3.1072 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=86 \[ \frac{3}{16 a c^3 (1-a x)}-\frac{1}{16 a c^3 (a x+1)}+\frac{1}{8 a c^3 (1-a x)^2}+\frac{1}{12 a c^3 (1-a x)^3}+\frac{\tanh ^{-1}(a x)}{4 a c^3} \]

[Out]

1/(12*a*c^3*(1 - a*x)^3) + 1/(8*a*c^3*(1 - a*x)^2) + 3/(16*a*c^3*(1 - a*x)) - 1/(16*a*c^3*(1 + a*x)) + ArcTanh
[a*x]/(4*a*c^3)

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Rubi [A]  time = 0.0676645, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6140, 44, 207} \[ \frac{3}{16 a c^3 (1-a x)}-\frac{1}{16 a c^3 (a x+1)}+\frac{1}{8 a c^3 (1-a x)^2}+\frac{1}{12 a c^3 (1-a x)^3}+\frac{\tanh ^{-1}(a x)}{4 a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(12*a*c^3*(1 - a*x)^3) + 1/(8*a*c^3*(1 - a*x)^2) + 3/(16*a*c^3*(1 - a*x)) - 1/(16*a*c^3*(1 + a*x)) + ArcTanh
[a*x]/(4*a*c^3)

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{e^{2 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx &=\frac{\int \frac{1}{(1-a x)^4 (1+a x)^2} \, dx}{c^3}\\ &=\frac{\int \left (\frac{1}{4 (-1+a x)^4}-\frac{1}{4 (-1+a x)^3}+\frac{3}{16 (-1+a x)^2}+\frac{1}{16 (1+a x)^2}-\frac{1}{4 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^3}\\ &=\frac{1}{12 a c^3 (1-a x)^3}+\frac{1}{8 a c^3 (1-a x)^2}+\frac{3}{16 a c^3 (1-a x)}-\frac{1}{16 a c^3 (1+a x)}-\frac{\int \frac{1}{-1+a^2 x^2} \, dx}{4 c^3}\\ &=\frac{1}{12 a c^3 (1-a x)^3}+\frac{1}{8 a c^3 (1-a x)^2}+\frac{3}{16 a c^3 (1-a x)}-\frac{1}{16 a c^3 (1+a x)}+\frac{\tanh ^{-1}(a x)}{4 a c^3}\\ \end{align*}

Mathematica [A]  time = 0.0313864, size = 63, normalized size = 0.73 \[ -\frac{3 a^3 x^3-6 a^2 x^2+a x-3 (a x-1)^3 (a x+1) \tanh ^{-1}(a x)+4}{12 a c^3 (a x-1)^3 (a x+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4 + a*x - 6*a^2*x^2 + 3*a^3*x^3 - 3*(-1 + a*x)^3*(1 + a*x)*ArcTanh[a*x])/(12*a*c^3*(-1 + a*x)^3*(1 + a*x))

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Maple [A]  time = 0.036, size = 90, normalized size = 1.1 \begin{align*} -{\frac{1}{16\,a{c}^{3} \left ( ax+1 \right ) }}+{\frac{\ln \left ( ax+1 \right ) }{8\,a{c}^{3}}}-{\frac{1}{12\,a{c}^{3} \left ( ax-1 \right ) ^{3}}}+{\frac{1}{8\,a{c}^{3} \left ( ax-1 \right ) ^{2}}}-{\frac{3}{16\,a{c}^{3} \left ( ax-1 \right ) }}-{\frac{\ln \left ( ax-1 \right ) }{8\,a{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a/c^3/(a*x+1)+1/8*ln(a*x+1)/a/c^3-1/12/c^3/a/(a*x-1)^3+1/8/c^3/a/(a*x-1)^2-3/16/c^3/a/(a*x-1)-1/8/c^3/a*
ln(a*x-1)

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Maxima [A]  time = 0.963124, size = 123, normalized size = 1.43 \begin{align*} -\frac{3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x + 4}{12 \,{\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} + \frac{\log \left (a x + 1\right )}{8 \, a c^{3}} - \frac{\log \left (a x - 1\right )}{8 \, a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

-1/12*(3*a^3*x^3 - 6*a^2*x^2 + a*x + 4)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3) + 1/8*log(a*x + 1)
/(a*c^3) - 1/8*log(a*x - 1)/(a*c^3)

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Fricas [A]  time = 2.11872, size = 267, normalized size = 3.1 \begin{align*} -\frac{6 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 2 \, a x - 3 \,{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \,{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) + 8}{24 \,{\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

-1/24*(6*a^3*x^3 - 12*a^2*x^2 + 2*a*x - 3*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x + 1) + 3*(a^4*x^4 - 2*a^3*
x^3 + 2*a*x - 1)*log(a*x - 1) + 8)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3)

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Sympy [A]  time = 0.657277, size = 83, normalized size = 0.97 \begin{align*} - \frac{3 a^{3} x^{3} - 6 a^{2} x^{2} + a x + 4}{12 a^{5} c^{3} x^{4} - 24 a^{4} c^{3} x^{3} + 24 a^{2} c^{3} x - 12 a c^{3}} + \frac{- \frac{\log{\left (x - \frac{1}{a} \right )}}{8} + \frac{\log{\left (x + \frac{1}{a} \right )}}{8}}{a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a**3*x**3 - 6*a**2*x**2 + a*x + 4)/(12*a**5*c**3*x**4 - 24*a**4*c**3*x**3 + 24*a**2*c**3*x - 12*a*c**3) +
(-log(x - 1/a)/8 + log(x + 1/a)/8)/(a*c**3)

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Giac [A]  time = 1.13387, size = 100, normalized size = 1.16 \begin{align*} \frac{\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{3}} - \frac{\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{3}} - \frac{3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x + 4}{12 \,{\left (a x + 1\right )}{\left (a x - 1\right )}^{3} a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

1/8*log(abs(a*x + 1))/(a*c^3) - 1/8*log(abs(a*x - 1))/(a*c^3) - 1/12*(3*a^3*x^3 - 6*a^2*x^2 + a*x + 4)/((a*x +
1)*(a*x - 1)^3*a*c^3)