3.1329 \(\int e^{n \tanh ^{-1}(a x)} x \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=173 \[ -\frac{2^{\frac{n+3}{2}} n \sqrt{c-a^2 c x^2} (1-a x)^{\frac{3-n}{2}} \text{Hypergeometric2F1}\left (\frac{1}{2} (-n-1),\frac{3-n}{2},\frac{5-n}{2},\frac{1}{2} (1-a x)\right )}{3 a^2 (3-n) \sqrt{1-a^2 x^2}}-\frac{\sqrt{c-a^2 c x^2} (a x+1)^{\frac{n+3}{2}} (1-a x)^{\frac{3-n}{2}}}{3 a^2 \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.187153, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {6153, 6150, 80, 69} \[ -\frac{2^{\frac{n+3}{2}} n \sqrt{c-a^2 c x^2} (1-a x)^{\frac{3-n}{2}} \, _2F_1\left (\frac{1}{2} (-n-1),\frac{3-n}{2};\frac{5-n}{2};\frac{1}{2} (1-a x)\right )}{3 a^2 (3-n) \sqrt{1-a^2 x^2}}-\frac{\sqrt{c-a^2 c x^2} (a x+1)^{\frac{n+3}{2}} (1-a x)^{\frac{3-n}{2}}}{3 a^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - a*x)^((3 - n)/2)*(1 + a*x)^((3 + n)/2)*Sqrt[c - a^2*c*x^2])/(3*a^2*Sqrt[1 - a^2*x^2]) - (2^((3 + n)/2)*
n*(1 - a*x)^((3 - n)/2)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[(-1 - n)/2, (3 - n)/2, (5 - n)/2, (1 - a*x)/2])/
(3*a^2*(3 - n)*Sqrt[1 - a^2*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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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 e^{n \tanh ^{-1}(a x)} x \sqrt{c-a^2 c x^2} \, dx &=\frac{\sqrt{c-a^2 c x^2} \int e^{n \tanh ^{-1}(a x)} x \sqrt{1-a^2 x^2} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\sqrt{c-a^2 c x^2} \int x (1-a x)^{\frac{1}{2}-\frac{n}{2}} (1+a x)^{\frac{1}{2}+\frac{n}{2}} \, dx}{\sqrt{1-a^2 x^2}}\\ &=-\frac{(1-a x)^{\frac{3-n}{2}} (1+a x)^{\frac{3+n}{2}} \sqrt{c-a^2 c x^2}}{3 a^2 \sqrt{1-a^2 x^2}}+\frac{\left (n \sqrt{c-a^2 c x^2}\right ) \int (1-a x)^{\frac{1}{2}-\frac{n}{2}} (1+a x)^{\frac{1}{2}+\frac{n}{2}} \, dx}{3 a \sqrt{1-a^2 x^2}}\\ &=-\frac{(1-a x)^{\frac{3-n}{2}} (1+a x)^{\frac{3+n}{2}} \sqrt{c-a^2 c x^2}}{3 a^2 \sqrt{1-a^2 x^2}}-\frac{2^{\frac{3+n}{2}} n (1-a x)^{\frac{3-n}{2}} \sqrt{c-a^2 c x^2} \, _2F_1\left (\frac{1}{2} (-1-n),\frac{3-n}{2};\frac{5-n}{2};\frac{1}{2} (1-a x)\right )}{3 a^2 (3-n) \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.12782, size = 125, normalized size = 0.72 \[ \frac{\sqrt{c-a^2 c x^2} (1-a x)^{\frac{3}{2}-\frac{n}{2}} \left (2^{\frac{n+3}{2}} n \text{Hypergeometric2F1}\left (\frac{1}{2} (-n-1),\frac{3-n}{2},\frac{5-n}{2},\frac{1}{2} (1-a x)\right )-(n-3) (a x+1)^{\frac{n+3}{2}}\right )}{3 a^2 (n-3) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^(3/2 - n/2)*Sqrt[c - a^2*c*x^2]*(-((-3 + n)*(1 + a*x)^((3 + n)/2)) + 2^((3 + n)/2)*n*Hypergeometric
2F1[(-1 - n)/2, (3 - n)/2, (5 - n)/2, (1 - a*x)/2]))/(3*a^2*(-3 + n)*Sqrt[1 - a^2*x^2])

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Maple [F]  time = 0.2, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\it Artanh} \left ( ax \right ) }}x\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 \sqrt{-a^{2} c x^{2} + c} x \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*arctanh(a*x))*x*(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(sqrt(-a^2*c*x^2 + c)*x*((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 x \sqrt{- c \left (a x - 1\right ) \left (a x + 1\right )} e^{n \operatorname{atanh}{\left (a x \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(x*sqrt(-c*(a*x - 1)*(a*x + 1))*exp(n*atanh(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} x \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*arctanh(a*x))*x*(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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