3.1.9 \(\int \frac {e^{n \cot ^{-1}(a x)}}{(c+a^2 c x^2)^{2/3}} \, dx\) [9]

Optimal. Leaf size=147 \[ -\frac {3 \left (1+\frac {1}{a^2 x^2}\right )^{2/3} \left (\frac {a-\frac {i}{x}}{a+\frac {i}{x}}\right )^{\frac {1}{6} (4-3 i n)} \left (1-\frac {i}{a x}\right )^{\frac {1}{6} (-4+3 i n)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (2-3 i n)} x \, _2F_1\left (\frac {1}{3},\frac {1}{6} (4-3 i n);\frac {4}{3};\frac {2 i}{\left (a+\frac {i}{x}\right ) x}\right )}{\left (c+a^2 c x^2\right )^{2/3}} \]

[Out]

-3*(1+1/a^2/x^2)^(2/3)*((a-I/x)/(a+I/x))^(2/3-1/2*I*n)*(1-I/a/x)^(-2/3+1/2*I*n)*(1+I/a/x)^(1/3-1/2*I*n)*x*hype
rgeom([1/3, 2/3-1/2*I*n],[4/3],2*I/(a+I/x)/x)/(a^2*c*x^2+c)^(2/3)

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Rubi [A]
time = 0.15, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5230, 5234, 134} \begin {gather*} -\frac {3 x \left (\frac {1}{a^2 x^2}+1\right )^{2/3} \left (\frac {a-\frac {i}{x}}{a+\frac {i}{x}}\right )^{\frac {1}{6} (4-3 i n)} \left (1-\frac {i}{a x}\right )^{\frac {1}{6} (-4+3 i n)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (2-3 i n)} \, _2F_1\left (\frac {1}{3},\frac {1}{6} (4-3 i n);\frac {4}{3};\frac {2 i}{\left (a+\frac {i}{x}\right ) x}\right )}{\left (a^2 c x^2+c\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcCot[a*x])/(c + a^2*c*x^2)^(2/3),x]

[Out]

(-3*(1 + 1/(a^2*x^2))^(2/3)*((a - I/x)/(a + I/x))^((4 - (3*I)*n)/6)*(1 - I/(a*x))^((-4 + (3*I)*n)/6)*(1 + I/(a
*x))^((2 - (3*I)*n)/6)*x*Hypergeometric2F1[1/3, (4 - (3*I)*n)/6, 4/3, (2*I)/((a + I/x)*x)])/(c + a^2*c*x^2)^(2
/3)

Rule 134

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

Rule 5230

Int[E^(ArcCot[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c + d*x^2)^p/(x^(2*p)*(1
+ 1/(a^2*x^2))^p), Int[u*x^(2*p)*(1 + 1/(a^2*x^2))^p*E^(n*ArcCot[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] &
& EqQ[d, a^2*c] &&  !IntegerQ[I*(n/2)] &&  !IntegerQ[p]

Rule 5234

Int[E^(ArcCot[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_)^2)^(p_.)*(x_)^(m_), x_Symbol] :> Dist[(-c^p)*x^m*(1/x)^m,
Subst[Int[(1 - I*(x/a))^(p + I*(n/2))*((1 + I*(x/a))^(p - I*(n/2))/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, c,
d, m, n, p}, x] && EqQ[c, a^2*d] &&  !IntegerQ[I*(n/2)] && (IntegerQ[p] || GtQ[c, 0]) &&  !(IntegerQ[2*p] &&
IntegerQ[p + I*(n/2)]) &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{2/3}} \, dx &=\frac {\left (\left (1+\frac {1}{a^2 x^2}\right )^{2/3} x^{4/3}\right ) \int \frac {e^{n \cot ^{-1}(a x)}}{\left (1+\frac {1}{a^2 x^2}\right )^{2/3} x^{4/3}} \, dx}{\left (c+a^2 c x^2\right )^{2/3}}\\ &=-\frac {\left (1+\frac {1}{a^2 x^2}\right )^{2/3} \text {Subst}\left (\int \frac {\left (1-\frac {i x}{a}\right )^{-\frac {2}{3}+\frac {i n}{2}} \left (1+\frac {i x}{a}\right )^{-\frac {2}{3}-\frac {i n}{2}}}{x^{2/3}} \, dx,x,\frac {1}{x}\right )}{\left (\frac {1}{x}\right )^{4/3} \left (c+a^2 c x^2\right )^{2/3}}\\ &=-\frac {3 \left (1+\frac {1}{a^2 x^2}\right )^{2/3} \left (\frac {a-\frac {i}{x}}{a+\frac {i}{x}}\right )^{\frac {1}{6} (4-3 i n)} \left (1-\frac {i}{a x}\right )^{\frac {1}{6} (-4+3 i n)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (2-3 i n)} x \, _2F_1\left (\frac {1}{3},\frac {1}{6} (4-3 i n);\frac {4}{3};\frac {2 i}{\left (a+\frac {i}{x}\right ) x}\right )}{\left (c+a^2 c x^2\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 89, normalized size = 0.61 \begin {gather*} -\frac {3 e^{(-2 i+n) \cot ^{-1}(a x)} \left (-1+e^{2 i \cot ^{-1}(a x)}\right ) \sqrt [3]{c+a^2 c x^2} \, _2F_1\left (1,\frac {2}{3}+\frac {i n}{2};\frac {4}{3}+\frac {i n}{2};e^{-2 i \cot ^{-1}(a x)}\right )}{a c (-2 i+3 n)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(n*ArcCot[a*x])/(c + a^2*c*x^2)^(2/3),x]

[Out]

(-3*E^((-2*I + n)*ArcCot[a*x])*(-1 + E^((2*I)*ArcCot[a*x]))*(c + a^2*c*x^2)^(1/3)*Hypergeometric2F1[1, 2/3 + (
I/2)*n, 4/3 + (I/2)*n, E^((-2*I)*ArcCot[a*x])])/(a*c*(-2*I + 3*n))

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{n \,\mathrm {arccot}\left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(2/3),x)

[Out]

int(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(2/3),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(n*arccot(a*x))/(a^2*c*x^2+c)^(2/3),x, algorithm="maxima")

[Out]

integrate(e^(n*arccot(a*x))/(a^2*c*x^2 + c)^(2/3), 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(n*arccot(a*x))/(a^2*c*x^2+c)^(2/3),x, algorithm="fricas")

[Out]

integral(e^(n*arccot(a*x))/(a^2*c*x^2 + c)^(2/3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{n \operatorname {acot}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {2}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acot(a*x))/(a**2*c*x**2+c)**(2/3),x)

[Out]

Integral(exp(n*acot(a*x))/(c*(a**2*x**2 + 1))**(2/3), 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(n*arccot(a*x))/(a^2*c*x^2+c)^(2/3),x, algorithm="giac")

[Out]

integrate(e^(n*arccot(a*x))/(a^2*c*x^2 + c)^(2/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {e}}^{n\,\mathrm {acot}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*acot(a*x))/(c + a^2*c*x^2)^(2/3),x)

[Out]

int(exp(n*acot(a*x))/(c + a^2*c*x^2)^(2/3), x)

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