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

Optimal. Leaf size=125 \[ \frac{2 (1-i a x)^{-n/2} (1+i a x)^{n/2} \, _2F_1\left (1,-\frac{n}{2};1-\frac{n}{2};\frac{1-i a x}{i a x+1}\right )}{n}-\frac{2^{\frac{n}{2}+1} (1-i a x)^{-n/2} \, _2F_1\left (-\frac{n}{2},-\frac{n}{2};1-\frac{n}{2};\frac{1}{2} (1-i a x)\right )}{n} \]

[Out]

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

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Rubi [A]  time = 0.0503532, antiderivative size = 125, 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, 105, 69, 131} \[ \frac{2 (1-i a x)^{-n/2} (1+i a x)^{n/2} \, _2F_1\left (1,-\frac{n}{2};1-\frac{n}{2};\frac{1-i a x}{i a x+1}\right )}{n}-\frac{2^{\frac{n}{2}+1} (1-i a x)^{-n/2} \, _2F_1\left (-\frac{n}{2},-\frac{n}{2};1-\frac{n}{2};\frac{1}{2} (1-i a x)\right )}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

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]))

Rule 131

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*c -
a*d)^n*(a + b*x)^(m + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

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

Mathematica [A]  time = 0.0257732, size = 106, normalized size = 0.85 \[ \frac{2 (1-i a x)^{-n/2} \left ((1+i a x)^{n/2} \, _2F_1\left (1,-\frac{n}{2};1-\frac{n}{2};\frac{a x+i}{i-a x}\right )-2^{n/2} \, _2F_1\left (-\frac{n}{2},-\frac{n}{2};1-\frac{n}{2};\frac{1}{2} (1-i a x)\right )\right )}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.166, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\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 \frac{e^{\left (i \, n \arctan \left (a x\right )\right )}}{x}\,{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(e^(I*n*arctan(a*x))/x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{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(1/(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 \frac{e^{i n \operatorname{atan}{\left (a x \right )}}}{x}\, 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(exp(I*n*atan(a*x))/x, x)

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