\(\int \frac {e^{n \coth ^{-1}(a x)}}{x^2} \, dx\) [153]

Optimal result

Integrand size = 12, antiderivative size = 70 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^2} \, dx=\frac {2^{1+\frac {n}{2}} a \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {a-\frac {1}{x}}{2 a}\right )}{2-n} \]

[Out]

2^(1+1/2*n)*a*(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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6306, 71} \[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^2} \, dx=\frac {a 2^{\frac {n}{2}+1} \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {a-\frac {1}{x}}{2 a}\right )}{2-n} \]

[In]

Int[E^(n*ArcCoth[a*x])/x^2,x]

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 6306

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*} \text {integral}& = -\text {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 {2^{1+\frac {n}{2}} a \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {a-\frac {1}{x}}{2 a}\right )}{2-n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.63 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {4 a e^{(2+n) \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (2,1+\frac {n}{2},2+\frac {n}{2},-e^{2 \coth ^{-1}(a x)}\right )}{2+n} \]

[In]

Integrate[E^(n*ArcCoth[a*x])/x^2,x]

[Out]

(-4*a*E^((2 + n)*ArcCoth[a*x])*Hypergeometric2F1[2, 1 + n/2, 2 + n/2, -E^(2*ArcCoth[a*x])])/(2 + n)

Maple [F]

\[\int \frac {{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}}{x^{2}}d x\]

[In]

int(exp(n*arccoth(a*x))/x^2,x)

[Out]

int(exp(n*arccoth(a*x))/x^2,x)

Fricas [F]

\[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^2} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{x^{2}} \,d x } \]

[In]

integrate(exp(n*arccoth(a*x))/x^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^2} \, dx=\int \frac {e^{n \operatorname {acoth}{\left (a x \right )}}}{x^{2}}\, dx \]

[In]

integrate(exp(n*acoth(a*x))/x**2,x)

[Out]

Integral(exp(n*acoth(a*x))/x**2, x)

Maxima [F]

\[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^2} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{x^{2}} \,d x } \]

[In]

integrate(exp(n*arccoth(a*x))/x^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^2} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{x^{2}} \,d x } \]

[In]

integrate(exp(n*arccoth(a*x))/x^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^2} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}}{x^2} \,d x \]

[In]

int(exp(n*acoth(a*x))/x^2,x)

[Out]

int(exp(n*acoth(a*x))/x^2, x)