3.745 \(\int e^{n \coth ^{-1}(a x)} \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=116 \[ \frac{8 \sqrt{c-a^2 c x^2} \left (1-\frac{1}{a x}\right )^{\frac{3-n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n-3}{2}} \text{Hypergeometric2F1}\left (3,\frac{3-n}{2},\frac{5-n}{2},\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a^2 (3-n) x \sqrt{1-\frac{1}{a^2 x^2}}} \]

[Out]

(8*(1 - 1/(a*x))^((3 - n)/2)*(1 + 1/(a*x))^((-3 + n)/2)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[3, (3 - n)/2, (5
- n)/2, (a - x^(-1))/(a + x^(-1))])/(a^2*(3 - n)*Sqrt[1 - 1/(a^2*x^2)]*x)

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Rubi [A]  time = 0.163098, antiderivative size = 116, 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 = {6192, 6195, 131} \[ \frac{8 \sqrt{c-a^2 c x^2} \left (1-\frac{1}{a x}\right )^{\frac{3-n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n-3}{2}} \, _2F_1\left (3,\frac{3-n}{2};\frac{5-n}{2};\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a^2 (3-n) x \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(1 - 1/(a*x))^((3 - n)/2)*(1 + 1/(a*x))^((-3 + n)/2)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[3, (3 - n)/2, (5
- n)/2, (a - x^(-1))/(a + x^(-1))])/(a^2*(3 - n)*Sqrt[1 - 1/(a^2*x^2)]*x)

Rule 6192

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

Rule 6195

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_)^2)^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^p, Subst[Int[((
1 - x/a)^(p - n/2)*(1 + x/a)^(p + n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2
*d, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) &&  !IntegersQ[2*p, p + n/2] && 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)} \sqrt{c-a^2 c x^2} \, dx &=\frac{\sqrt{c-a^2 c x^2} \int e^{n \coth ^{-1}(a x)} \sqrt{1-\frac{1}{a^2 x^2}} x \, dx}{\sqrt{1-\frac{1}{a^2 x^2}} x}\\ &=-\frac{\sqrt{c-a^2 c x^2} \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^{\frac{1}{2}-\frac{n}{2}} \left (1+\frac{x}{a}\right )^{\frac{1}{2}+\frac{n}{2}}}{x^3} \, dx,x,\frac{1}{x}\right )}{\sqrt{1-\frac{1}{a^2 x^2}} x}\\ &=\frac{8 \left (1-\frac{1}{a x}\right )^{\frac{3-n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{1}{2} (-3+n)} \sqrt{c-a^2 c x^2} \, _2F_1\left (3,\frac{3-n}{2};\frac{5-n}{2};\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a^2 (3-n) \sqrt{1-\frac{1}{a^2 x^2}} x}\\ \end{align*}

Mathematica [A]  time = 0.553351, size = 101, normalized size = 0.87 \[ -\frac{c x \sqrt{1-\frac{1}{a^2 x^2}} e^{n \coth ^{-1}(a x)} \left (2 (n-1) e^{\coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (1,\frac{n+1}{2},\frac{n+3}{2},e^{2 \coth ^{-1}(a x)}\right )+a x \sqrt{1-\frac{1}{a^2 x^2}} (a x+n)\right )}{2 \sqrt{c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.31, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\rm arccoth} \left (ax\right )}}\sqrt{-{a}^{2}c{x}^{2}+c}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*((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 (\sqrt{-a^{2} c x^{2} + 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^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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