3.1337 \(\int \frac{e^{n \tanh ^{-1}(a x)}}{\sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{2^{\frac{n+1}{2}} \sqrt{1-a^2 x^2} (1-a x)^{\frac{1-n}{2}} \text{Hypergeometric2F1}\left (\frac{1-n}{2},\frac{1-n}{2},\frac{3-n}{2},\frac{1}{2} (1-a x)\right )}{a (1-n) \sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.105508, antiderivative size = 104, 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 = {6143, 6140, 69} \[ -\frac{2^{\frac{n+1}{2}} \sqrt{1-a^2 x^2} (1-a x)^{\frac{1-n}{2}} \, _2F_1\left (\frac{1-n}{2},\frac{1-n}{2};\frac{3-n}{2};\frac{1}{2} (1-a x)\right )}{a (1-n) \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6143

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

Rule 6140

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.0352407, size = 101, normalized size = 0.97 \[ \frac{2^{\frac{n+1}{2}} \sqrt{1-a^2 x^2} (1-a x)^{\frac{1}{2}-\frac{n}{2}} \text{Hypergeometric2F1}\left (\frac{1}{2}-\frac{n}{2},\frac{1}{2}-\frac{n}{2},\frac{3}{2}-\frac{n}{2},\frac{1}{2}-\frac{a x}{2}\right )}{a (n-1) \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.203, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{n{\it Artanh} \left ( ax \right ) }}{\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))/(-a^2*c*x^2+c)^(1/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*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)

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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^{2} - c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(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)/(a^2*c*x^2 - c), 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 )}}}{\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))/(-a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(exp(n*atanh(a*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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*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)