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

Optimal. Leaf size=167 \[ \frac{a x^2 \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{n+3}{2}} \left (1-\frac{1}{a x}\right )^{\frac{2-n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+3}{2},\frac{3}{2},\frac{2}{x \left (a+\frac{1}{x}\right )}\right )}{(n+3) (c-a c x)^{5/2}}-\frac{a x^2 \left (1-\frac{1}{a x}\right )^{\frac{2-n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}}}{(n+3) (c-a c x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.228236, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6176, 6181, 94, 132} \[ \frac{a x^2 \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{n+3}{2}} \left (1-\frac{1}{a x}\right )^{\frac{2-n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \, _2F_1\left (\frac{1}{2},\frac{n+3}{2};\frac{3}{2};\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{(n+3) (c-a c x)^{5/2}}-\frac{a x^2 \left (1-\frac{1}{a x}\right )^{\frac{2-n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}}}{(n+3) (c-a c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 94

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

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)}}{(c-a c x)^{5/2}} \, dx &=\frac{\left (\left (1-\frac{1}{a x}\right )^{5/2} x^{5/2}\right ) \int \frac{e^{n \coth ^{-1}(a x)}}{\left (1-\frac{1}{a x}\right )^{5/2} x^{5/2}} \, dx}{(c-a c x)^{5/2}}\\ &=-\frac{\left (1-\frac{1}{a x}\right )^{5/2} \operatorname{Subst}\left (\int \sqrt{x} \left (1-\frac{x}{a}\right )^{-\frac{5}{2}-\frac{n}{2}} \left (1+\frac{x}{a}\right )^{n/2} \, dx,x,\frac{1}{x}\right )}{\left (\frac{1}{x}\right )^{5/2} (c-a c x)^{5/2}}\\ &=-\frac{a \left (1-\frac{1}{a x}\right )^{\frac{2-n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}} x^2}{(3+n) (c-a c x)^{5/2}}+\frac{\left (a \left (1-\frac{1}{a x}\right )^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^{-\frac{3}{2}-\frac{n}{2}} \left (1+\frac{x}{a}\right )^{n/2}}{\sqrt{x}} \, dx,x,\frac{1}{x}\right )}{2 (3+n) \left (\frac{1}{x}\right )^{5/2} (c-a c x)^{5/2}}\\ &=-\frac{a \left (1-\frac{1}{a x}\right )^{\frac{2-n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}} x^2}{(3+n) (c-a c x)^{5/2}}+\frac{a \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{3+n}{2}} \left (1-\frac{1}{a x}\right )^{\frac{2-n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}} x^2 \, _2F_1\left (\frac{1}{2},\frac{3+n}{2};\frac{3}{2};\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{(3+n) (c-a c x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.107925, size = 117, normalized size = 0.7 \[ \frac{\left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{n/2} \left ((a x-1) \left (\frac{a x-1}{a x+1}\right )^{\frac{n+1}{2}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+3}{2},\frac{3}{2},\frac{2}{a x+1}\right )-a x-1\right )}{a c^2 (n+3) (a x-1) \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.329, 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 \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{{\left (-a c x + c\right )}^{\frac{5}{2}}}\,{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*x - 1)/(a*x + 1))^(1/2*n)/(-a*c*x + c)^(5/2), 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^{3} c^{3} x^{3} - 3 \, a^{2} c^{3} x^{2} + 3 \, a c^{3} x - c^{3}}, 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(-sqrt(-a*c*x + c)*((a*x - 1)/(a*x + 1))^(1/2*n)/(a^3*c^3*x^3 - 3*a^2*c^3*x^2 + 3*a*c^3*x - c^3), 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 \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{{\left (-a c x + c\right )}^{\frac{5}{2}}}\,{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*x - 1)/(a*x + 1))^(1/2*n)/(-a*c*x + c)^(5/2), x)