3.1009 \(\int \frac{e^{\tanh ^{-1}(a x)} (1-a^2 x^2)^p}{x} \, dx\)

Optimal. Leaf size=72 \[ a x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2}-p,\frac{3}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}} \text{Hypergeometric2F1}\left (1,p+\frac{1}{2},p+\frac{3}{2},1-a^2 x^2\right )}{2 p+1} \]

[Out]

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

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Rubi [A]  time = 0.0927399, antiderivative size = 72, 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, 266, 65, 245} \[ a x \, _2F_1\left (\frac{1}{2},\frac{1}{2}-p;\frac{3}{2};a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}} \, _2F_1\left (1,p+\frac{1}{2};p+\frac{3}{2};1-a^2 x^2\right )}{2 p+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^p}{x} \, dx &=\int \frac{(1+a x) \left (1-a^2 x^2\right )^{-\frac{1}{2}+p}}{x} \, dx\\ &=a \int \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx+\int \frac{\left (1-a^2 x^2\right )^{-\frac{1}{2}+p}}{x} \, dx\\ &=a x \, _2F_1\left (\frac{1}{2},\frac{1}{2}-p;\frac{3}{2};a^2 x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (1-a^2 x\right )^{-\frac{1}{2}+p}}{x} \, dx,x,x^2\right )\\ &=a x \, _2F_1\left (\frac{1}{2},\frac{1}{2}-p;\frac{3}{2};a^2 x^2\right )-\frac{\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 = 0.0251872, size = 74, normalized size = 1.03 \[ a x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2}-p,\frac{3}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}} \text{Hypergeometric2F1}\left (1,p+\frac{1}{2},p+\frac{3}{2},1-a^2 x^2\right )}{2 \left (p+\frac{1}{2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.343, size = 90, normalized size = 1.3 \begin{align*} ax{\mbox{$_2$F$_1$}({\frac{1}{2}},{\frac{1}{2}}-p;\,{\frac{3}{2}};\,{a}^{2}{x}^{2})}+{\frac{1}{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}} \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,x)

[Out]

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

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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}\,{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,x, algorithm="maxima")

[Out]

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

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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^{2} - x}, 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,x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 14.5145, size = 286, normalized size = 3.97 \begin{align*} - \frac{a a^{2 p} x x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac{1}{2}\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} \frac{1}{2}, 1, p + \frac{1}{2} \\ p + 1, p + \frac{3}{2} \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt{\pi } \Gamma \left (\frac{1}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac{a a^{2 p} x x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac{1}{2}\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} 1, - p, - p - \frac{1}{2} \\ \frac{1}{2}, \frac{1}{2} - p \end{matrix}\middle |{\frac{1}{a^{2} x^{2}}} \right )}}{2 \sqrt{\pi } \Gamma \left (\frac{1}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac{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^{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 )} \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,x)

[Out]

-a*a**(2*p)*x*x**(2*p)*exp(I*pi*p)*gamma(-p - 1/2)*gamma(p + 1/2)*hyper((1/2, 1, p + 1/2), (p + 1, p + 3/2), a
**2*x**2*exp_polar(2*I*pi))/(2*sqrt(pi)*gamma(1/2 - p)*gamma(p + 1)) - a*a**(2*p)*x*x**(2*p)*exp(I*pi*p)*gamma
(-p - 1/2)*gamma(p + 1/2)*hyper((1, -p, -p - 1/2), (1/2, 1/2 - p), 1/(a**2*x**2))/(2*sqrt(pi)*gamma(1/2 - p)*g
amma(p + 1)) - 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_polar(2*I*pi))/(2*sqrt(pi)*gamma(1 - p)*gamma(p + 1)) - 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))

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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}\,{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,x, algorithm="giac")

[Out]

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