3.742 \(\int \frac{e^{n \coth ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=127 \[ -\frac{(n-4 a x) e^{n \coth ^{-1}(a x)}}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}-\frac{12 (n-2 a x) e^{n \coth ^{-1}(a x)}}{a c^3 \left (4-n^2\right ) \left (16-n^2\right ) \left (1-a^2 x^2\right )}+\frac{24 e^{n \coth ^{-1}(a x)}}{a c^3 n \left (n^4-20 n^2+64\right )} \]

[Out]

(24*E^(n*ArcCoth[a*x]))/(a*c^3*n*(64 - 20*n^2 + n^4)) - (E^(n*ArcCoth[a*x])*(n - 4*a*x))/(a*c^3*(16 - n^2)*(1
- a^2*x^2)^2) - (12*E^(n*ArcCoth[a*x])*(n - 2*a*x))/(a*c^3*(4 - n^2)*(16 - n^2)*(1 - a^2*x^2))

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Rubi [A]  time = 0.129511, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {6185, 6183} \[ -\frac{(n-4 a x) e^{n \coth ^{-1}(a x)}}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}-\frac{12 (n-2 a x) e^{n \coth ^{-1}(a x)}}{a c^3 \left (4-n^2\right ) \left (16-n^2\right ) \left (1-a^2 x^2\right )}+\frac{24 e^{n \coth ^{-1}(a x)}}{a c^3 n \left (n^4-20 n^2+64\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcCoth[a*x])/(c - a^2*c*x^2)^3,x]

[Out]

(24*E^(n*ArcCoth[a*x]))/(a*c^3*n*(64 - 20*n^2 + n^4)) - (E^(n*ArcCoth[a*x])*(n - 4*a*x))/(a*c^3*(16 - n^2)*(1
- a^2*x^2)^2) - (12*E^(n*ArcCoth[a*x])*(n - 2*a*x))/(a*c^3*(4 - n^2)*(16 - n^2)*(1 - a^2*x^2))

Rule 6185

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((n + 2*a*(p + 1)*x)*(c + d*x^
2)^(p + 1)*E^(n*ArcCoth[a*x]))/(a*c*(n^2 - 4*(p + 1)^2)), x] - Dist[(2*(p + 1)*(2*p + 3))/(c*(n^2 - 4*(p + 1)^
2)), Int[(c + d*x^2)^(p + 1)*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !In
tegerQ[n/2] && LtQ[p, -1] && NeQ[p, -3/2] && NeQ[n^2 - 4*(p + 1)^2, 0] && (IntegerQ[p] ||  !IntegerQ[n])

Rule 6183

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[E^(n*ArcCoth[a*x])/(a*c*n), x] /; F
reeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2]

Rubi steps

\begin{align*} \int \frac{e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx &=-\frac{e^{n \coth ^{-1}(a x)} (n-4 a x)}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}+\frac{12 \int \frac{e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx}{c \left (16-n^2\right )}\\ &=-\frac{e^{n \coth ^{-1}(a x)} (n-4 a x)}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}-\frac{12 e^{n \coth ^{-1}(a x)} (n-2 a x)}{a c^3 \left (4-n^2\right ) \left (16-n^2\right ) \left (1-a^2 x^2\right )}+\frac{24 \int \frac{e^{n \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{c^2 \left (64-20 n^2+n^4\right )}\\ &=\frac{24 e^{n \coth ^{-1}(a x)}}{a c^3 n \left (64-20 n^2+n^4\right )}-\frac{e^{n \coth ^{-1}(a x)} (n-4 a x)}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}-\frac{12 e^{n \coth ^{-1}(a x)} (n-2 a x)}{a c^3 \left (4-n^2\right ) \left (16-n^2\right ) \left (1-a^2 x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.216089, size = 97, normalized size = 0.76 \[ \frac{\left (4 n^2 \left (3 a^2 x^2-4\right )-8 a n x \left (3 a^2 x^2-5\right )+24 \left (a^2 x^2-1\right )^2-4 a n^3 x+n^4\right ) e^{n \coth ^{-1}(a x)}}{a c^3 n \left (n^2-16\right ) \left (n^2-4\right ) \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcCoth[a*x])/(c - a^2*c*x^2)^3,x]

[Out]

(E^(n*ArcCoth[a*x])*(n^4 - 4*a*n^3*x + 24*(-1 + a^2*x^2)^2 - 8*a*n*x*(-5 + 3*a^2*x^2) + 4*n^2*(-4 + 3*a^2*x^2)
))/(a*c^3*n*(-16 + n^2)*(-4 + n^2)*(-1 + a^2*x^2)^2)

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Maple [A]  time = 0.043, size = 101, normalized size = 0.8 \begin{align*}{\frac{ \left ( 24\,{x}^{4}{a}^{4}-24\,{x}^{3}{a}^{3}n+12\,{a}^{2}{n}^{2}{x}^{2}-4\,a{n}^{3}x-48\,{a}^{2}{x}^{2}+{n}^{4}+40\,nax-16\,{n}^{2}+24 \right ){{\rm e}^{n{\rm arccoth} \left (ax\right )}}}{ \left ({a}^{2}{x}^{2}-1 \right ) ^{2}{c}^{3}a \left ({n}^{2}-16 \right ) \left ({n}^{2}-4 \right ) n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))/(-a^2*c*x^2+c)^3,x)

[Out]

(24*a^4*x^4-24*a^3*n*x^3+12*a^2*n^2*x^2-4*a*n^3*x-48*a^2*x^2+n^4+40*a*n*x-16*n^2+24)*exp(n*arccoth(a*x))/(a^2*
x^2-1)^2/c^3/a/(n^2-16)/(n^2-4)/n

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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}}{{\left (a^{2} c x^{2} - c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

-integrate(((a*x - 1)/(a*x + 1))^(1/2*n)/(a^2*c*x^2 - c)^3, x)

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Fricas [A]  time = 1.66716, size = 373, normalized size = 2.94 \begin{align*} -\frac{{\left (24 \, a^{4} x^{4} + 24 \, a^{3} n x^{3} + n^{4} + 12 \,{\left (a^{2} n^{2} - 4 \, a^{2}\right )} x^{2} - 16 \, n^{2} + 4 \,{\left (a n^{3} - 10 \, a n\right )} x + 24\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{a c^{3} n^{5} - 20 \, a c^{3} n^{3} + 64 \, a c^{3} n +{\left (a^{5} c^{3} n^{5} - 20 \, a^{5} c^{3} n^{3} + 64 \, a^{5} c^{3} n\right )} x^{4} - 2 \,{\left (a^{3} c^{3} n^{5} - 20 \, a^{3} c^{3} n^{3} + 64 \, a^{3} c^{3} n\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

-(24*a^4*x^4 + 24*a^3*n*x^3 + n^4 + 12*(a^2*n^2 - 4*a^2)*x^2 - 16*n^2 + 4*(a*n^3 - 10*a*n)*x + 24)*((a*x - 1)/
(a*x + 1))^(1/2*n)/(a*c^3*n^5 - 20*a*c^3*n^3 + 64*a*c^3*n + (a^5*c^3*n^5 - 20*a^5*c^3*n^3 + 64*a^5*c^3*n)*x^4
- 2*(a^3*c^3*n^5 - 20*a^3*c^3*n^3 + 64*a^3*c^3*n)*x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acoth(a*x))/(-a**2*c*x**2+c)**3,x)

[Out]

Timed out

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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}}{{\left (a^{2} c x^{2} - c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

integrate(-((a*x - 1)/(a*x + 1))^(1/2*n)/(a^2*c*x^2 - c)^3, x)