3.1218 \(\int e^{-\tanh ^{-1}(a x)} x (1-a^2 x^2)^p \, dx\)

Optimal. Leaf size=58 \[ -\frac{1}{3} a x^3 \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{1}{2}-p,\frac{5}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{a^2 (2 p+1)} \]

[Out]

-((1 - a^2*x^2)^(1/2 + p)/(a^2*(1 + 2*p))) - (a*x^3*Hypergeometric2F1[3/2, 1/2 - p, 5/2, a^2*x^2])/3

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Rubi [A]  time = 0.0648003, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6149, 764, 261, 364} \[ -\frac{1}{3} a x^3 \, _2F_1\left (\frac{3}{2},\frac{1}{2}-p;\frac{5}{2};a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{a^2 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - a^2*x^2)^(1/2 + p)/(a^2*(1 + 2*p))) - (a*x^3*Hypergeometric2F1[3/2, 1/2 - p, 5/2, a^2*x^2])/3

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 261

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

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)} x \left (1-a^2 x^2\right )^p \, dx &=\int x (1-a x) \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=-\left (a \int x^2 \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\right )+\int x \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=-\frac{\left (1-a^2 x^2\right )^{\frac{1}{2}+p}}{a^2 (1+2 p)}-\frac{1}{3} a x^3 \, _2F_1\left (\frac{3}{2},\frac{1}{2}-p;\frac{5}{2};a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0247705, size = 60, normalized size = 1.03 \[ -\frac{1}{3} a x^3 \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{1}{2}-p,\frac{5}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{2 a^2 \left (p+\frac{1}{2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - a^2*x^2)^(1/2 + p)/(2*a^2*(1/2 + p)) - (a*x^3*Hypergeometric2F1[3/2, 1/2 - p, 5/2, a^2*x^2])/3

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Maple [F]  time = 0.429, size = 0, normalized size = 0. \begin{align*} \int{\frac{x \left ( -{a}^{2}{x}^{2}+1 \right ) ^{p}}{ax+1}\sqrt{-{a}^{2}{x}^{2}+1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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