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

Optimal. Leaf size=39 \[ -\frac {2 i \log (a x+i)}{a^3}+\frac {2 x}{a^2}+\frac {i x^2}{a}-\frac {x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 39, 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, 77} \[ \frac {2 x}{a^2}-\frac {2 i \log (a x+i)}{a^3}+\frac {i x^2}{a}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

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

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Mathematica [A]  time = 0.01, size = 39, normalized size = 1.00 \[ -\frac {2 i \log (a x+i)}{a^3}+\frac {2 x}{a^2}+\frac {i x^2}{a}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 37, normalized size = 0.95 \[ -\frac {a^{3} x^{3} - 3 i \, a^{2} x^{2} - 6 \, a x + 6 i \, \log \left (\frac {a x + i}{a}\right )}{3 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a^3*x^3 - 3*I*a^2*x^2 - 6*a*x + 6*I*log((a*x + I)/a))/a^3

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giac [A]  time = 0.15, size = 39, normalized size = 1.00 \[ -\frac {2 \, i \log \left (a x + i\right )}{a^{3}} - \frac {a^{3} x^{3} - 3 \, a^{2} i x^{2} - 6 \, a x}{3 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*i*log(a*x + i)/a^3 - 1/3*(a^3*x^3 - 3*a^2*i*x^2 - 6*a*x)/a^3

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maple [A]  time = 0.04, size = 47, normalized size = 1.21 \[ -\frac {x^{3}}{3}+\frac {i x^{2}}{a}+\frac {2 x}{a^{2}}-\frac {i \ln \left (a^{2} x^{2}+1\right )}{a^{3}}-\frac {2 \arctan \left (a x \right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 47, normalized size = 1.21 \[ -\frac {a^{2} x^{3} - 3 i \, a x^{2} - 6 \, x}{3 \, a^{2}} - \frac {2 \, \arctan \left (a x\right )}{a^{3}} - \frac {i \, \log \left (a^{2} x^{2} + 1\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.42, size = 36, normalized size = 0.92 \[ \frac {2\,x}{a^2}-\frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,2{}\mathrm {i}}{a^3}-\frac {x^3}{3}+\frac {x^2\,1{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 32, normalized size = 0.82 \[ - \frac {x^{3}}{3} + \frac {i x^{2}}{a} + \frac {2 x}{a^{2}} - \frac {2 i \log {\left (i a x - 1 \right )}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**3/3 + I*x**2/a + 2*x/a**2 - 2*I*log(I*a*x - 1)/a**3

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