3.11 \(\int \frac{e^{n \cot ^{-1}(a x)}}{(c+a^2 c x^2)^{5/3}} \, dx\)

Optimal. Leaf size=207 \[ -\frac{3 (3 n-4 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}}-\frac{12 x \left (\frac{1}{a^2 x^2}+1\right )^{2/3} \left (\frac{a-\frac{i}{x}}{a+\frac{i}{x}}\right )^{\frac{1}{6} (4-3 i n)} \left (1-\frac{i}{a x}\right )^{\frac{1}{6} (-4+3 i n)} \left (1+\frac{i}{a x}\right )^{\frac{1}{6} (2-3 i n)} \, _2F_1\left (\frac{1}{3},\frac{1}{6} (4-3 i n);\frac{4}{3};\frac{2 i}{\left (a+\frac{i}{x}\right ) x}\right )}{c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}} \]

[Out]

(-3*E^(n*ArcCot[a*x])*(3*n - 4*a*x))/(a*c*(16 + 9*n^2)*(c + a^2*c*x^2)^(2/3)) - (12*(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*Hyper
geometric2F1[1/3, (4 - (3*I)*n)/6, 4/3, (2*I)/((a + I/x)*x)])/(c*(16 + 9*n^2)*(c + a^2*c*x^2)^(2/3))

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Rubi [A]  time = 0.249964, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {5115, 5122, 5126, 132} \[ -\frac{3 (3 n-4 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}}-\frac{12 x \left (\frac{1}{a^2 x^2}+1\right )^{2/3} \left (\frac{a-\frac{i}{x}}{a+\frac{i}{x}}\right )^{\frac{1}{6} (4-3 i n)} \left (1-\frac{i}{a x}\right )^{\frac{1}{6} (-4+3 i n)} \left (1+\frac{i}{a x}\right )^{\frac{1}{6} (2-3 i n)} \, _2F_1\left (\frac{1}{3},\frac{1}{6} (4-3 i n);\frac{4}{3};\frac{2 i}{\left (a+\frac{i}{x}\right ) x}\right )}{c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcCot[a*x])/(c + a^2*c*x^2)^(5/3),x]

[Out]

(-3*E^(n*ArcCot[a*x])*(3*n - 4*a*x))/(a*c*(16 + 9*n^2)*(c + a^2*c*x^2)^(2/3)) - (12*(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*Hyper
geometric2F1[1/3, (4 - (3*I)*n)/6, 4/3, (2*I)/((a + I/x)*x)])/(c*(16 + 9*n^2)*(c + a^2*c*x^2)^(2/3))

Rule 5115

Int[E^(ArcCot[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((n + 2*a*(p + 1)*x)*(c + d*x^
2)^(p + 1)*E^(n*ArcCot[a*x]))/(a*c*(n^2 + 4*(p + 1)^2)), x] + Dist[(2*(p + 1)*(2*p + 3))/(c*(n^2 + 4*(p + 1)^2
)), Int[(c + d*x^2)^(p + 1)*E^(n*ArcCot[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] && LtQ[p, -1]
&& NeQ[p, -3/2] && NeQ[n^2 + 4*(p + 1)^2, 0] &&  !(IntegerQ[p] && IntegerQ[(I*n)/2]) &&  !( !IntegerQ[p] && I
ntegerQ[(I*n - 1)/2])

Rule 5122

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 5126

Int[E^(ArcCot[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_)^2)^(p_.)*(x_)^(m_), x_Symbol] :> -Dist[c^p*x^m*(1/x)^m, Su
bst[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] && In
tegerQ[p + (I*n)/2]) &&  !IntegerQ[m]

Rule 132

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)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((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]

Rubi steps

\begin{align*} \int \frac{e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/3}} \, dx &=-\frac{3 e^{n \cot ^{-1}(a x)} (3 n-4 a x)}{a c \left (16+9 n^2\right ) \left (c+a^2 c x^2\right )^{2/3}}+\frac{4 \int \frac{e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{2/3}} \, dx}{c \left (16+9 n^2\right )}\\ &=-\frac{3 e^{n \cot ^{-1}(a x)} (3 n-4 a x)}{a c \left (16+9 n^2\right ) \left (c+a^2 c x^2\right )^{2/3}}+\frac{\left (4 \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}{c \left (16+9 n^2\right ) \left (c+a^2 c x^2\right )^{2/3}}\\ &=-\frac{3 e^{n \cot ^{-1}(a x)} (3 n-4 a x)}{a c \left (16+9 n^2\right ) \left (c+a^2 c x^2\right )^{2/3}}-\frac{\left (4 \left (1+\frac{1}{a^2 x^2}\right )^{2/3}\right ) \operatorname{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 )}{c \left (16+9 n^2\right ) \left (\frac{1}{x}\right )^{4/3} \left (c+a^2 c x^2\right )^{2/3}}\\ &=-\frac{3 e^{n \cot ^{-1}(a x)} (3 n-4 a x)}{a c \left (16+9 n^2\right ) \left (c+a^2 c x^2\right )^{2/3}}-\frac{12 \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 )}{c \left (16+9 n^2\right ) \left (c+a^2 c x^2\right )^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.179111, size = 89, normalized size = 0.43 \[ -\frac{3 \left (-1+e^{2 i \cot ^{-1}(a x)}\right ) e^{(n-2 i) \cot ^{-1}(a x)} \, _2F_1\left (1,\frac{i n}{2}+\frac{5}{3};\frac{i n}{2}+\frac{1}{3};e^{-2 i \cot ^{-1}(a x)}\right )}{a c (3 n+4 i) \left (a^2 c x^2+c\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(n*ArcCot[a*x])/(c + a^2*c*x^2)^(5/3),x]

[Out]

(-3*E^((-2*I + n)*ArcCot[a*x])*(-1 + E^((2*I)*ArcCot[a*x]))*Hypergeometric2F1[1, 5/3 + (I/2)*n, 1/3 + (I/2)*n,
E^((-2*I)*ArcCot[a*x])])/(a*c*(4*I + 3*n)*(c + a^2*c*x^2)^(2/3))

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Maple [F]  time = 0.293, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{n{\rm arccot} \left (ax\right )}} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x)

[Out]

int(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (n \operatorname{arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x, algorithm="maxima")

[Out]

integrate(e^(n*arccot(a*x))/(a^2*c*x^2 + c)^(5/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{2} c x^{2} + c\right )}^{\frac{1}{3}} e^{\left (n \operatorname{arccot}\left (a x\right )\right )}}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x, algorithm="fricas")

[Out]

integral((a^2*c*x^2 + c)^(1/3)*e^(n*arccot(a*x))/(a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acot(a*x))/(a**2*c*x**2+c)**(5/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (n \operatorname{arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x, algorithm="giac")

[Out]

integrate(e^(n*arccot(a*x))/(a^2*c*x^2 + c)^(5/3), x)