3.1314 \(\int \frac{e^{n \tanh ^{-1}(a x)} x}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=94 \[ \frac{2^{\frac{n}{2}+1} (1-a x)^{-n/2} \text{Hypergeometric2F1}\left (-\frac{n}{2},-\frac{n}{2},1-\frac{n}{2},\frac{1}{2} (1-a x)\right )}{a^2 c n}-\frac{(1-a x)^{-n/2} (a x+1)^{n/2}}{a^2 c n} \]

[Out]

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

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Rubi [A]  time = 0.0828106, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6150, 79, 69} \[ \frac{2^{\frac{n}{2}+1} (1-a x)^{-n/2} \, _2F_1\left (-\frac{n}{2},-\frac{n}{2};1-\frac{n}{2};\frac{1}{2} (1-a x)\right )}{a^2 c n}-\frac{(1-a x)^{-n/2} (a x+1)^{n/2}}{a^2 c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6150

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

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

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

Mathematica [A]  time = 0.055578, size = 74, normalized size = 0.79 \[ \frac{(1-a x)^{-n/2} \left (2^{\frac{n}{2}+1} \text{Hypergeometric2F1}\left (-\frac{n}{2},-\frac{n}{2},1-\frac{n}{2},\frac{1}{2} (1-a x)\right )-(a x+1)^{n/2}\right )}{a^2 c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.184, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{n{\it Artanh} \left ( ax \right ) }}x}{-{a}^{2}c{x}^{2}+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{x e^{n \operatorname{atanh}{\left (a x \right )}}}{a^{2} x^{2} - 1}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x*exp(n*atanh(a*x))/(a**2*x**2 - 1), x)/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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