3.1018 \(\int \frac{e^{\tanh ^{-1}(a x)} (c-a^2 c x^2)^p}{x^3} \, dx\)

Optimal. Leaf size=116 \[ -\frac{a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{2}-p,\frac{1}{2},a^2 x^2\right )}{x}-\frac{a^2 \sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (2,p+\frac{1}{2},p+\frac{3}{2},1-a^2 x^2\right )}{2 p+1} \]

[Out]

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

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Rubi [A]  time = 0.160676, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {6153, 6148, 764, 266, 65, 364} \[ -\frac{a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-\frac{1}{2},\frac{1}{2}-p;\frac{1}{2};a^2 x^2\right )}{x}-\frac{a^2 \sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p \, _2F_1\left (2,p+\frac{1}{2};p+\frac{3}{2};1-a^2 x^2\right )}{2 p+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[n/2]

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

Mathematica [A]  time = 0.0217302, size = 108, normalized size = 0.93 \[ \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (-\frac{a^2 \left (1-a^2 x^2\right )^{p+\frac{1}{2}} \text{Hypergeometric2F1}\left (2,p+\frac{1}{2},p+\frac{3}{2},1-a^2 x^2\right )}{2 \left (p+\frac{1}{2}\right )}-\frac{a \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{2}-p,\frac{1}{2},a^2 x^2\right )}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - a^2*c*x^2)^p*(-((a*Hypergeometric2F1[-1/2, 1/2 - p, 1/2, a^2*x^2])/x) - (a^2*(1 - a^2*x^2)^(1/2 + p)*Hyp
ergeometric2F1[2, 1/2 + p, 3/2 + p, 1 - a^2*x^2])/(2*(1/2 + p))))/(1 - a^2*x^2)^p

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Maple [F]  time = 0.319, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ) \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p}}{{x}^{3}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}\, dx \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*c*x^2+c)^p/x^3,x)

[Out]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^p/x^3,x)

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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} c x^{2} + c\right )}^{p}}{\sqrt{-a^{2} x^{2} + 1} x^{3}}\,{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*c*x^2+c)^p/x^3,x, algorithm="maxima")

[Out]

integrate((a*x + 1)*(-a^2*c*x^2 + c)^p/(sqrt(-a^2*x^2 + 1)*x^3), 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} c x^{2} + c\right )}^{p}}{a x^{4} - x^{3}}, 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*c*x^2+c)^p/x^3,x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*(-a^2*c*x^2 + c)^p/(a*x^4 - x^3), x)

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Sympy [C]  time = 24.6242, size = 301, normalized size = 2.59 \begin{align*} - \frac{a a^{2 p} c^{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 a^{2 p} c^{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 )} - \frac{a^{2 p} c^{p} x^{2 p} e^{i \pi p} \Gamma \left (1 - p\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} \frac{1}{2}, 1, p - 1 \\ p, p + 1 \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt{\pi } x^{2} \Gamma \left (2 - p\right ) \Gamma \left (p + 1\right )} - \frac{a^{2 p} c^{p} x^{2 p} e^{i \pi p} \Gamma \left (1 - p\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} 1, - p, 1 - p \\ \frac{1}{2}, 2 - p \end{matrix}\middle |{\frac{1}{a^{2} x^{2}}} \right )}}{2 \sqrt{\pi } x^{2} \Gamma \left (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*c*x**2+c)**p/x**3,x)

[Out]

-a*a**(2*p)*c**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*a**(2*p)*c**p*x**(2*p)*exp(I*pi*p
)*gamma(1/2 - p)*gamma(p + 1/2)*hyper((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)) - a**(2*p)*c**p*x**(2*p)*exp(I*pi*p)*gamma(1 - p)*gamma(p + 1/2)*hyper((1/2, 1, p - 1), (p
, p + 1), a**2*x**2*exp_polar(2*I*pi))/(2*sqrt(pi)*x**2*gamma(2 - p)*gamma(p + 1)) - a**(2*p)*c**p*x**(2*p)*ex
p(I*pi*p)*gamma(1 - p)*gamma(p + 1/2)*hyper((1, -p, 1 - p), (1/2, 2 - p), 1/(a**2*x**2))/(2*sqrt(pi)*x**2*gamm
a(2 - 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} c x^{2} + c\right )}^{p}}{\sqrt{-a^{2} x^{2} + 1} x^{3}}\,{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*c*x^2+c)^p/x^3,x, algorithm="giac")

[Out]

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