3.1010 \(\int \frac{e^{\tanh ^{-1}(a x)} (1-a^2 x^2)^p}{x^2} \, dx\)

Optimal. Leaf size=75 \[ -\frac{a \left (1-a^2 x^2\right )^{p+\frac{1}{2}} \text{Hypergeometric2F1}\left (1,p+\frac{1}{2},p+\frac{3}{2},1-a^2 x^2\right )}{2 p+1}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{2}-p,\frac{1}{2},a^2 x^2\right )}{x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.100196, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6148, 764, 364, 266, 65} \[ -\frac{a \left (1-a^2 x^2\right )^{p+\frac{1}{2}} \, _2F_1\left (1,p+\frac{1}{2};p+\frac{3}{2};1-a^2 x^2\right )}{2 p+1}-\frac{\, _2F_1\left (-\frac{1}{2},\frac{1}{2}-p;\frac{1}{2};a^2 x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

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

Mathematica [A]  time = 0.026374, size = 77, normalized size = 1.03 \[ -\frac{a \left (1-a^2 x^2\right )^{p+\frac{1}{2}} \text{Hypergeometric2F1}\left (1,p+\frac{1}{2},p+\frac{3}{2},1-a^2 x^2\right )}{2 \left (p+\frac{1}{2}\right )}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{2}-p,\frac{1}{2},a^2 x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.331, size = 93, normalized size = 1.2 \begin{align*}{\frac{a}{2} \left ( \left ( \Psi \left ({\frac{1}{2}}-p \right ) +\gamma +2\,\ln \left ( x \right ) +\ln \left ( -{a}^{2} \right ) \right ) \Gamma \left ({\frac{1}{2}}-p \right ) +\Gamma \left ({\frac{3}{2}}-p \right ){a}^{2}{x}^{2}{\mbox{$_3$F$_2$}(1,1,{\frac{3}{2}}-p;\,2,2;\,{a}^{2}{x}^{2})} \right ) \left ( \Gamma \left ({\frac{1}{2}}-p \right ) \right ) ^{-1}}-{\frac{1}{x}{\mbox{$_2$F$_1$}(-{\frac{1}{2}},{\frac{1}{2}}-p;\,{\frac{1}{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)*(-a^2*x^2+1)^p/x^2,x)

[Out]

1/2*a*((Psi(1/2-p)+gamma+2*ln(x)+ln(-a^2))*GAMMA(1/2-p)+GAMMA(3/2-p)*a^2*x^2*hypergeom([1,1,3/2-p],[2,2],a^2*x
^2))/GAMMA(1/2-p)-hypergeom([-1/2,1/2-p],[1/2],a^2*x^2)/x

________________________________________________________________________________________

Maxima [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 - \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)*(-a^2*x^2+1)^p/x^2,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}}{a x^{3} - x^{2}}, 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)*(-a^2*x^2+1)^p/x^2,x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*(-a^2*x^2 + 1)^p/(a*x^3 - x^2), x)

________________________________________________________________________________________

Sympy [C]  time = 15.5307, size = 280, normalized size = 3.73 \begin{align*} - \frac{a a^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} \frac{1}{2}, 1, p \\ p + 1, p + 1 \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt{\pi } \Gamma \left (1 - p\right ) \Gamma \left (p + 1\right )} - \frac{a a^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} 1, - p, - p \\ \frac{1}{2}, 1 - p \end{matrix}\middle |{\frac{1}{a^{2} x^{2}}} \right )}}{2 \sqrt{\pi } \Gamma \left (1 - p\right ) \Gamma \left (p + 1\right )} - \frac{a^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (\frac{1}{2} - p\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} \frac{1}{2}, 1, p - \frac{1}{2} \\ p + \frac{1}{2}, p + 1 \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt{\pi } x \Gamma \left (\frac{3}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac{a^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (\frac{1}{2} - p\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} 1, - p, \frac{1}{2} - p \\ \frac{1}{2}, \frac{3}{2} - p \end{matrix}\middle |{\frac{1}{a^{2} x^{2}}} \right )}}{2 \sqrt{\pi } x \Gamma \left (\frac{3}{2} - 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)*(-a**2*x**2+1)**p/x**2,x)

[Out]

-a*a**(2*p)*x**(2*p)*exp(I*pi*p)*gamma(-p)*gamma(p + 1/2)*hyper((1/2, 1, p), (p + 1, p + 1), a**2*x**2*exp_pol
ar(2*I*pi))/(2*sqrt(pi)*gamma(1 - p)*gamma(p + 1)) - a*a**(2*p)*x**(2*p)*exp(I*pi*p)*gamma(-p)*gamma(p + 1/2)*
hyper((1, -p, -p), (1/2, 1 - p), 1/(a**2*x**2))/(2*sqrt(pi)*gamma(1 - p)*gamma(p + 1)) - a**(2*p)*x**(2*p)*exp
(I*pi*p)*gamma(1/2 - p)*gamma(p + 1/2)*hyper((1/2, 1, p - 1/2), (p + 1/2, p + 1), a**2*x**2*exp_polar(2*I*pi))
/(2*sqrt(pi)*x*gamma(3/2 - p)*gamma(p + 1)) - a**(2*p)*x**(2*p)*exp(I*pi*p)*gamma(1/2 - p)*gamma(p + 1/2)*hype
r((1, -p, 1/2 - p), (1/2, 3/2 - p), 1/(a**2*x**2))/(2*sqrt(pi)*x*gamma(3/2 - 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}}{\sqrt{-a^{2} x^{2} + 1} 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)*(-a^2*x^2+1)^p/x^2,x, algorithm="giac")

[Out]

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