\(\int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x^m \, dx\) [489]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 60 \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x^m \, dx=\frac {\sqrt {c-\frac {c}{a x}} x^{1+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-1-m,-m,-\frac {1}{a x}\right )}{(1+m) \sqrt {1-\frac {1}{a x}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6317, 6316, 66} \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x^m \, dx=\frac {x^{m+1} \sqrt {c-\frac {c}{a x}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m-1,-m,-\frac {1}{a x}\right )}{(m+1) \sqrt {1-\frac {1}{a x}}} \]

[In]

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

[Out]

(Sqrt[c - c/(a*x)]*x^(1 + m)*Hypergeometric2F1[-1/2, -1 - m, -m, -(1/(a*x))])/((1 + m)*Sqrt[1 - 1/(a*x)])

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 6316

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

Rule 6317

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x^m \, dx=\frac {\sqrt {c-\frac {c}{a x}} x^{1+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-1-m,-m,-\frac {1}{a x}\right )}{(1+m) \sqrt {1-\frac {1}{a x}}} \]

[In]

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

[Out]

(Sqrt[c - c/(a*x)]*x^(1 + m)*Hypergeometric2F1[-1/2, -1 - m, -m, -(1/(a*x))])/((1 + m)*Sqrt[1 - 1/(a*x)])

Maple [F]

\[\int \frac {x^{m} \sqrt {c -\frac {c}{a x}}}{\sqrt {\frac {a x -1}{a x +1}}}d x\]

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)*x^m*(c-c/a/x)^(1/2),x)

[Out]

int(1/((a*x-1)/(a*x+1))^(1/2)*x^m*(c-c/a/x)^(1/2),x)

Fricas [F]

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

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^m*(c-c/a/x)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x^m \, dx=\text {Timed out} \]

[In]

integrate(1/((a*x-1)/(a*x+1))**(1/2)*x**m*(c-c/a/x)**(1/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^m*(c-c/a/x)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^m*(c-c/a/x)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x^m \, dx=\int \frac {x^m\,\sqrt {c-\frac {c}{a\,x}}}{\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

[In]

int((x^m*(c - c/(a*x))^(1/2))/((a*x - 1)/(a*x + 1))^(1/2),x)

[Out]

int((x^m*(c - c/(a*x))^(1/2))/((a*x - 1)/(a*x + 1))^(1/2), x)