3.1221 \(\int \frac{e^{-\tanh ^{-1}(a x)} (1-a^2 x^2)^p}{x^2} \, dx\)

Optimal. Leaf size=74 \[ \frac{a \left (1-a^2 x^2\right )^{p+\frac{1}{2}} \text{Hypergeometric2F1}\left (1,p+\frac{1}{2},p+\frac{3}{2},1-a^2 x^2\right )}{2 p+1}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{2}-p,\frac{1}{2},a^2 x^2\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.105648, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {6149, 764, 364, 266, 65} \[ \frac{a \left (1-a^2 x^2\right )^{p+\frac{1}{2}} \, _2F_1\left (1,p+\frac{1}{2};p+\frac{3}{2};1-a^2 x^2\right )}{2 p+1}-\frac{\, _2F_1\left (-\frac{1}{2},\frac{1}{2}-p;\frac{1}{2};a^2 x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^p/(E^ArcTanh[a*x]*x^2),x]

[Out]

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

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 764

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

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])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 65

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

Rubi steps

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

Mathematica [A]  time = 0.0264227, size = 74, normalized size = 1. \[ \frac{a \left (1-a^2 x^2\right )^{p+\frac{1}{2}} \text{Hypergeometric2F1}\left (1,p+\frac{1}{2},p+\frac{3}{2},1-a^2 x^2\right )}{2 p+1}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{2}-p,\frac{1}{2},a^2 x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - a^2*x^2)^p/(E^ArcTanh[a*x]*x^2),x]

[Out]

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

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Maple [F]  time = 0.424, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{a}^{2}{x}^{2}+1 \right ) ^{p}}{ \left ( ax+1 \right ){x}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-a^2*x^2 + 1)^(p + 1/2)/((a*x + 1)*x^2), 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 (-a^{2} x^{2} + 1\right )}^{p}}{a x^{3} + x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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