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

Optimal. Leaf size=37 \[ \frac{2}{a c (1-a x)}+\frac{3 \log (1-a x)}{a c}+\frac{x}{c} \]

[Out]

x/c + 2/(a*c*(1 - a*x)) + (3*Log[1 - a*x])/(a*c)

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Rubi [A]  time = 0.124074, antiderivative size = 37, 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 = {6167, 6131, 6129, 77} \[ \frac{2}{a c (1-a x)}+\frac{3 \log (1-a x)}{a c}+\frac{x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/c + 2/(a*c*(1 - a*x)) + (3*Log[1 - a*x])/(a*c)

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 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]))))

Rubi steps

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

Mathematica [A]  time = 0.0283811, size = 30, normalized size = 0.81 \[ \frac{a x+\frac{2}{1-a x}+3 \log (1-a x)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x + 2/(1 - a*x) + 3*Log[1 - a*x])/(a*c)

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Maple [A]  time = 0.044, size = 36, normalized size = 1. \begin{align*}{\frac{x}{c}}+3\,{\frac{\ln \left ( ax-1 \right ) }{ac}}-2\,{\frac{1}{ac \left ( ax-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02992, size = 47, normalized size = 1.27 \begin{align*} \frac{x}{c} - \frac{2}{a^{2} c x - a c} + \frac{3 \, \log \left (a x - 1\right )}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.54618, size = 86, normalized size = 2.32 \begin{align*} \frac{a^{2} x^{2} - a x + 3 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{a^{2} c x - a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.362151, size = 26, normalized size = 0.7 \begin{align*} - \frac{2}{a^{2} c x - a c} + \frac{x}{c} + \frac{3 \log{\left (a x - 1 \right )}}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14996, size = 49, normalized size = 1.32 \begin{align*} \frac{x}{c} + \frac{3 \, \log \left ({\left | a x - 1 \right |}\right )}{a c} - \frac{2}{{\left (a x - 1\right )} a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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