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

Optimal. Leaf size=81 \[ \frac{16 c^2 \left (1-\frac{1}{a x}\right )^{3-\frac{n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n-6}{2}} \text{Hypergeometric2F1}\left (4,3-\frac{n}{2},4-\frac{n}{2},\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a (6-n)} \]

[Out]

(16*c^2*(1 - 1/(a*x))^(3 - n/2)*(1 + 1/(a*x))^((-6 + n)/2)*Hypergeometric2F1[4, 3 - n/2, 4 - n/2, (a - x^(-1))
/(a + x^(-1))])/(a*(6 - n))

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Rubi [A]  time = 0.128074, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6175, 6180, 131} \[ \frac{16 c^2 \left (1-\frac{1}{a x}\right )^{3-\frac{n}{2}} \left (\frac{1}{a x}+1\right )^{\frac{n-6}{2}} \, _2F_1\left (4,3-\frac{n}{2};4-\frac{n}{2};\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a (6-n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*c^2*(1 - 1/(a*x))^(3 - n/2)*(1 + 1/(a*x))^((-6 + n)/2)*Hypergeometric2F1[4, 3 - n/2, 4 - n/2, (a - x^(-1))
/(a + x^(-1))])/(a*(6 - n))

Rule 6175

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

Rule 6180

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^p, Subst[Int[((1
+ (d*x)/c)^p*(1 + x/a)^(n/2))/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ
[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && 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)} (c-a c x)^2 \, dx &=\left (a^2 c^2\right ) \int e^{n \coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right )^2 x^2 \, dx\\ &=-\left (\left (a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^{2-\frac{n}{2}} \left (1+\frac{x}{a}\right )^{n/2}}{x^4} \, dx,x,\frac{1}{x}\right )\right )\\ &=\frac{16 c^2 \left (1-\frac{1}{a x}\right )^{3-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{1}{2} (-6+n)} \, _2F_1\left (4,3-\frac{n}{2};4-\frac{n}{2};\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )}{a (6-n)}\\ \end{align*}

Mathematica [A]  time = 1.04044, size = 144, normalized size = 1.78 \[ \frac{c^2 e^{n \coth ^{-1}(a x)} \left ((n+2) \left (\left (n^2-6 n+8\right ) \text{Hypergeometric2F1}\left (1,\frac{n}{2},\frac{n}{2}+1,e^{2 \coth ^{-1}(a x)}\right )+n \left (a^2 x^2-6 a x-1\right )+2 a^3 x^3-6 a^2 x^2+a n^2 x+6 a x+6\right )+n \left (n^2-6 n+8\right ) e^{2 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (1,\frac{n}{2}+1,\frac{n}{2}+2,e^{2 \coth ^{-1}(a x)}\right )\right )}{6 a (n+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*E^(n*ArcCoth[a*x])*(E^(2*ArcCoth[a*x])*n*(8 - 6*n + n^2)*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, E^(2*ArcC
oth[a*x])] + (2 + n)*(6 + 6*a*x + a*n^2*x - 6*a^2*x^2 + 2*a^3*x^3 + n*(-1 - 6*a*x + a^2*x^2) + (8 - 6*n + n^2)
*Hypergeometric2F1[1, n/2, 1 + n/2, E^(2*ArcCoth[a*x])])))/(6*a*(2 + n))

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Maple [F]  time = 0.228, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\rm arccoth} \left (ax\right )}} \left ( -acx+c \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((a*c*x - c)^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^{2} x^{2} - 2 \, a c^{2} x + c^{2}\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*c*x+c)^2,x, algorithm="fricas")

[Out]

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

[Out]

c**2*(Integral(-2*a*x*exp(n*acoth(a*x)), x) + Integral(a**2*x**2*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 c x - c\right )}^{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*arccoth(a*x))*(-a*c*x+c)^2,x, algorithm="giac")

[Out]

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