3.789 \(\int e^{-2 p \tanh ^{-1}(a x)} (c-\frac{c}{a^2 x^2})^p \, dx\)

Optimal. Leaf size=53 \[ \frac{x \left (1-a^2 x^2\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \text{Hypergeometric2F1}(1-2 p,-2 p,2-2 p,a x)}{1-2 p} \]

[Out]

((c - c/(a^2*x^2))^p*x*Hypergeometric2F1[1 - 2*p, -2*p, 2 - 2*p, a*x])/((1 - 2*p)*(1 - a^2*x^2)^p)

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Rubi [A]  time = 0.115895, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6160, 6150, 64} \[ \frac{x \left (1-a^2 x^2\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \, _2F_1(1-2 p,-2 p;2-2 p;a x)}{1-2 p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - c/(a^2*x^2))^p*x*Hypergeometric2F1[1 - 2*p, -2*p, 2 - 2*p, a*x])/((1 - 2*p)*(1 - a^2*x^2)^p)

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 64

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

Rubi steps

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

Mathematica [A]  time = 0.0193244, size = 53, normalized size = 1. \[ \frac{x \left (1-a^2 x^2\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \text{Hypergeometric2F1}(1-2 p,-2 p,2-2 p,a x)}{1-2 p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - c/(a^2*x^2))^p*x*Hypergeometric2F1[1 - 2*p, -2*p, 2 - 2*p, a*x])/((1 - 2*p)*(1 - a^2*x^2)^p)

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Maple [F]  time = 0.227, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{{\rm e}^{2\,p{\it Artanh} \left ( ax \right ) }}} \left ( c-{\frac{c}{{a}^{2}{x}^{2}}} \right ) ^{p}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a^2/x^2)^p/exp(2*p*arctanh(a*x)),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((a^2*c*x^2 - c)/(a^2*x^2))^p/((a*x + 1)/(a*x - 1))^p, 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((c-c/a**2/x**2)**p/exp(2*p*atanh(a*x)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a^2/x^2)^p/exp(2*p*arctanh(a*x)),x, algorithm="giac")

[Out]

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