3.2.61 \(\int \frac {e^{i n \text {ArcTan}(a x)}}{x^4} \, dx\) [161]

Optimal. Leaf size=171 \[ -\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{3 x^3}-\frac {i a n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{6 x^2}+\frac {2 i a^3 \left (2+n^2\right ) (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {1}{2} (-2+n)} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {1-i a x}{1+i a x}\right )}{3 (2-n)} \]

[Out]

-1/3*(1-I*a*x)^(1-1/2*n)*(1+I*a*x)^(1+1/2*n)/x^3-1/6*I*a*n*(1-I*a*x)^(1-1/2*n)*(1+I*a*x)^(1+1/2*n)/x^2+2/3*I*a
^3*(n^2+2)*(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 = 171, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5170, 105, 156, 12, 133} \begin {gather*} \frac {2 i a^3 \left (n^2+2\right ) (1+i a x)^{\frac {n-2}{2}} (1-i a x)^{1-\frac {n}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {1-i a x}{i a x+1}\right )}{3 (2-n)}-\frac {(1+i a x)^{\frac {n+2}{2}} (1-i a x)^{1-\frac {n}{2}}}{3 x^3}-\frac {i a n (1+i a x)^{\frac {n+2}{2}} (1-i a x)^{1-\frac {n}{2}}}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(I*n*ArcTan[a*x])/x^4,x]

[Out]

-1/3*((1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((2 + n)/2))/x^3 - ((I/6)*a*n*(1 - I*a*x)^(1 - n/2)*(1 + I*a*x)^((2 +
n)/2))/x^2 + (((2*I)/3)*a^3*(2 + n^2)*(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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 105

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[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*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
 c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

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 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

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^4} \, dx &=\int \frac {(1-i a x)^{-n/2} (1+i a x)^{n/2}}{x^4} \, dx\\ &=-\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{3 x^3}-\frac {1}{3} \int \frac {(1-i a x)^{-n/2} (1+i a x)^{n/2} \left (-i a n+a^2 x\right )}{x^3} \, dx\\ &=-\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{3 x^3}-\frac {i a n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{6 x^2}-\frac {1}{6} \int \frac {a^2 \left (2+n^2\right ) (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}}}{3 x^3}-\frac {i a n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{6 x^2}-\frac {1}{6} \left (a^2 \left (2+n^2\right )\right ) \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}}}{3 x^3}-\frac {i a n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{6 x^2}+\frac {2 i a^3 \left (2+n^2\right ) (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 )}{3 (2-n)}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 119, normalized size = 0.70 \begin {gather*} -\frac {(1-i a x)^{-n/2} (1+i a x)^{\frac {1}{2} (-2+n)} (i+a x) \left (-\left ((-2+n) (-i+a x)^2 (-2 i+a n x)\right )+4 a^3 \left (2+n^2\right ) x^3 \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {i+a x}{i-a x}\right )\right )}{6 (-2+n) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(I*n*ArcTan[a*x])/x^4,x]

[Out]

-1/6*((1 + I*a*x)^((-2 + n)/2)*(I + a*x)*(-((-2 + n)*(-I + a*x)^2*(-2*I + a*n*x)) + 4*a^3*(2 + n^2)*x^3*Hyperg
eometric2F1[2, 1 - n/2, 2 - n/2, (I + a*x)/(I - a*x)]))/((-2 + n)*x^3*(1 - I*a*x)^(n/2))

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{i n \arctan \left (a x \right )}}{x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(I*n*arctan(a*x))/x^4,x)

[Out]

int(exp(I*n*arctan(a*x))/x^4,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^4,x, algorithm="maxima")

[Out]

integrate(e^(I*n*arctan(a*x))/x^4, 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^4,x, algorithm="fricas")

[Out]

integral(1/(x^4*(-(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^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(I*n*atan(a*x))/x**4,x)

[Out]

Integral(exp(I*n*atan(a*x))/x**4, 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^4,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^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atan(a*x)*1i)/x^4,x)

[Out]

int(exp(n*atan(a*x)*1i)/x^4, x)

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