3.1362 \(\int e^{n \tanh ^{-1}(a x)} (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=103 \[ -\frac{2^{\frac{n}{2}+p+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p (1-a x)^{-\frac{n}{2}+p+1} \text{Hypergeometric2F1}\left (-\frac{n}{2}-p,-\frac{n}{2}+p+1,-\frac{n}{2}+p+2,\frac{1}{2} (1-a x)\right )}{a (-n+2 p+2)} \]

[Out]

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

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Rubi [A]  time = 0.085118, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6143, 6140, 69} \[ -\frac{2^{\frac{n}{2}+p+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p (1-a x)^{-\frac{n}{2}+p+1} \, _2F_1\left (-\frac{n}{2}-p,-\frac{n}{2}+p+1;-\frac{n}{2}+p+2;\frac{1}{2} (1-a x)\right )}{a (-n+2 p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(1 + n/2 + p)*(1 - a*x)^(1 - n/2 + p)*(c - a^2*c*x^2)^p*Hypergeometric2F1[-n/2 - p, 1 - n/2 + p, 2 - n/2
+ p, (1 - a*x)/2])/(a*(2 - n + 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^{n \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^{n \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{n}{2}+p} (1+a x)^{\frac{n}{2}+p} \, dx\\ &=-\frac{2^{1+\frac{n}{2}+p} (1-a x)^{1-\frac{n}{2}+p} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-\frac{n}{2}-p,1-\frac{n}{2}+p;2-\frac{n}{2}+p;\frac{1}{2} (1-a x)\right )}{a (2-n+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.0250776, size = 102, normalized size = 0.99 \[ -\frac{2^{\frac{n}{2}+p} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p (1-a x)^{-\frac{n}{2}+p+1} \text{Hypergeometric2F1}\left (-\frac{n}{2}-p,-\frac{n}{2}+p+1,-\frac{n}{2}+p+2,\frac{1}{2} (1-a x)\right )}{a \left (-\frac{n}{2}+p+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.24, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\it Artanh} \left ( ax \right ) }} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))*(-a^2*c*x^2+c)^p,x)

[Out]

int(exp(n*arctanh(a*x))*(-a^2*c*x^2+c)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a^{2} c x^{2} + c\right )}^{p} \left (\frac{a x + 1}{a x - 1}\right )^{\frac{1}{2} \, n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*(-a^2*c*x^2+c)^p,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (-a^{2} c x^{2} + c\right )}^{p} \left (\frac{a x + 1}{a x - 1}\right )^{\frac{1}{2} \, n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*(-a^2*c*x^2+c)^p,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{p} e^{n \operatorname{atanh}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))*(-a**2*c*x**2+c)**p,x)

[Out]

Integral((-c*(a*x - 1)*(a*x + 1))**p*exp(n*atanh(a*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a^{2} c x^{2} + c\right )}^{p} \left (\frac{a x + 1}{a x - 1}\right )^{\frac{1}{2} \, n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*(-a^2*c*x^2+c)^p,x, algorithm="giac")

[Out]

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