3.2.60 \(\int \frac {e^{i n \text {ArcTan}(a x)}}{x^3} \, dx\) [160]

Optimal. Leaf size=120 \[ -\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{2 x^2}+\frac {2 a^2 n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {1}{2} (-2+n)} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {1-i a x}{1+i a x}\right )}{2-n} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5170, 98, 133} \begin {gather*} \frac {2 a^2 n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {n-2}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {1-i a x}{i a x+1}\right )}{2-n}-\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {n+2}{2}}}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(I*n*ArcTan[a*x])/x^3,x]

[Out]

-1/2*((1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((2 + n)/2))/x^2 + (2*a^2*n*(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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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

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Mathematica [A]
time = 0.03, size = 114, normalized size = 0.95 \begin {gather*} \frac {(1-i a x)^{-n/2} (1+i a x)^{n/2} (i+a x) \left (-\left ((-2+n) (-i+a x)^2\right )+4 a^2 n x^2 \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {i+a x}{i-a x}\right )\right )}{2 (-2+n) x^2 (-i+a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(I*n*ArcTan[a*x])/x^3,x]

[Out]

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

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{i n \arctan \left (a x \right )}}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(I*n*arctan(a*x))/x^3,x)

[Out]

int(exp(I*n*arctan(a*x))/x^3,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(I*n*arctan(a*x))/x^3,x, algorithm="maxima")

[Out]

integrate(e^(I*n*arctan(a*x))/x^3, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(I*n*arctan(a*x))/x^3,x, algorithm="fricas")

[Out]

integral(1/(x^3*(-(a*x + I)/(a*x - I))^(1/2*n)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{i n \operatorname {atan}{\left (a x \right )}}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(I*n*atan(a*x))/x**3,x)

[Out]

Integral(exp(I*n*atan(a*x))/x**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(I*n*arctan(a*x))/x^3,x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )\,1{}\mathrm {i}}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atan(a*x)*1i)/x^3,x)

[Out]

int(exp(n*atan(a*x)*1i)/x^3, x)

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