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

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

[Out]

3*(1+1/a^2/x^2)^(1/3)*((a-I/x)/(a+I/x))^(1/3-1/2*I*n)*(1-I/a/x)^(-1/3+1/2*I*n)*(1+I/a/x)^(2/3-1/2*I*n)*x*hyper
geom([-1/3, 1/3-1/2*I*n],[2/3],2*I/(a+I/x)/x)/(a^2*c*x^2+c)^(1/3)

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Rubi [A]  time = 0.18, 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 = {5122, 5126, 132} \[ \frac {3 x \sqrt [3]{\frac {1}{a^2 x^2}+1} \left (\frac {a-\frac {i}{x}}{a+\frac {i}{x}}\right )^{\frac {1}{6} (2-3 i n)} \left (1-\frac {i}{a x}\right )^{\frac {1}{6} (-2+3 i n)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (4-3 i n)} \, _2F_1\left (-\frac {1}{3},\frac {1}{6} (2-3 i n);\frac {2}{3};\frac {2 i}{\left (a+\frac {i}{x}\right ) x}\right )}{\sqrt [3]{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcCot[a*x])/(c + a^2*c*x^2)^(1/3),x]

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 89, normalized size = 0.61 \[ -\frac {3 \left (a^2 c x^2+c\right )^{2/3} \left (-1+e^{2 i \cot ^{-1}(a x)}\right ) e^{(n-2 i) \cot ^{-1}(a x)} \, _2F_1\left (1,\frac {i n}{2}+\frac {1}{3};\frac {i n}{2}+\frac {5}{3};e^{-2 i \cot ^{-1}(a x)}\right )}{a c (3 n-4 i)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(n*ArcCot[a*x])/(c + a^2*c*x^2)^(1/3),x]

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{\left (n \operatorname {arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {1}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(1/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\left (n \operatorname {arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \,\mathrm {arccot}\left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(1/3),x)

[Out]

int(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(1/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\left (n \operatorname {arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(1/3),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*acot(a*x))/(c + a^2*c*x^2)^(1/3),x)

[Out]

int(exp(n*acot(a*x))/(c + a^2*c*x^2)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acot(a*x))/(a**2*c*x**2+c)**(1/3),x)

[Out]

Integral(exp(n*acot(a*x))/(c*(a**2*x**2 + 1))**(1/3), x)

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