3.1224 \(\int e^{-\tanh ^{-1}(a x)} x (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=96 \[ -\frac{1}{3} a x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{1}{2}-p,\frac{5}{2},a^2 x^2\right )-\frac{\sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^2 (2 p+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.112623, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {6153, 6149, 764, 261, 364} \[ -\frac{1}{3} a x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{3}{2},\frac{1}{2}-p;\frac{5}{2};a^2 x^2\right )-\frac{\sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^2 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !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 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 (c-a^2 c x^2\right )^p \, dx &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int e^{-\tanh ^{-1}(a x)} x \left (1-a^2 x^2\right )^p \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x (1-a x) \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx-\left (a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^2 \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=-\frac{\sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^2 (1+2 p)}-\frac{1}{3} a x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{3}{2},\frac{1}{2}-p;\frac{5}{2};a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0295533, size = 88, normalized size = 0.92 \[ \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (-\frac{1}{3} a x^3 \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{1}{2}-p,\frac{5}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{2 a^2 \left (p+\frac{1}{2}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.43, size = 0, normalized size = 0. \begin{align*} \int{\frac{x \left ( -{a}^{2}c{x}^{2}+c \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*c*x^2+c)^p/(a*x+1)*(-a^2*x^2+1)^(1/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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