\(\int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx\) [932]

Optimal result

Integrand size = 24, antiderivative size = 183 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1+n}{2}} x}{\sqrt {c-\frac {c}{a^2 x^2}}}+\frac {2 n \sqrt {1-\frac {1}{a^2 x^2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1}{2} (-1+n)} \operatorname {Hypergeometric2F1}\left (1,\frac {1-n}{2},\frac {3-n}{2},\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a (1-n) \sqrt {c-\frac {c}{a^2 x^2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6332, 6329, 98, 133} \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=\frac {2 n \sqrt {1-\frac {1}{a^2 x^2}} \left (\frac {1}{a x}+1\right )^{\frac {n-1}{2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \operatorname {Hypergeometric2F1}\left (1,\frac {1-n}{2},\frac {3-n}{2},\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a (1-n) \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left (\frac {1}{a x}+1\right )^{\frac {n+1}{2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}}}{\sqrt {c-\frac {c}{a^2 x^2}}} \]

[In]

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

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*(1 - 1/(a*x))^((1 - n)/2)*(1 + 1/(a*x))^((1 + n)/2)*x)/Sqrt[c - c/(a^2*x^2)] + (2*n*Sqr
t[1 - 1/(a^2*x^2)]*(1 - 1/(a*x))^((1 - n)/2)*(1 + 1/(a*x))^((-1 + n)/2)*Hypergeometric2F1[1, (1 - n)/2, (3 - n
)/2, (a - x^(-1))/(a + x^(-1))])/(a*(1 - n)*Sqrt[c - c/(a^2*x^2)])

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 133

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

Rule 6329

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

Rule 6332

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1-\frac {1}{a^2 x^2}} \int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {1-\frac {1}{a^2 x^2}}} \, dx}{\sqrt {c-\frac {c}{a^2 x^2}}} \\ & = -\frac {\sqrt {1-\frac {1}{a^2 x^2}} \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{-\frac {1}{2}+\frac {n}{2}}}{x^2} \, dx,x,\frac {1}{x}\right )}{\sqrt {c-\frac {c}{a^2 x^2}}} \\ & = \frac {\sqrt {1-\frac {1}{a^2 x^2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1+n}{2}} x}{\sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\left (n \sqrt {1-\frac {1}{a^2 x^2}}\right ) \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{-\frac {1}{2}+\frac {n}{2}}}{x} \, dx,x,\frac {1}{x}\right )}{a \sqrt {c-\frac {c}{a^2 x^2}}} \\ & = \frac {\sqrt {1-\frac {1}{a^2 x^2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1+n}{2}} x}{\sqrt {c-\frac {c}{a^2 x^2}}}+\frac {2 n \sqrt {1-\frac {1}{a^2 x^2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1}{2} (-1+n)} \operatorname {Hypergeometric2F1}\left (1,\frac {1-n}{2},\frac {3-n}{2},\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a (1-n) \sqrt {c-\frac {c}{a^2 x^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.61 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=\frac {e^{n \coth ^{-1}(a x)} \left (-1+a^2 x^2\right ) \left (a (1+n) \sqrt {1-\frac {1}{a^2 x^2}} x+2 e^{\coth ^{-1}(a x)} n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},e^{2 \coth ^{-1}(a x)}\right )\right )}{a^3 (1+n) \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a^2 x^2}} x^2} \]

[In]

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

[Out]

(E^(n*ArcCoth[a*x])*(-1 + a^2*x^2)*(a*(1 + n)*Sqrt[1 - 1/(a^2*x^2)]*x + 2*E^ArcCoth[a*x]*n*Hypergeometric2F1[1
, (1 + n)/2, (3 + n)/2, E^(2*ArcCoth[a*x])]))/(a^3*(1 + n)*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a^2*x^2)]*x^2)

Maple [F]

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

[In]

int(exp(n*arccoth(a*x))/(c-c/a^2/x^2)^(1/2),x)

[Out]

int(exp(n*arccoth(a*x))/(c-c/a^2/x^2)^(1/2),x)

Fricas [F]

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

[In]

integrate(exp(n*arccoth(a*x))/(c-c/a^2/x^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(exp(n*acoth(a*x))/(c-c/a**2/x**2)**(1/2),x)

[Out]

Integral(exp(n*acoth(a*x))/sqrt(-c*(-1 + 1/(a*x))*(1 + 1/(a*x))), x)

Maxima [F]

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

[In]

integrate(exp(n*arccoth(a*x))/(c-c/a^2/x^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(exp(n*arccoth(a*x))/(c-c/a^2/x^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(exp(n*acoth(a*x))/(c - c/(a^2*x^2))^(1/2),x)

[Out]

int(exp(n*acoth(a*x))/(c - c/(a^2*x^2))^(1/2), x)