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

Optimal. Leaf size=125 \[ -\frac{2^{\frac{n}{2}+1} (1-a x)^{-n/2} \text{Hypergeometric2F1}\left (-\frac{n}{2},-\frac{n}{2},1-\frac{n}{2},\frac{1}{2} (1-a x)\right )}{a c}+\frac{x (a x+1)^{n/2} (1-a x)^{-n/2}}{c}-\frac{(1-n) (a x+1)^{n/2} (1-a x)^{-n/2}}{a c n} \]

[Out]

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

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Rubi [A]  time = 0.171389, antiderivative size = 125, normalized size of antiderivative = 1., 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 = {6157, 6150, 90, 79, 69} \[ -\frac{2^{\frac{n}{2}+1} (1-a x)^{-n/2} \, _2F_1\left (-\frac{n}{2},-\frac{n}{2};1-\frac{n}{2};\frac{1}{2} (1-a x)\right )}{a c}+\frac{x (a x+1)^{n/2} (1-a x)^{-n/2}}{c}-\frac{(1-n) (a x+1)^{n/2} (1-a x)^{-n/2}}{a c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6157

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

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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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)}}{c-\frac{c}{a^2 x^2}} \, dx &=-\frac{a^2 \int \frac{e^{n \tanh ^{-1}(a x)} x^2}{1-a^2 x^2} \, dx}{c}\\ &=-\frac{a^2 \int x^2 (1-a x)^{-1-\frac{n}{2}} (1+a x)^{-1+\frac{n}{2}} \, dx}{c}\\ &=\frac{x (1-a x)^{-n/2} (1+a x)^{n/2}}{c}+\frac{\int (1-a x)^{-1-\frac{n}{2}} (1+a x)^{-1+\frac{n}{2}} (-1-a n x) \, dx}{c}\\ &=-\frac{(1-n) (1-a x)^{-n/2} (1+a x)^{n/2}}{a c n}+\frac{x (1-a x)^{-n/2} (1+a x)^{n/2}}{c}-\frac{n \int (1-a x)^{-1-\frac{n}{2}} (1+a x)^{n/2} \, dx}{c}\\ &=-\frac{(1-n) (1-a x)^{-n/2} (1+a x)^{n/2}}{a c n}+\frac{x (1-a x)^{-n/2} (1+a x)^{n/2}}{c}-\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 c}\\ \end{align*}

Mathematica [A]  time = 0.104928, size = 82, normalized size = 0.66 \[ \frac{(1-a x)^{-n/2} \left ((a x+1)^{n/2} (a n x+n-1)-2^{\frac{n}{2}+1} n \text{Hypergeometric2F1}\left (-\frac{n}{2},-\frac{n}{2},1-\frac{n}{2},\frac{1}{2} (1-a x)\right )\right )}{a c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.044, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{n{\it Artanh} \left ( ax \right ) }} \left ( c-{\frac{c}{{a}^{2}{x}^{2}}} \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a^{2} x^{2} \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))/(c-c/a^2/x^2),x, algorithm="fricas")

[Out]

integral(a^2*x^2*((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{a^{2} \int \frac{x^{2} 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))/(c-c/a**2/x**2),x)

[Out]

a**2*Integral(x**2*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{\left (\frac{a x + 1}{a x - 1}\right )^{\frac{1}{2} \, n}}{c - \frac{c}{a^{2} x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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