3.1217 \(\int e^{-\tanh ^{-1}(a x)} x^2 (1-a^2 x^2)^p \, dx\)

Optimal. Leaf size=84 \[ \frac {1}{3} x^3 \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )+\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^3 (2 p+1)}-\frac {\left (1-a^2 x^2\right )^{p+\frac {3}{2}}}{a^3 (2 p+3)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 84, normalized size of antiderivative = 1.00, 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, 364, 266, 43} \[ \frac {1}{3} x^3 \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )+\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^3 (2 p+1)}-\frac {\left (1-a^2 x^2\right )^{p+\frac {3}{2}}}{a^3 (2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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]

Rubi steps

\begin {align*} \int e^{-\tanh ^{-1}(a x)} x^2 \left (1-a^2 x^2\right )^p \, dx &=\int x^2 (1-a x) \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=-\left (a \int x^3 \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\right )+\int x^2 \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=\frac {1}{3} x^3 \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )-\frac {1}{2} a \operatorname {Subst}\left (\int x \left (1-a^2 x\right )^{-\frac {1}{2}+p} \, dx,x,x^2\right )\\ &=\frac {1}{3} x^3 \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )-\frac {1}{2} a \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^3 (1+2 p)}-\frac {\left (1-a^2 x^2\right )^{\frac {3}{2}+p}}{a^3 (3+2 p)}+\frac {1}{3} 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 = 0.07, size = 75, normalized size = 0.89 \[ \frac {1}{3} x^3 \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )+\frac {\left (a^2 (2 p+1) x^2+2\right ) \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^3 \left (4 p^2+8 p+3\right )} \]

Antiderivative was successfully verified.

[In]

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

________________________________________________________________________________________

fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{2}}{a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-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^2/(a*x + 1), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{2}}{a x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-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^2/(a*x + 1), x)

________________________________________________________________________________________

maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (-a^{2} x^{2}+1\right )^{p} \sqrt {-a^{2} x^{2}+1}}{a x +1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{p + \frac {1}{2}} x^{2}}{a x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-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^2/(a*x + 1), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\left (1-a^2\,x^2\right )}^p\,\sqrt {1-a^2\,x^2}}{a\,x+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{p}}{a x + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________