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

Optimal. Leaf size=159 \[ -\frac{i 2^{n/2} \left (n^2+2\right ) (1-i a x)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (1-i a x)\right )}{3 a^3 (2-n)}-\frac{i n (1+i a x)^{\frac{n+2}{2}} (1-i a x)^{1-\frac{n}{2}}}{6 a^3}+\frac{x (1+i a x)^{\frac{n+2}{2}} (1-i a x)^{1-\frac{n}{2}}}{3 a^2} \]

[Out]

((-I/6)*n*(1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((2 + n)/2))/a^3 + (x*(1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((2 + n)/2
))/(3*a^2) - ((I/3)*2^(n/2)*(2 + n^2)*(1 - I*a*x)^(1 - n/2)*Hypergeometric2F1[1 - n/2, -n/2, 2 - n/2, (1 - I*a
*x)/2])/(a^3*(2 - n))

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Rubi [A]  time = 0.0857674, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {5062, 90, 80, 69} \[ -\frac{i 2^{n/2} \left (n^2+2\right ) (1-i a x)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (1-i a x)\right )}{3 a^3 (2-n)}-\frac{i n (1+i a x)^{\frac{n+2}{2}} (1-i a x)^{1-\frac{n}{2}}}{6 a^3}+\frac{x (1+i a x)^{\frac{n+2}{2}} (1-i a x)^{1-\frac{n}{2}}}{3 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/6)*n*(1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((2 + n)/2))/a^3 + (x*(1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((2 + n)/2
))/(3*a^2) - ((I/3)*2^(n/2)*(2 + n^2)*(1 - I*a*x)^(1 - n/2)*Hypergeometric2F1[1 - n/2, -n/2, 2 - n/2, (1 - I*a
*x)/2])/(a^3*(2 - n))

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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int e^{i n \tan ^{-1}(a x)} x^2 \, dx &=\int x^2 (1-i a x)^{-n/2} (1+i a x)^{n/2} \, dx\\ &=\frac{x (1-i a x)^{1-\frac{n}{2}} (1+i a x)^{\frac{2+n}{2}}}{3 a^2}+\frac{\int (1-i a x)^{-n/2} (1+i a x)^{n/2} (-1-i a n x) \, dx}{3 a^2}\\ &=-\frac{i n (1-i a x)^{1-\frac{n}{2}} (1+i a x)^{\frac{2+n}{2}}}{6 a^3}+\frac{x (1-i a x)^{1-\frac{n}{2}} (1+i a x)^{\frac{2+n}{2}}}{3 a^2}-\frac{\left (2+n^2\right ) \int (1-i a x)^{-n/2} (1+i a x)^{n/2} \, dx}{6 a^2}\\ &=-\frac{i n (1-i a x)^{1-\frac{n}{2}} (1+i a x)^{\frac{2+n}{2}}}{6 a^3}+\frac{x (1-i a x)^{1-\frac{n}{2}} (1+i a x)^{\frac{2+n}{2}}}{3 a^2}-\frac{i 2^{n/2} \left (2+n^2\right ) (1-i a x)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (1-i a x)\right )}{3 a^3 (2-n)}\\ \end{align*}

Mathematica [A]  time = 0.0494213, size = 116, normalized size = 0.73 \[ \frac{(a x+i) (1-i a x)^{-n/2} \left (2^{\frac{n}{2}+1} \left (n^2+2\right ) \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (1-i a x)\right )+(n-2) (a x-i) (2 a x-i n) (1+i a x)^{n/2}\right )}{6 a^3 (n-2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I + a*x)*((-2 + n)*(1 + I*a*x)^(n/2)*(-I + a*x)*((-I)*n + 2*a*x) + 2^(1 + n/2)*(2 + n^2)*Hypergeometric2F1[1
- n/2, -n/2, 2 - n/2, (1 - I*a*x)/2]))/(6*a^3*(-2 + n)*(1 - I*a*x)^(n/2))

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Maple [F]  time = 0.082, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{in\arctan \left ( ax \right ) }}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(I*n*arctan(a*x))*x^2,x)

[Out]

int(exp(I*n*arctan(a*x))*x^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} e^{\left (i \, n \arctan \left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(I*n*arctan(a*x))*x^2,x, algorithm="maxima")

[Out]

integrate(x^2*e^(I*n*arctan(a*x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2}}{\left (-\frac{a x + i}{a x - i}\right )^{\frac{1}{2} \, n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(I*n*arctan(a*x))*x^2,x, algorithm="fricas")

[Out]

integral(x^2/(-(a*x + I)/(a*x - I))^(1/2*n), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} e^{i n \operatorname{atan}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(I*n*atan(a*x))*x**2,x)

[Out]

Integral(x**2*exp(I*n*atan(a*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} e^{\left (i \, n \arctan \left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(I*n*arctan(a*x))*x^2,x, algorithm="giac")

[Out]

integrate(x^2*e^(I*n*arctan(a*x)), x)