### 3.375 $$\int \frac{e^{n \coth ^{-1}(a x)}}{\sqrt{c-a c x}} \, dx$$

Optimal. Leaf size=96 $\frac{2 x \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{n+1}{2}} \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{n+1}{2},\frac{1}{2},\frac{2}{x \left (a+\frac{1}{x}\right )}\right )}{\sqrt{c-a c x}}$

[Out]

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

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Rubi [A]  time = 0.179679, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {6176, 6181, 132} $\frac{2 x \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{n+1}{2}} \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \, _2F_1\left (-\frac{1}{2},\frac{n+1}{2};\frac{1}{2};\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{\sqrt{c-a c x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6176

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

Rule 6181

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> -Dist[c^p*x^m*(1/x)^m, Sub
st[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 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]

Rubi steps

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

Mathematica [A]  time = 0.0534178, size = 96, normalized size = 1. $\frac{2 (a x+1) \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{n/2} \left (\frac{a x-1}{a x+1}\right )^{\frac{n+1}{2}} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{n+1}{2},\frac{1}{2},\frac{2}{a x+1}\right )}{a \sqrt{c-a c x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 0.341, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{n{\rm arccoth} \left (ax\right )}}{\frac{1}{\sqrt{-acx+c}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{\sqrt{-a c x + c}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a c x + c} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{a c x - c}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{n \operatorname{acoth}{\left (a x \right )}}}{\sqrt{- c \left (a x - 1\right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{\sqrt{-a c x + c}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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