3.160 \(\int \frac {e^{i n \tan ^{-1}(a x)}}{x^3} \, dx\)

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

[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 = 0.05, 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 = {5062, 96, 131} \[ \frac {2 a^2 n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {n-2}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {1-i a x}{i a x+1}\right )}{2-n}-\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {n+2}{2}}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((2 + n)/2))/(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 96

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 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]

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]

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 \[ \frac {(a x+i) (1-i a x)^{-n/2} (1+i a x)^{n/2} \left (4 a^2 n x^2 \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {a x+i}{i-a x}\right )-(n-2) (a x-i)^2\right )}{2 (n-2) x^2 (a x-i)} \]

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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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{x^{3} \left (-\frac {a x + i}{a x - i}\right )^{\frac {1}{2} \, n}}, x\right ) \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.16, 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 \[ \int \frac {e^{\left (i \, n \arctan \left (a x\right )\right )}}{x^{3}}\,{d x} \]

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

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

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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