i x^{2} - 24 \, a^{4} x}{3 \, a^{6}} \]

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

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maple [A]  time = 0.06, size = 60, normalized size = 1.13 \[ \frac {x^{3}}{3}-\frac {2 i x^{2}}{a}-\frac {8 x}{a^{2}}-\frac {4}{a^{3} \left (a x +i\right )}+\frac {6 i \ln \left (a^{2} x^{2}+1\right )}{a^{3}}+\frac {12 \arctan \left (a x \right )}{a^{3}} \]

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

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maxima [A]  time = 0.42, size = 68, normalized size = 1.28 \[ -\frac {8 \, a x - 8 i}{2 \, {\left (a^{5} x^{2} + a^{3}\right )}} + \frac {a^{2} x^{3} - 6 i \, a x^{2} - 24 \, x}{3 \, a^{2}} + \frac {12 \, \arctan \left (a x\right )}{a^{3}} + \frac {6 i \, \log \left (a^{2} x^{2} + 1\right )}{a^{3}} \]

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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mupad [B]  time = 0.06, size = 51, normalized size = 0.96 \[ \frac {x^3}{3}+\frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,12{}\mathrm {i}}{a^3}-\frac {8\,x}{a^2}-\frac {4}{a^4\,\left (x+\frac {1{}\mathrm {i}}{a}\right )}-\frac {x^2\,2{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.24, size = 44, normalized size = 0.83 \[ \frac {x^{3}}{3} - \frac {4}{a^{4} x + i a^{3}} - \frac {2 i x^{2}}{a} - \frac {8 x}{a^{2}} + \frac {12 i \log {\left (a x + i \right )}}{a^{3}} \]

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]

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

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