3.154 \(\int \frac {e^{n \coth ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=114 \[ \frac {a^2 2^{n/2} n \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {a-\frac {1}{x}}{2 a}\right )}{2-n}+\frac {1}{2} a^2 \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \]

[Out]

1/2*a^2*(1-1/a/x)^(1-1/2*n)*(1+1/a/x)^(1+1/2*n)+2^(1/2*n)*a^2*n*(1-1/a/x)^(1-1/2*n)*hypergeom([-1/2*n, 1-1/2*n
],[2-1/2*n],1/2*(a-1/x)/a)/(2-n)

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Rubi [A]  time = 0.06, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6171, 80, 69} \[ \frac {a^2 2^{n/2} n \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {a-\frac {1}{x}}{2 a}\right )}{2-n}+\frac {1}{2} a^2 \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 - 1/(a*x))^(1 - n/2)*(1 + 1/(a*x))^((2 + n)/2))/2 + (2^(n/2)*a^2*n*(1 - 1/(a*x))^(1 - n/2)*Hypergeomet
ric2F1[1 - n/2, -n/2, 2 - n/2, (a - x^(-1))/(2*a)])/(2 - n)

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 80

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

Rule 6171

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x/a)^(n/2)/(x^(m + 2)*(1 - x/a)^(n/2
)), x], x, 1/x] /; FreeQ[{a, n}, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {e^{n \coth ^{-1}(a x)}}{x^3} \, dx &=-\operatorname {Subst}\left (\int x \left (1-\frac {x}{a}\right )^{-n/2} \left (1+\frac {x}{a}\right )^{n/2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}}-\frac {1}{2} (a n) \operatorname {Subst}\left (\int \left (1-\frac {x}{a}\right )^{-n/2} \left (1+\frac {x}{a}\right )^{n/2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}}+\frac {2^{n/2} a^2 n \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {a-\frac {1}{x}}{2 a}\right )}{2-n}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 107, normalized size = 0.94 \[ -\frac {a^2 e^{n \coth ^{-1}(a x)} \left ((n+2) \left (\frac {1}{a^2 x^2}+n \, _2F_1\left (1,\frac {n}{2};\frac {n}{2}+1;-e^{2 \coth ^{-1}(a x)}\right )+\frac {n}{a x}-1\right )-n^2 e^{2 \coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n}{2}+1;\frac {n}{2}+2;-e^{2 \coth ^{-1}(a x)}\right )\right )}{2 (n+2)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(n*ArcCoth[a*x])/x^3,x]

[Out]

-1/2*(a^2*E^(n*ArcCoth[a*x])*(-(E^(2*ArcCoth[a*x])*n^2*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, -E^(2*ArcCoth[a*
x])]) + (2 + n)*(-1 + 1/(a^2*x^2) + n/(a*x) + n*Hypergeometric2F1[1, n/2, 1 + n/2, -E^(2*ArcCoth[a*x])])))/(2
+ n)

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/x^3,x, algorithm="fricas")

[Out]

integral(((a*x - 1)/(a*x + 1))^(1/2*n)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((a*x - 1)/(a*x + 1))^(1/2*n)/x^3, x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))/x^3,x)

[Out]

int(exp(n*arccoth(a*x))/x^3,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/x^3,x, algorithm="maxima")

[Out]

integrate(((a*x - 1)/(a*x + 1))^(1/2*n)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*acoth(a*x))/x^3,x)

[Out]

int(exp(n*acoth(a*x))/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acoth(a*x))/x**3,x)

[Out]

Integral(exp(n*acoth(a*x))/x**3, x)

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