3.1007 \(\int e^{\tanh ^{-1}(a x)} x (1-a^2 x^2)^p \, dx\)

Optimal. Leaf size=58 \[ \frac {1}{3} a x^3 \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2 (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6148, 764, 261, 364} \[ \frac {1}{3} a x^3 \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - a^2*x^2)^(1/2 + p)/(a^2*(1 + 2*p))) + (a*x^3*Hypergeometric2F1[3/2, 1/2 - p, 5/2, a^2*x^2])/3

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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 e^{\tanh ^{-1}(a x)} x \left (1-a^2 x^2\right )^p \, dx &=\int x (1+a x) \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=a \int x^2 \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx+\int x \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p}}{a^2 (1+2 p)}+\frac {1}{3} a x^3 \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 60, normalized size = 1.03 \[ \frac {1}{3} a x^3 \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{2 a^2 \left (p+\frac {1}{2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(1 - a^2*x^2)^(1/2 + p)/(a^2*(1/2 + p)) + (a*x^3*Hypergeometric2F1[3/2, 1/2 - p, 5/2, a^2*x^2])/3

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} x^{2} + 1\right )}^{p} x}{a x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.34, size = 47, normalized size = 0.81 \[ \frac {a \,x^{3} \hypergeom \left (\left [\frac {3}{2}, \frac {1}{2}-p \right ], \left [\frac {5}{2}\right ], a^{2} x^{2}\right )}{3}+\frac {x^{2} \hypergeom \left (\left [1, \frac {1}{2}-p \right ], \relax [2], a^{2} x^{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x\,{\left (1-a^2\,x^2\right )}^p\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 12.10, size = 301, normalized size = 5.19 \[ - \frac {a a^{2 p} x^{3} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p + \frac {3}{2} \\ p + 1, p + \frac {5}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {1}{2}\right ) \Gamma \left (p + 1\right )} - \frac {a a^{2 p} x^{3} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, - p - \frac {3}{2} \\ \frac {1}{2}, - p - \frac {1}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {1}{2}\right ) \Gamma \left (p + 1\right )} - \frac {a^{2 p} x^{2} x^{2 p} e^{i \pi p} \Gamma \left (- p - 1\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, 1 \\ p + 2 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} - \frac {a^{2 p} x^{2} x^{2 p} e^{i \pi p} \Gamma \left (- p - 1\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} 1, - p - 1 \\ \frac {1}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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