3.1015 \(\int e^{\tanh ^{-1}(a x)} (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=86 \[ -\frac{2^{p+\frac{3}{2}} (1-a x)^{p+\frac{1}{2}} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (-p-\frac{1}{2},p+\frac{1}{2},p+\frac{3}{2},\frac{1}{2} (1-a x)\right )}{a (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0644312, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6143, 6140, 69} \[ -\frac{2^{p+\frac{3}{2}} (1-a x)^{p+\frac{1}{2}} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-p-\frac{1}{2},p+\frac{1}{2};p+\frac{3}{2};\frac{1}{2} (1-a x)\right )}{a (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6143

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

Rule 6140

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

Rule 69

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

Rubi steps

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

Mathematica [A]  time = 0.0173702, size = 86, normalized size = 1. \[ -\frac{2^{p+\frac{1}{2}} (1-a x)^{p+\frac{1}{2}} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (-p-\frac{1}{2},p+\frac{1}{2},p+\frac{3}{2},\frac{1}{2} (1-a x)\right )}{a \left (p+\frac{1}{2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.322, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ax+1 \right ) \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p}{\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)

[Out]

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

[Out]

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

[Out]

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

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Sympy [C]  time = 10.8214, size = 306, normalized size = 3.56 \begin{align*} - \frac{a a^{2 p} c^{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 a^{2 p} c^{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 )} - \frac{a^{2 p} c^{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^{2 p} c^{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 )} \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)

[Out]

-a*a**(2*p)*c**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*ex
p_polar(2*I*pi))/(2*sqrt(pi)*gamma(-p)*gamma(p + 1)) - a*a**(2*p)*c**p*x**2*x**(2*p)*exp(I*pi*p)*gamma(-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)) - a**(2*p)*c**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**(2*p)*c**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)*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}}\,{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, algorithm="giac")

[Out]

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