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

Optimal. Leaf size=18 \[ \frac{\tanh ^{-1}(a x)}{a c^2}-\frac{x}{c^2} \]

[Out]

-(x/c^2) + ArcTanh[a*x]/(a*c^2)

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Rubi [A]  time = 0.109727, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6131, 6129, 72, 207} \[ \frac{\tanh ^{-1}(a x)}{a c^2}-\frac{x}{c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - c/(a*x))^2),x]

[Out]

-(x/c^2) + ArcTanh[a*x]/(a*c^2)

Rule 6131

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

Rule 6129

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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

Mathematica [B]  time = 0.0728216, size = 40, normalized size = 2.22 \[ -\frac{\log (1-a x)}{2 a c^2}+\frac{\log (a x+1)}{2 a c^2}-\frac{x}{c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - c/(a*x))^2),x]

[Out]

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

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Maple [A]  time = 0.036, size = 36, normalized size = 2. \begin{align*} -{\frac{x}{{c}^{2}}}+{\frac{\ln \left ( ax+1 \right ) }{2\,a{c}^{2}}}-{\frac{\ln \left ( ax-1 \right ) }{2\,a{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.964698, size = 47, normalized size = 2.61 \begin{align*} -\frac{x}{c^{2}} + \frac{\log \left (a x + 1\right )}{2 \, a c^{2}} - \frac{\log \left (a x - 1\right )}{2 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.20241, size = 70, normalized size = 3.89 \begin{align*} -\frac{2 \, a x - \log \left (a x + 1\right ) + \log \left (a x - 1\right )}{2 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.353467, size = 36, normalized size = 2. \begin{align*} - a^{2} \left (\frac{x}{a^{2} c^{2}} + \frac{\frac{\log{\left (x - \frac{1}{a} \right )}}{2} - \frac{\log{\left (x + \frac{1}{a} \right )}}{2}}{a^{3} c^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21909, size = 47, normalized size = 2.61 \begin{align*} -\frac{a x + 1}{a c^{2}} - \frac{\log \left ({\left | -\frac{2}{a x + 1} + 1 \right |}\right )}{2 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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