3.1338 \(\int \frac{e^{n \tanh ^{-1}(a x)}}{x \sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=101 \[ -\frac{2 \sqrt{1-a^2 x^2} (1-a x)^{\frac{1-n}{2}} (a x+1)^{\frac{n-1}{2}} \text{Hypergeometric2F1}\left (1,\frac{1-n}{2},\frac{3-n}{2},\frac{1-a x}{a x+1}\right )}{(1-n) \sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.230984, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6153, 6150, 131} \[ -\frac{2 \sqrt{1-a^2 x^2} (1-a x)^{\frac{1-n}{2}} (a x+1)^{\frac{n-1}{2}} \, _2F_1\left (1,\frac{1-n}{2};\frac{3-n}{2};\frac{1-a x}{a x+1}\right )}{(1-n) \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6153

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

Rule 6150

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

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

Mathematica [A]  time = 0.0482765, size = 99, normalized size = 0.98 \[ \frac{2 \sqrt{1-a^2 x^2} (1-a x)^{\frac{1}{2}-\frac{n}{2}} (a x+1)^{\frac{n-1}{2}} \text{Hypergeometric2F1}\left (1,\frac{1}{2}-\frac{n}{2},\frac{3}{2}-\frac{n}{2},\frac{1-a x}{a x+1}\right )}{(n-1) \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.205, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{n{\it Artanh} \left ( ax \right ) }}}{x}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(n*arctanh(a*x))/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 \frac{\left (\frac{a x + 1}{a x - 1}\right )^{\frac{1}{2} \, n}}{\sqrt{-a^{2} c x^{2} + c} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/x/(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} c x^{2} + c} \left (\frac{a x + 1}{a x - 1}\right )^{\frac{1}{2} \, n}}{a^{2} c x^{3} - c x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/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)/(a^2*c*x^3 - c*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{\sqrt{-a^{2} c x^{2} + c} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/x/(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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