3.1061 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x}{(c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=51 \[ -\frac{1}{4 a^2 c^2 (1-a x)}+\frac{1}{4 a^2 c^2 (1-a x)^2}-\frac{\tanh ^{-1}(a x)}{4 a^2 c^2} \]

[Out]

1/(4*a^2*c^2*(1 - a*x)^2) - 1/(4*a^2*c^2*(1 - a*x)) - ArcTanh[a*x]/(4*a^2*c^2)

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Rubi [A]  time = 0.0795022, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6150, 77, 207} \[ -\frac{1}{4 a^2 c^2 (1-a x)}+\frac{1}{4 a^2 c^2 (1-a x)^2}-\frac{\tanh ^{-1}(a x)}{4 a^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(4*a^2*c^2*(1 - a*x)^2) - 1/(4*a^2*c^2*(1 - a*x)) - ArcTanh[a*x]/(4*a^2*c^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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A]  time = 0.0247767, size = 34, normalized size = 0.67 \[ \frac{a x-(a x-1)^2 \tanh ^{-1}(a x)}{4 a^2 c^2 (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 60, normalized size = 1.2 \begin{align*} -{\frac{\ln \left ( ax+1 \right ) }{8\,{a}^{2}{c}^{2}}}+{\frac{1}{4\,{a}^{2}{c}^{2} \left ( ax-1 \right ) ^{2}}}+{\frac{1}{4\,{a}^{2}{c}^{2} \left ( ax-1 \right ) }}+{\frac{\ln \left ( ax-1 \right ) }{8\,{a}^{2}{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/c^2/a^2*ln(a*x+1)+1/4/c^2/a^2/(a*x-1)^2+1/4/c^2/a^2/(a*x-1)+1/8/c^2/a^2*ln(a*x-1)

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Maxima [A]  time = 0.962135, size = 80, normalized size = 1.57 \begin{align*} \frac{x}{4 \,{\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} - \frac{\log \left (a x + 1\right )}{8 \, a^{2} c^{2}} + \frac{\log \left (a x - 1\right )}{8 \, a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.32214, size = 169, normalized size = 3.31 \begin{align*} \frac{2 \, a x -{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) +{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right )}{8 \,{\left (a^{4} c^{2} x^{2} - 2 \, a^{3} c^{2} x + a^{2} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.449353, size = 53, normalized size = 1.04 \begin{align*} \frac{x}{4 a^{3} c^{2} x^{2} - 8 a^{2} c^{2} x + 4 a c^{2}} - \frac{- \frac{\log{\left (x - \frac{1}{a} \right )}}{8} + \frac{\log{\left (x + \frac{1}{a} \right )}}{8}}{a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(4*a**3*c**2*x**2 - 8*a**2*c**2*x + 4*a*c**2) - (-log(x - 1/a)/8 + log(x + 1/a)/8)/(a**2*c**2)

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Giac [A]  time = 1.14136, size = 63, normalized size = 1.24 \begin{align*} -\frac{\log \left ({\left | a x + 1 \right |}\right )}{8 \, a^{2} c^{2}} + \frac{\log \left ({\left | a x - 1 \right |}\right )}{8 \, a^{2} c^{2}} + \frac{x}{4 \,{\left (a x - 1\right )}^{2} a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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