3.1333 \(\int e^{n \tanh ^{-1}(a x)} (c-a^2 c x^2)^{3/2} \, dx\)

Optimal. Leaf size=105 \[ -\frac{c 2^{\frac{n+5}{2}} \sqrt{c-a^2 c x^2} (1-a x)^{\frac{5-n}{2}} \text{Hypergeometric2F1}\left (\frac{1}{2} (-n-3),\frac{5-n}{2},\frac{7-n}{2},\frac{1}{2} (1-a x)\right )}{a (5-n) \sqrt{1-a^2 x^2}} \]

[Out]

-((2^((5 + n)/2)*c*(1 - a*x)^((5 - n)/2)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[(-3 - n)/2, (5 - n)/2, (7 - n)/
2, (1 - a*x)/2])/(a*(5 - n)*Sqrt[1 - a^2*x^2]))

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Rubi [A]  time = 0.112971, antiderivative size = 105, 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} \[ -\frac{c 2^{\frac{n+5}{2}} \sqrt{c-a^2 c x^2} (1-a x)^{\frac{5-n}{2}} \, _2F_1\left (\frac{1}{2} (-n-3),\frac{5-n}{2};\frac{7-n}{2};\frac{1}{2} (1-a x)\right )}{a (5-n) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a*x])*(c - a^2*c*x^2)^(3/2),x]

[Out]

-((2^((5 + n)/2)*c*(1 - a*x)^((5 - n)/2)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[(-3 - n)/2, (5 - n)/2, (7 - n)/
2, (1 - a*x)/2])/(a*(5 - n)*Sqrt[1 - a^2*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 e^{n \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx &=\frac{\left (c \sqrt{c-a^2 c x^2}\right ) \int e^{n \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^{3/2} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (c \sqrt{c-a^2 c x^2}\right ) \int (1-a x)^{\frac{3}{2}-\frac{n}{2}} (1+a x)^{\frac{3}{2}+\frac{n}{2}} \, dx}{\sqrt{1-a^2 x^2}}\\ &=-\frac{2^{\frac{5+n}{2}} c (1-a x)^{\frac{5-n}{2}} \sqrt{c-a^2 c x^2} \, _2F_1\left (\frac{1}{2} (-3-n),\frac{5-n}{2};\frac{7-n}{2};\frac{1}{2} (1-a x)\right )}{a (5-n) \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0741913, size = 102, normalized size = 0.97 \[ \frac{c 2^{\frac{n+5}{2}} \sqrt{c-a^2 c x^2} (1-a x)^{\frac{5}{2}-\frac{n}{2}} \text{Hypergeometric2F1}\left (-\frac{n}{2}-\frac{3}{2},\frac{5}{2}-\frac{n}{2},\frac{7}{2}-\frac{n}{2},\frac{1}{2}-\frac{a x}{2}\right )}{a (n-5) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a*x])*(c - a^2*c*x^2)^(3/2),x]

[Out]

(2^((5 + n)/2)*c*(1 - a*x)^(5/2 - n/2)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[-3/2 - n/2, 5/2 - n/2, 7/2 - n/2,
1/2 - (a*x)/2])/(a*(-5 + n)*Sqrt[1 - a^2*x^2])

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Maple [F]  time = 0.21, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\it Artanh} \left ( ax \right ) }} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}\, 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)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} \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))*(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((-a^2*c*x^2 + c)^(3/2)*((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 (-{\left (a^{2} c x^{2} - c\right )} \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*arctanh(a*x))*(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} \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))*(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^(3/2)*((a*x + 1)/(a*x - 1))^(1/2*n), x)