3.426 $$\int \frac{e^{-2 \coth ^{-1}(a x)}}{(c-\frac{c}{a x})^2} \, dx$$

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

[Out]

x/c^2 - ArcTanh[a*x]/(a*c^2)

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Rubi [A]  time = 0.136995, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.227, Rules used = {6167, 6131, 6129, 72, 207} $\frac{x}{c^2}-\frac{\tanh ^{-1}(a x)}{a c^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/c^2 - ArcTanh[a*x]/(a*c^2)

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/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 \coth ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^2} \, dx &=-\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 = 0.0825677, size = 39, normalized size = 2.17 $\frac{\log (1-a x)}{2 a c^2}-\frac{\log (a x+1)}{2 a c^2}+\frac{x}{c^2}$

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(2*ArcCoth[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 = 0.046, size = 35, normalized size = 1.9 \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)*(a*x-1)/(c-c/a/x)^2,x)

[Out]

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

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Maxima [A]  time = 1.01405, size = 46, normalized size = 2.56 \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((a*x-1)/(a*x+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 = 1.80369, size = 69, normalized size = 3.83 \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((a*x-1)/(a*x+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 = 0.328685, size = 34, normalized size = 1.89 \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((a*x-1)/(a*x+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 [B]  time = 1.16819, size = 49, normalized size = 2.72 \begin{align*} \frac{x}{c^{2}} - \frac{\log \left ({\left | a x + 1 \right |}\right )}{2 \, a c^{2}} + \frac{\log \left ({\left | a x - 1 \right |}\right )}{2 \, a c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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