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

Optimal. Leaf size=137 \[ \frac{x \left (1-a^2 x^2\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \text{Hypergeometric2F1}\left (\frac{1}{2} (1-2 p),\frac{1}{2}-p,\frac{1}{2} (3-2 p),a^2 x^2\right )}{1-2 p}-\frac{a x^2 \left (1-a^2 x^2\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \text{Hypergeometric2F1}\left (\frac{1}{2}-p,1-p,2-p,a^2 x^2\right )}{2 (1-p)} \]

[Out]

((c - c/(a^2*x^2))^p*x*Hypergeometric2F1[(1 - 2*p)/2, 1/2 - p, (3 - 2*p)/2, a^2*x^2])/((1 - 2*p)*(1 - a^2*x^2)
^p) - (a*(c - c/(a^2*x^2))^p*x^2*Hypergeometric2F1[1/2 - p, 1 - p, 2 - p, a^2*x^2])/(2*(1 - p)*(1 - a^2*x^2)^p
)

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Rubi [A]  time = 0.13672, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6160, 6149, 808, 364} \[ \frac{x \left (1-a^2 x^2\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \, _2F_1\left (\frac{1}{2} (1-2 p),\frac{1}{2}-p;\frac{1}{2} (3-2 p);a^2 x^2\right )}{1-2 p}-\frac{a x^2 \left (1-a^2 x^2\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \, _2F_1\left (\frac{1}{2}-p,1-p;2-p;a^2 x^2\right )}{2 (1-p)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - c/(a^2*x^2))^p*x*Hypergeometric2F1[(1 - 2*p)/2, 1/2 - p, (3 - 2*p)/2, a^2*x^2])/((1 - 2*p)*(1 - a^2*x^2)
^p) - (a*(c - c/(a^2*x^2))^p*x^2*Hypergeometric2F1[1/2 - p, 1 - p, 2 - p, a^2*x^2])/(2*(1 - 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 6149

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

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.0291181, size = 112, normalized size = 0.82 \[ \frac{x \left (1-a^2 x^2\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \left (a (2 p-1) x \text{Hypergeometric2F1}\left (\frac{1}{2}-p,1-p,2-p,a^2 x^2\right )-2 (p-1) \text{Hypergeometric2F1}\left (\frac{1}{2}-p,\frac{1}{2}-p,\frac{3}{2}-p,a^2 x^2\right )\right )}{2 (p-1) (2 p-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - c/(a^2*x^2))^p*x*(-2*(-1 + p)*Hypergeometric2F1[1/2 - p, 1/2 - p, 3/2 - p, a^2*x^2] + a*(-1 + 2*p)*x*Hyp
ergeometric2F1[1/2 - p, 1 - p, 2 - p, a^2*x^2]))/(2*(-1 + p)*(-1 + 2*p)*(1 - a^2*x^2)^p)

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Maple [F]  time = 0.324, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ax+1} \left ( c-{\frac{c}{{a}^{2}{x}^{2}}} \right ) ^{p}\sqrt{-{a}^{2}{x}^{2}+1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))**p/(a*x + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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