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

Optimal. Leaf size=430 \[ -\frac{a^2 \left (3-n^2\right ) x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac{n-3}{2}} (1-a x)^{\frac{3-n}{2}} \text{Hypergeometric2F1}\left (1,\frac{n-3}{2},\frac{n-1}{2},\frac{a x+1}{1-a x}\right )}{(3-n) \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 2^{\frac{n-1}{2}} n x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} (1-a x)^{\frac{5-n}{2}} \text{Hypergeometric2F1}\left (\frac{3-n}{2},\frac{5-n}{2},\frac{7-n}{2},\frac{1}{2} (1-a x)\right )}{(3-n) (5-n) \left (1-a^2 x^2\right )^{3/2}}-\frac{3 a^2 x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac{n-3}{2}} (1-a x)^{\frac{5-n}{2}}}{(3-n) \left (1-a^2 x^2\right )^{3/2}}-\frac{a (n+4) x^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac{n-3}{2}} (1-a x)^{\frac{5-n}{2}}}{2 \left (1-a^2 x^2\right )^{3/2}}-\frac{x \left (c-\frac{c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac{n-3}{2}} (1-a x)^{\frac{5-n}{2}}}{2 \left (1-a^2 x^2\right )^{3/2}} \]

[Out]

-((c - c/(a^2*x^2))^(3/2)*x*(1 - a*x)^((5 - n)/2)*(1 + a*x)^((-3 + n)/2))/(2*(1 - a^2*x^2)^(3/2)) - (a*(4 + n)
*(c - c/(a^2*x^2))^(3/2)*x^2*(1 - a*x)^((5 - n)/2)*(1 + a*x)^((-3 + n)/2))/(2*(1 - a^2*x^2)^(3/2)) - (3*a^2*(c
- c/(a^2*x^2))^(3/2)*x^3*(1 - a*x)^((5 - n)/2)*(1 + a*x)^((-3 + n)/2))/((3 - n)*(1 - a^2*x^2)^(3/2)) - (a^2*(
3 - n^2)*(c - c/(a^2*x^2))^(3/2)*x^3*(1 - a*x)^((3 - n)/2)*(1 + a*x)^((-3 + n)/2)*Hypergeometric2F1[1, (-3 + n
)/2, (-1 + n)/2, (1 + a*x)/(1 - a*x)])/((3 - n)*(1 - a^2*x^2)^(3/2)) + (2^((-1 + n)/2)*a^2*n*(c - c/(a^2*x^2))
^(3/2)*x^3*(1 - a*x)^((5 - n)/2)*Hypergeometric2F1[(3 - n)/2, (5 - n)/2, (7 - n)/2, (1 - a*x)/2])/((3 - n)*(5
- n)*(1 - a^2*x^2)^(3/2))

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Rubi [C]  time = 0.206055, antiderivative size = 103, normalized size of antiderivative = 0.24, 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 = {6160, 6150, 136} \[ -\frac{a^2 2^{\frac{5}{2}-\frac{n}{2}} x^3 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac{n+5}{2}} F_1\left (\frac{n+5}{2};\frac{n-3}{2},3;\frac{n+7}{2};\frac{1}{2} (a x+1),a x+1\right )}{(n+5) \left (1-a^2 x^2\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2^(5/2 - n/2)*a^2*(c - c/(a^2*x^2))^(3/2)*x^3*(1 + a*x)^((5 + n)/2)*AppellF1[(5 + n)/2, (-3 + n)/2, 3, (7 +
n)/2, (1 + a*x)/2, 1 + a*x])/((5 + n)*(1 - a^2*x^2)^(3/2)))

Rule 6160

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[(x^(2*p)*(c + d/x^2)^p)/
(1 + (c*x^2)/d)^p, Int[(u*(1 + (c*x^2)/d)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] &
& EqQ[c + a^2*d, 0] &&  !IntegerQ[p] &&  !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 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin{align*} \int e^{n \tanh ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right )^{3/2} \, dx &=\frac{\left (\left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3\right ) \int \frac{e^{n \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^{3/2}}{x^3} \, dx}{\left (1-a^2 x^2\right )^{3/2}}\\ &=\frac{\left (\left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3\right ) \int \frac{(1-a x)^{\frac{3}{2}-\frac{n}{2}} (1+a x)^{\frac{3}{2}+\frac{n}{2}}}{x^3} \, dx}{\left (1-a^2 x^2\right )^{3/2}}\\ &=-\frac{2^{\frac{5}{2}-\frac{n}{2}} a^2 \left (c-\frac{c}{a^2 x^2}\right )^{3/2} x^3 (1+a x)^{\frac{5+n}{2}} F_1\left (\frac{5+n}{2};\frac{1}{2} (-3+n),3;\frac{7+n}{2};\frac{1}{2} (1+a x),1+a x\right )}{(5+n) \left (1-a^2 x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.68723, size = 190, normalized size = 0.44 \[ \frac{c x \sqrt{c-\frac{c}{a^2 x^2}} e^{n \tanh ^{-1}(a x)} \text{csch}\left (\frac{1}{2} \tanh ^{-1}(a x)\right ) \text{sech}\left (\frac{1}{2} \tanh ^{-1}(a x)\right ) \left (-4 a \left (n^2-3\right ) x e^{\tanh ^{-1}(a x)} \text{Hypergeometric2F1}\left (1,\frac{n+1}{2},\frac{n+3}{2},e^{2 \tanh ^{-1}(a x)}\right )+8 a n x e^{\tanh ^{-1}(a x)} \text{Hypergeometric2F1}\left (1,\frac{n+1}{2},\frac{n+3}{2},-e^{2 \tanh ^{-1}(a x)}\right )-(n+1) \text{csch}\left (\frac{1}{2} \tanh ^{-1}(a x)\right ) \text{sech}\left (\frac{1}{2} \tanh ^{-1}(a x)\right ) \left (\left (1-a^2 x^2\right ) \cosh \left (2 \tanh ^{-1}(a x)\right )+a x (a x+n)\right )\right )}{8 (n+1) \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*E^(n*ArcTanh[a*x])*Sqrt[c - c/(a^2*x^2)]*x*Csch[ArcTanh[a*x]/2]*Sech[ArcTanh[a*x]/2]*(8*a*E^ArcTanh[a*x]*n*
x*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -E^(2*ArcTanh[a*x])] - 4*a*E^ArcTanh[a*x]*(-3 + n^2)*x*Hypergeome
tric2F1[1, (1 + n)/2, (3 + n)/2, E^(2*ArcTanh[a*x])] - (1 + n)*(a*x*(n + a*x) + (1 - a^2*x^2)*Cosh[2*ArcTanh[a
*x]])*Csch[ArcTanh[a*x]/2]*Sech[ArcTanh[a*x]/2]))/(8*(1 + n)*(-1 + a^2*x^2))

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Maple [F]  time = 0.133, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\it Artanh} \left ( ax \right ) }} \left ( c-{\frac{c}{{a}^{2}{x}^{2}}} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

[Out]

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