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

Optimal. Leaf size=118 \[ \frac{x \left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{-p-\frac{3}{2}} \left (1-\frac{1}{a x}\right )^{p+\frac{3}{2}} \left (\frac{1}{a x}+1\right )^{p-\frac{1}{2}} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (-2 p-1,-p-\frac{3}{2},-2 p,\frac{2}{x \left (a+\frac{1}{x}\right )}\right )}{2 p+1} \]

[Out]

(((a - x^(-1))/(a + x^(-1)))^(-3/2 - p)*(1 - 1/(a*x))^(3/2 + p)*(1 + 1/(a*x))^(-1/2 + p)*x*(c - a^2*c*x^2)^p*H
ypergeometric2F1[-1 - 2*p, -3/2 - p, -2*p, 2/((a + x^(-1))*x)])/((1 + 2*p)*(1 - 1/(a^2*x^2))^p)

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Rubi [A]  time = 0.144269, antiderivative size = 118, 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 = {6192, 6196, 132} \[ \frac{x \left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{-p-\frac{3}{2}} \left (1-\frac{1}{a x}\right )^{p+\frac{3}{2}} \left (\frac{1}{a x}+1\right )^{p-\frac{1}{2}} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-2 p-1,-p-\frac{3}{2};-2 p;\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{2 p+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((a - x^(-1))/(a + x^(-1)))^(-3/2 - p)*(1 - 1/(a*x))^(3/2 + p)*(1 + 1/(a*x))^(-1/2 + p)*x*(c - a^2*c*x^2)^p*H
ypergeometric2F1[-1 - 2*p, -3/2 - p, -2*p, 2/((a + x^(-1))*x)])/((1 + 2*p)*(1 - 1/(a^2*x^2))^p)

Rule 6192

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

Rule 6196

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

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n), x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rubi steps

\begin{align*} \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx &=\left (\left (1-\frac{1}{a^2 x^2}\right )^{-p} x^{-2 p} \left (c-a^2 c x^2\right )^p\right ) \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac{1}{a^2 x^2}\right )^p x^{2 p} \, dx\\ &=-\left (\left (\left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (\frac{1}{x}\right )^{2 p} \left (c-a^2 c x^2\right )^p\right ) \operatorname{Subst}\left (\int x^{-2-2 p} \left (1-\frac{x}{a}\right )^{\frac{3}{2}+p} \left (1+\frac{x}{a}\right )^{-\frac{3}{2}+p} \, dx,x,\frac{1}{x}\right )\right )\\ &=\frac{\left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{-\frac{3}{2}-p} \left (1-\frac{1}{a x}\right )^{\frac{3}{2}+p} \left (1+\frac{1}{a x}\right )^{-\frac{1}{2}+p} x \left (c-a^2 c x^2\right )^p \, _2F_1\left (-1-2 p,-\frac{3}{2}-p;-2 p;\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{1+2 p}\\ \end{align*}

Mathematica [A]  time = 0.267677, size = 119, normalized size = 1.01 \[ \frac{4^{p+1} \left (a x \sqrt{1-\frac{1}{a^2 x^2}}\right )^{-2 p} e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \left (1-e^{2 \coth ^{-1}(a x)}\right )^{2 p} \left (\frac{e^{\coth ^{-1}(a x)}}{e^{2 \coth ^{-1}(a x)}-1}\right )^{2 p} \text{Hypergeometric2F1}\left (p-\frac{1}{2},2 p+2,p+\frac{1}{2},e^{2 \coth ^{-1}(a x)}\right )}{a-2 a p} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4^(1 + p)*(1 - E^(2*ArcCoth[a*x]))^(2*p)*(E^ArcCoth[a*x]/(-1 + E^(2*ArcCoth[a*x])))^(2*p)*(c - a^2*c*x^2)^p*H
ypergeometric2F1[-1/2 + p, 2 + 2*p, 1/2 + p, E^(2*ArcCoth[a*x])])/(E^ArcCoth[a*x]*(a - 2*a*p)*(a*Sqrt[1 - 1/(a
^2*x^2)]*x)^(2*p))

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Maple [F]  time = 0.374, size = 0, normalized size = 0. \begin{align*} \int \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*c*x^2+c)^p*((a*x-1)/(a*x+1))^(3/2),x)

[Out]

int((-a^2*c*x^2+c)^p*((a*x-1)/(a*x+1))^(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 )}^{p} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^p*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a x - 1\right )}{\left (-a^{2} c x^{2} + c\right )}^{p} \sqrt{\frac{a x - 1}{a x + 1}}}{a x + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^p*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas")

[Out]

integral((a*x - 1)*(-a^2*c*x^2 + c)^p*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1), 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((-a**2*c*x**2+c)**p*((a*x-1)/(a*x+1))**(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 )}^{p} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^p*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")

[Out]

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