### 3.372 $$\int e^{n \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx$$

Optimal. Leaf size=98 $\frac{2}{7} x (c-a c x)^{5/2} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{n-5}{2}} \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \text{Hypergeometric2F1}\left (-\frac{7}{2},\frac{n-5}{2},-\frac{5}{2},\frac{2}{x \left (a+\frac{1}{x}\right )}\right )$

[Out]

(2*((a - x^(-1))/(a + x^(-1)))^((-5 + n)/2)*(1 + 1/(a*x))^((2 + n)/2)*x*(c - a*c*x)^(5/2)*Hypergeometric2F1[-7
/2, (-5 + n)/2, -5/2, 2/((a + x^(-1))*x)])/(7*(1 - 1/(a*x))^(n/2))

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Rubi [A]  time = 0.199031, antiderivative size = 98, 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} $\frac{2}{7} x (c-a c x)^{5/2} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{n-5}{2}} \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \, _2F_1\left (-\frac{7}{2},\frac{n-5}{2};-\frac{5}{2};\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(n*ArcCoth[a*x])*(c - a*c*x)^(5/2),x]

[Out]

(2*((a - x^(-1))/(a + x^(-1)))^((-5 + n)/2)*(1 + 1/(a*x))^((2 + n)/2)*x*(c - a*c*x)^(5/2)*Hypergeometric2F1[-7
/2, (-5 + n)/2, -5/2, 2/((a + x^(-1))*x)])/(7*(1 - 1/(a*x))^(n/2))

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 e^{n \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx &=\frac{(c-a c x)^{5/2} \int e^{n \coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right )^{5/2} x^{5/2} \, dx}{\left (1-\frac{1}{a x}\right )^{5/2} x^{5/2}}\\ &=-\frac{\left (\left (\frac{1}{x}\right )^{5/2} (c-a c x)^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^{\frac{5}{2}-\frac{n}{2}} \left (1+\frac{x}{a}\right )^{n/2}}{x^{9/2}} \, dx,x,\frac{1}{x}\right )}{\left (1-\frac{1}{a x}\right )^{5/2}}\\ &=\frac{2}{7} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{1}{2} (-5+n)} \left (1-\frac{1}{a x}\right )^{-n/2} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}} x (c-a c x)^{5/2} \, _2F_1\left (-\frac{7}{2},\frac{1}{2} (-5+n);-\frac{5}{2};\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )\\ \end{align*}

Mathematica [A]  time = 0.096458, size = 103, normalized size = 1.05 $\frac{2 c^2 (a x+1)^3 \sqrt{c-a c x} \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{n/2} \left (\frac{a x-1}{a x+1}\right )^{\frac{n-1}{2}} \text{Hypergeometric2F1}\left (-\frac{7}{2},\frac{n-5}{2},-\frac{5}{2},\frac{2}{a x+1}\right )}{7 a}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(n*ArcCoth[a*x])*(c - a*c*x)^(5/2),x]

[Out]

(2*c^2*(1 + 1/(a*x))^(n/2)*((-1 + a*x)/(1 + a*x))^((-1 + n)/2)*(1 + a*x)^3*Sqrt[c - a*c*x]*Hypergeometric2F1[-
7/2, (-5 + n)/2, -5/2, 2/(1 + a*x)])/(7*a*(1 - 1/(a*x))^(n/2))

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Maple [F]  time = 0.404, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\rm arccoth} \left (ax\right )}} \left ( -acx+c \right ) ^{{\frac{5}{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)^(5/2),x)

[Out]

int(exp(n*arccoth(a*x))*(-a*c*x+c)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a c x + c\right )}^{\frac{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((-a*c*x + c)^(5/2)*((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^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}\right )} \sqrt{-a c x + c} \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)^(5/2),x, algorithm="fricas")

[Out]

integral((a^2*c^2*x^2 - 2*a*c^2*x + c^2)*sqrt(-a*c*x + c)*((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)**(5/2),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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((-a*c*x + c)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2*n), x)