3.1011 \(\int \frac {e^{\tanh ^{-1}(a x)} (1-a^2 x^2)^p}{x^3} \, dx\)

Optimal. Leaf size=78 \[ -\frac {a^2 \left (1-a^2 x^2\right )^{p+\frac {1}{2}} \, _2F_1\left (2,p+\frac {1}{2};p+\frac {3}{2};1-a^2 x^2\right )}{2 p+1}-\frac {a \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-p;\frac {1}{2};a^2 x^2\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 78, normalized size of antiderivative = 1.00, 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, 364} \[ -\frac {a^2 \left (1-a^2 x^2\right )^{p+\frac {1}{2}} \, _2F_1\left (2,p+\frac {1}{2};p+\frac {3}{2};1-a^2 x^2\right )}{2 p+1}-\frac {a \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-p;\frac {1}{2};a^2 x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(1 - a^2*x^2)^p)/x^3,x]

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 80, normalized size = 1.03 \[ -\frac {a^2 \left (1-a^2 x^2\right )^{p+\frac {1}{2}} \, _2F_1\left (2,p+\frac {1}{2};p+\frac {3}{2};1-a^2 x^2\right )}{2 \left (p+\frac {1}{2}\right )}-\frac {a \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-p;\frac {1}{2};a^2 x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*(1 - a^2*x^2)^p)/x^3,x]

[Out]

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

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} x^{2} + 1\right )}^{p}}{a x^{4} - x^{3}}, x\right ) \]

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^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} {\left (-a^{2} x^{2} + 1\right )}^{p}}{\sqrt {-a^{2} x^{2} + 1} x^{3}}\,{d x} \]

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^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.40, size = 112, normalized size = 1.44 \[ -\frac {a \hypergeom \left (\left [-\frac {1}{2}, \frac {1}{2}-p \right ], \left [\frac {1}{2}\right ], a^{2} x^{2}\right )}{x}-\frac {a^{2} \left (-\frac {\Gamma \left (\frac {5}{2}-p \right ) a^{2} x^{2} \hypergeom \left (\left [1, 1, \frac {5}{2}-p \right ], \left [2, 3\right ], a^{2} x^{2}\right )}{2}-\left (\Psi \left (\frac {3}{2}-p \right )+\gamma -1+2 \ln \relax (x )+\ln \left (-a^{2}\right )\right ) \Gamma \left (\frac {3}{2}-p \right )+\frac {\Gamma \left (\frac {1}{2}-p \right )}{x^{2} a^{2}}\right )}{2 \Gamma \left (\frac {1}{2}-p \right )} \]

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^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} {\left (-a^{2} x^{2} + 1\right )}^{p - \frac {1}{2}}}{x^{3}}\,{d x} \]

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^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-a^2\,x^2\right )}^p\,\left (a\,x+1\right )}{x^3\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 56.40, size = 287, normalized size = 3.68 \[ - \frac {a a^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (\frac {1}{2} - p\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p - \frac {1}{2} \\ p + \frac {1}{2}, p + 1 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } x \Gamma \left (\frac {3}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac {a a^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (\frac {1}{2} - p\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, \frac {1}{2} - p \\ \frac {1}{2}, \frac {3}{2} - p \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } x \Gamma \left (\frac {3}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac {a^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (1 - p\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p - 1 \\ p, p + 1 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } x^{2} \Gamma \left (2 - p\right ) \Gamma \left (p + 1\right )} - \frac {a^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (1 - p\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, 1 - p \\ \frac {1}{2}, 2 - p \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } x^{2} \Gamma \left (2 - p\right ) \Gamma \left (p + 1\right )} \]

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**3,x)

[Out]

-a*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*a**(2*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)*gamm
a(p + 1)) - a**(2*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)*x**(2*p)*exp(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*gamma(2 - p)*gamma(p + 1
))

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