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

Optimal. Leaf size=107 \[ \frac{2^{n/2} n (1-i a x)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (1-i a x)\right )}{a^2 (2-n)}+\frac{(1+i a x)^{\frac{n+2}{2}} (1-i a x)^{1-\frac{n}{2}}}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.0458557, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {5062, 80, 69} \[ \frac{2^{n/2} n (1-i a x)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (1-i a x)\right )}{a^2 (2-n)}+\frac{(1+i a x)^{\frac{n+2}{2}} (1-i a x)^{1-\frac{n}{2}}}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((2 + n)/2))/(2*a^2) + (2^(n/2)*n*(1 - I*a*x)^(1 - n/2)*Hypergeometric2F1[1
- n/2, -n/2, 2 - n/2, (1 - I*a*x)/2])/(a^2*(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 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 \, dx &=\int x (1-i a x)^{-n/2} (1+i a x)^{n/2} \, dx\\ &=\frac{(1-i a x)^{1-\frac{n}{2}} (1+i a x)^{\frac{2+n}{2}}}{2 a^2}-\frac{(i n) \int (1-i a x)^{-n/2} (1+i a x)^{n/2} \, dx}{2 a}\\ &=\frac{(1-i a x)^{1-\frac{n}{2}} (1+i a x)^{\frac{2+n}{2}}}{2 a^2}+\frac{2^{n/2} n (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 )}{a^2 (2-n)}\\ \end{align*}

Mathematica [A]  time = 0.0245457, size = 105, normalized size = 0.98 \[ \frac{(a x+i) (1-i a x)^{-n/2} \left (i 2^{\frac{n}{2}+1} n \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (1-i a x)\right )+(n-2) (a x-i) (1+i a x)^{n/2}\right )}{2 a^2 (n-2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.074, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{in\arctan \left ( ax \right ) }}x\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x 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,x, algorithm="maxima")

[Out]

integrate(x*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}{\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,x, algorithm="fricas")

[Out]

integral(x/(-(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 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,x)

[Out]

Integral(x*exp(I*n*atan(a*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x 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,x, algorithm="giac")

[Out]

integrate(x*e^(I*n*arctan(a*x)), x)