3.1006 \(\int e^{\tanh ^{-1}(a x)} x^2 (1-a^2 x^2)^p \, dx\)

Optimal. Leaf size=84 \[ \frac{1}{3} x^3 \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{1}{2}-p,\frac{5}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{a^3 (2 p+1)}+\frac{\left (1-a^2 x^2\right )^{p+\frac{3}{2}}}{a^3 (2 p+3)} \]

[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

________________________________________________________________________________________

Rubi [A]  time = 0.115129, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6148, 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[E^ArcTanh[a*x]*x^2*(1 - a^2*x^2)^p,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 6148

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] || Gt
Q[c, 0]) && IGtQ[(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 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])

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\\ &=a \int x^3 \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx+\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 = 0.0483701, size = 76, normalized size = 0.9 \[ \frac{1}{3} x^3 \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{1}{2}-p,\frac{5}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}} \left (a^2 (2 p+1) x^2+2\right )}{a^3 \left (4 p^2+8 p+3\right )} \]

Antiderivative was successfully verified.

[In]

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

________________________________________________________________________________________

Maple [A]  time = 0.322, size = 47, normalized size = 0.6 \begin{align*}{\frac{{x}^{4}a}{4}{\mbox{$_2$F$_1$}(2,{\frac{1}{2}}-p;\,3;\,{a}^{2}{x}^{2})}}+{\frac{{x}^{3}}{3}{\mbox{$_2$F$_1$}({\frac{3}{2}},{\frac{1}{2}}-p;\,{\frac{5}{2}};\,{a}^{2}{x}^{2})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 19.2566, size = 255, normalized size = 3.04 \begin{align*} - \frac{a^{2 p} x^{3} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac{3}{2}\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} \frac{1}{2}, 1, p + \frac{3}{2} \\ p + 1, p + \frac{5}{2} \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt{\pi } \Gamma \left (- p - \frac{1}{2}\right ) \Gamma \left (p + 1\right )} - \frac{a^{2 p} x^{3} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac{3}{2}\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} 1, - p, - p - \frac{3}{2} \\ \frac{1}{2}, - p - \frac{1}{2} \end{matrix}\middle |{\frac{1}{a^{2} x^{2}}} \right )}}{2 \sqrt{\pi } \Gamma \left (- p - \frac{1}{2}\right ) \Gamma \left (p + 1\right )} - \frac{{G_{3, 3}^{2, 2}\left (\begin{matrix} - p - 1, 1 & -1 \\- p - \frac{3}{2}, - p - 1 & 0 \end{matrix} \middle |{\frac{e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac{1}{2}\right )}{2 \pi a^{3}} - \frac{{G_{3, 3}^{1, 3}\left (\begin{matrix} -1, - p - 2, 1 & \\- p - 2 & - p - \frac{3}{2}, 0 \end{matrix} \middle |{\frac{e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac{1}{2}\right )}{2 a^{3} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(2*p)*x**3*x**(2*p)*exp(I*pi*p)*gamma(-p - 3/2)*gamma(p + 1/2)*hyper((1/2, 1, p + 3/2), (p + 1, p + 5/2),
a**2*x**2*exp_polar(2*I*pi))/(2*sqrt(pi)*gamma(-p - 1/2)*gamma(p + 1)) - a**(2*p)*x**3*x**(2*p)*exp(I*pi*p)*ga
mma(-p - 3/2)*gamma(p + 1/2)*hyper((1, -p, -p - 3/2), (1/2, -p - 1/2), 1/(a**2*x**2))/(2*sqrt(pi)*gamma(-p - 1
/2)*gamma(p + 1)) - meijerg(((-p - 1, 1), (-1,)), ((-p - 3/2, -p - 1), (0,)), exp_polar(-I*pi)/(a**2*x**2))*ga
mma(p + 1/2)/(2*pi*a**3) - meijerg(((-1, -p - 2, 1), ()), ((-p - 2,), (-p - 3/2, 0)), exp_polar(-I*pi)/(a**2*x
**2))*gamma(p + 1/2)/(2*a**3*gamma(-p)*gamma(p + 1))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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