### 3.543 $$\int \frac{e^{n \coth ^{-1}(a x)}}{c-\frac{c}{a x}} \, dx$$

Optimal. Leaf size=113 $\frac{x \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}}}{c}-\frac{2 (n+1) \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{n/2} \text{Hypergeometric2F1}\left (1,-\frac{n}{2},1-\frac{n}{2},\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a c n}$

[Out]

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

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Rubi [A]  time = 0.0849104, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.136, Rules used = {6179, 96, 131} $\frac{x \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}}}{c}-\frac{2 (n+1) \left (1-\frac{1}{a x}\right )^{-n/2} \left (\frac{1}{a x}+1\right )^{n/2} \, _2F_1\left (1,-\frac{n}{2};1-\frac{n}{2};\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a c n}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6179

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> -Dist[c^p, Subst[Int[((1 + (d*x)/c)^
p*(1 + x/a)^(n/2))/(x^2*(1 - x/a)^(n/2)), x], x, 1/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])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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

Mathematica [A]  time = 0.401441, size = 97, normalized size = 0.86 $\frac{e^{n \coth ^{-1}(a x)} \left (n (n+1) e^{2 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (1,\frac{n}{2}+1,\frac{n}{2}+2,e^{2 \coth ^{-1}(a x)}\right )+(n+2) \left ((n+1) \text{Hypergeometric2F1}\left (1,\frac{n}{2},\frac{n}{2}+1,e^{2 \coth ^{-1}(a x)}\right )+a n x-1\right )\right )}{a c n (n+2)}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.062, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{n{\rm arccoth} \left (ax\right )}} \left ( c-{\frac{c}{ax}} \right ) ^{-1}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))/(c-c/a/x),x)

[Out]

int(exp(n*arccoth(a*x))/(c-c/a/x),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}}{c - \frac{c}{a x}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral(a*x*((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*} \frac{a \int \frac{x e^{n \operatorname{acoth}{\left (a x \right )}}}{a x - 1}\, dx}{c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{c - \frac{c}{a x}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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