3.736 \(\int e^{n \coth ^{-1}(a x)} (c-a^2 c x^2)^3 \, dx\)

Optimal. Leaf size=81 \[ -\frac{256 c^3 \left (1-\frac{1}{a x}\right )^{4-\frac{n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n-8}{2}} \text{Hypergeometric2F1}\left (8,4-\frac{n}{2},5-\frac{n}{2},\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a (8-n)} \]

[Out]

(-256*c^3*(1 - 1/(a*x))^(4 - n/2)*(1 + 1/(a*x))^((-8 + n)/2)*Hypergeometric2F1[8, 4 - n/2, 5 - n/2, (a - x^(-1
))/(a + x^(-1))])/(a*(8 - n))

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Rubi [A]  time = 0.135166, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6191, 6195, 131} \[ -\frac{256 c^3 \left (1-\frac{1}{a x}\right )^{4-\frac{n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n-8}{2}} \, _2F_1\left (8,4-\frac{n}{2};5-\frac{n}{2};\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a (8-n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*c^3*(1 - 1/(a*x))^(4 - n/2)*(1 + 1/(a*x))^((-8 + n)/2)*Hypergeometric2F1[8, 4 - n/2, 5 - n/2, (a - x^(-1
))/(a + x^(-1))])/(a*(8 - n))

Rule 6191

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

Rule 6195

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

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 e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx &=-\left (\left (a^6 c^3\right ) \int e^{n \coth ^{-1}(a x)} \left (1-\frac{1}{a^2 x^2}\right )^3 x^6 \, dx\right )\\ &=\left (a^6 c^3\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^{3-\frac{n}{2}} \left (1+\frac{x}{a}\right )^{3+\frac{n}{2}}}{x^8} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{256 c^3 \left (1-\frac{1}{a x}\right )^{4-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{1}{2} (-8+n)} \, _2F_1\left (8,4-\frac{n}{2};5-\frac{n}{2};\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a (8-n)}\\ \end{align*}

Mathematica [B]  time = 2.28098, size = 267, normalized size = 3.3 \[ -\frac{c^3 e^{n \coth ^{-1}(a x)} \left (n \left (n^5-2 n^4-52 n^3+104 n^2+576 n-1152\right ) e^{2 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (1,\frac{n}{2}+1,\frac{n}{2}+2,e^{2 \coth ^{-1}(a x)}\right )+\left (n^6-56 n^4+784 n^2-2304\right ) \text{Hypergeometric2F1}\left (1,\frac{n}{2},\frac{n}{2}+1,e^{2 \coth ^{-1}(a x)}\right )+24 a^5 n^2 x^5+6 a^4 n^3 x^4+2 a^3 n^4 x^3-152 a^3 n^2 x^3+a^2 n^5 x^2-64 a^2 n^3 x^2+120 a^6 n x^6-576 a^4 n x^4+1368 a^2 n x^2+720 a^7 x^7-3024 a^5 x^5+5040 a^3 x^3+a n^6 x-58 a n^4 x+912 a n^2 x-5040 a x-n^5+58 n^3-912 n\right )}{5040 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(c^3*E^(n*ArcCoth[a*x])*(-912*n + 58*n^3 - n^5 - 5040*a*x + 912*a*n^2*x - 58*a*n^4*x + a*n^6*x + 1368*a^2*n*x
^2 - 64*a^2*n^3*x^2 + a^2*n^5*x^2 + 5040*a^3*x^3 - 152*a^3*n^2*x^3 + 2*a^3*n^4*x^3 - 576*a^4*n*x^4 + 6*a^4*n^3
*x^4 - 3024*a^5*x^5 + 24*a^5*n^2*x^5 + 120*a^6*n*x^6 + 720*a^7*x^7 + E^(2*ArcCoth[a*x])*n*(-1152 + 576*n + 104
*n^2 - 52*n^3 - 2*n^4 + n^5)*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, E^(2*ArcCoth[a*x])] + (-2304 + 784*n^2 - 5
6*n^4 + n^6)*Hypergeometric2F1[1, n/2, 1 + n/2, E^(2*ArcCoth[a*x])]))/(5040*a)

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Maple [F]  time = 0.22, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\rm arccoth} \left (ax\right )}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(n*arccoth(a*x))*(-a^2*c*x^2+c)^3,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 )}^{3} \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*arccoth(a*x))*(-a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

-integrate((a^2*c*x^2 - c)^3*((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^{6} c^{3} x^{6} - 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} - c^{3}\right )} \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*arccoth(a*x))*(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

integral(-(a^6*c^3*x^6 - 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 - c^3)*((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*} - c^{3} \left (\int 3 a^{2} x^{2} e^{n \operatorname{acoth}{\left (a x \right )}}\, dx + \int - 3 a^{4} x^{4} e^{n \operatorname{acoth}{\left (a x \right )}}\, dx + \int a^{6} x^{6} e^{n \operatorname{acoth}{\left (a x \right )}}\, dx + \int - e^{n \operatorname{acoth}{\left (a x \right )}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acoth(a*x))*(-a**2*c*x**2+c)**3,x)

[Out]

-c**3*(Integral(3*a**2*x**2*exp(n*acoth(a*x)), x) + Integral(-3*a**4*x**4*exp(n*acoth(a*x)), x) + Integral(a**
6*x**6*exp(n*acoth(a*x)), x) + Integral(-exp(n*acoth(a*x)), x))

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

[Out]

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