3.1216 \(\int e^{-\tanh ^{-1}(a x)} x^3 (1-a^2 x^2)^p \, dx\)

Optimal. Leaf size=85 \[ -\frac{1}{5} a x^5 \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{1}{2}-p,\frac{7}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{a^4 (2 p+1)}+\frac{\left (1-a^2 x^2\right )^{p+\frac{3}{2}}}{a^4 (2 p+3)} \]

[Out]

-((1 - a^2*x^2)^(1/2 + p)/(a^4*(1 + 2*p))) + (1 - a^2*x^2)^(3/2 + p)/(a^4*(3 + 2*p)) - (a*x^5*Hypergeometric2F
1[5/2, 1/2 - p, 7/2, a^2*x^2])/5

________________________________________________________________________________________

Rubi [A]  time = 0.11718, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {6149, 764, 266, 43, 364} \[ -\frac{1}{5} a x^5 \, _2F_1\left (\frac{5}{2},\frac{1}{2}-p;\frac{7}{2};a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{a^4 (2 p+1)}+\frac{\left (1-a^2 x^2\right )^{p+\frac{3}{2}}}{a^4 (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(1 - a^2*x^2)^p)/E^ArcTanh[a*x],x]

[Out]

-((1 - a^2*x^2)^(1/2 + p)/(a^4*(1 + 2*p))) + (1 - a^2*x^2)^(3/2 + p)/(a^4*(3 + 2*p)) - (a*x^5*Hypergeometric2F
1[5/2, 1/2 - p, 7/2, a^2*x^2])/5

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 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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)} x^3 \left (1-a^2 x^2\right )^p \, dx &=\int x^3 (1-a x) \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=-\left (a \int x^4 \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\right )+\int x^3 \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=-\frac{1}{5} a x^5 \, _2F_1\left (\frac{5}{2},\frac{1}{2}-p;\frac{7}{2};a^2 x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int x \left (1-a^2 x\right )^{-\frac{1}{2}+p} \, dx,x,x^2\right )\\ &=-\frac{1}{5} a x^5 \, _2F_1\left (\frac{5}{2},\frac{1}{2}-p;\frac{7}{2};a^2 x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{\left (1-a^2 x\right )^{-\frac{1}{2}+p}}{a^2}-\frac{\left (1-a^2 x\right )^{\frac{1}{2}+p}}{a^2}\right ) \, dx,x,x^2\right )\\ &=-\frac{\left (1-a^2 x^2\right )^{\frac{1}{2}+p}}{a^4 (1+2 p)}+\frac{\left (1-a^2 x^2\right )^{\frac{3}{2}+p}}{a^4 (3+2 p)}-\frac{1}{5} a x^5 \, _2F_1\left (\frac{5}{2},\frac{1}{2}-p;\frac{7}{2};a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0723435, size = 77, normalized size = 0.91 \[ -\frac{1}{5} a x^5 \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{1}{2}-p,\frac{7}{2},a^2 x^2\right )-\frac{\left (a^2 (2 p+1) x^2+2\right ) \left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{a^4 \left (4 p^2+8 p+3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(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)*x^2))/(a^4*(3 + 8*p + 4*p^2))) - (a*x^5*Hypergeometric2F1[5/2, 1
/2 - p, 7/2, a^2*x^2])/5

________________________________________________________________________________________

Maple [F]  time = 0.431, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3} \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^3*(-a^2*x^2+1)^p/(a*x+1)*(-a^2*x^2+1)^(1/2),x)

[Out]

int(x^3*(-a^2*x^2+1)^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{{\left (-a^{2} x^{2} + 1\right )}^{p + \frac{1}{2}} x^{3}}{a x + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-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^3/(a*x + 1), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \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**3*(-a**2*x**2+1)**p/(a*x+1)*(-a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**3*sqrt(-(a*x - 1)*(a*x + 1))*(-(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} x^{2} + 1\right )}^{p} x^{3}}{a x + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-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^3/(a*x + 1), x)