\(\int \frac {e^{i n \arctan (a x)}}{x^2} \, dx\) [159]

Optimal result

Integrand size = 15, antiderivative size = 79 \[ \int \frac {e^{i n \arctan (a x)}}{x^2} \, dx=-\frac {4 i a (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {1}{2} (-2+n)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {n}{2},2-\frac {n}{2},\frac {1-i a x}{1+i a x}\right )}{2-n} \]

[Out]

-4*I*a*(1-I*a*x)^(1-1/2*n)*(1+I*a*x)^(-1+1/2*n)*hypergeom([2, 1-1/2*n],[2-1/2*n],(1-I*a*x)/(1+I*a*x))/(2-n)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5170, 133} \[ \int \frac {e^{i n \arctan (a x)}}{x^2} \, dx=-\frac {4 i a (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {n-2}{2}} \operatorname {Hypergeometric2F1}\left (2,1-\frac {n}{2},2-\frac {n}{2},\frac {1-i a x}{i a x+1}\right )}{2-n} \]

[In]

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

[Out]

((-4*I)*a*(1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((-2 + n)/2)*Hypergeometric2F1[2, 1 - n/2, 2 - n/2, (1 - I*a*x)/(1
+ I*a*x)])/(2 - n)

Rule 133

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)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rule 5170

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*} \text {integral}& = \int \frac {(1-i a x)^{-n/2} (1+i a x)^{n/2}}{x^2} \, dx \\ & = -\frac {4 i a (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {1}{2} (-2+n)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {n}{2},2-\frac {n}{2},\frac {1-i a x}{1+i a x}\right )}{2-n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {e^{i n \arctan (a x)}}{x^2} \, dx=-\frac {2 i a (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{-1+\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (2,1-\frac {n}{2},2-\frac {n}{2},-\frac {1-i a x}{-1-i a x}\right )}{1-\frac {n}{2}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {{\mathrm e}^{i n \arctan \left (a x \right )}}{x^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {e^{i n \arctan (a x)}}{x^2} \, dx=\int { \frac {e^{\left (i \, n \arctan \left (a x\right )\right )}}{x^{2}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {e^{i n \arctan (a x)}}{x^2} \, dx=\int \frac {e^{i n \operatorname {atan}{\left (a x \right )}}}{x^{2}}\, dx \]

[In]

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

[Out]

Integral(exp(I*n*atan(a*x))/x**2, x)

Maxima [F]

\[ \int \frac {e^{i n \arctan (a x)}}{x^2} \, dx=\int { \frac {e^{\left (i \, n \arctan \left (a x\right )\right )}}{x^{2}} \,d x } \]

[In]

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

[Out]

integrate(e^(I*n*arctan(a*x))/x^2, x)

Giac [F]

\[ \int \frac {e^{i n \arctan (a x)}}{x^2} \, dx=\int { \frac {e^{\left (i \, n \arctan \left (a x\right )\right )}}{x^{2}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{i n \arctan (a x)}}{x^2} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )\,1{}\mathrm {i}}}{x^2} \,d x \]

[In]

int(exp(n*atan(a*x)*1i)/x^2,x)

[Out]

int(exp(n*atan(a*x)*1i)/x^2, x)