### 3.363 $$\int e^{n \coth ^{-1}(a x)} (c-a c x)^{-2+\frac{n}{2}} \, dx$$

Optimal. Leaf size=88 $-\frac{2 x \left (1-\frac{1}{a x}\right )^{2-\frac{n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n-2}{2}} (c-a c x)^{\frac{n-4}{2}} \text{Hypergeometric2F1}\left (2,1-\frac{n}{2},2-\frac{n}{2},\frac{2}{x \left (a+\frac{1}{x}\right )}\right )}{2-n}$

[Out]

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

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Rubi [A]  time = 0.147318, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {6176, 6181, 131} $-\frac{2 x \left (1-\frac{1}{a x}\right )^{2-\frac{n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n-2}{2}} (c-a c x)^{\frac{n-4}{2}} \, _2F_1\left (2,1-\frac{n}{2};2-\frac{n}{2};\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{2-n}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 131

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)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

\begin{align*} \int e^{n \coth ^{-1}(a x)} (c-a c x)^{-2+\frac{n}{2}} \, dx &=\left (\left (1-\frac{1}{a x}\right )^{2-\frac{n}{2}} x^{2-\frac{n}{2}} (c-a c x)^{-2+\frac{n}{2}}\right ) \int e^{n \coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right )^{-2+\frac{n}{2}} x^{-2+\frac{n}{2}} \, dx\\ &=-\left (\left (\left (1-\frac{1}{a x}\right )^{2-\frac{n}{2}} \left (\frac{1}{x}\right )^{-2+\frac{n}{2}} (c-a c x)^{-2+\frac{n}{2}}\right ) \operatorname{Subst}\left (\int \frac{x^{-n/2} \left (1+\frac{x}{a}\right )^{n/2}}{\left (1-\frac{x}{a}\right )^2} \, dx,x,\frac{1}{x}\right )\right )\\ &=-\frac{2 \left (1-\frac{1}{a x}\right )^{2-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{1}{2} (-2+n)} x (c-a c x)^{\frac{1}{2} (-4+n)} \, _2F_1\left (2,1-\frac{n}{2};2-\frac{n}{2};\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{2-n}\\ \end{align*}

Mathematica [A]  time = 0.0390477, size = 89, normalized size = 1.01 $\frac{2 \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{n/2} (c-a c x)^{n/2} \text{Hypergeometric2F1}\left (2,1-\frac{n}{2},2-\frac{n}{2},\frac{2}{a x+1}\right )}{a c^2 (n-2) (a x+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 0.352, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\rm arccoth} \left (ax\right )}} \left ( -acx+c \right ) ^{-2+{\frac{n}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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