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

Optimal. Leaf size=20 \[ \frac{x}{c}-\frac{\log (a x+1)}{a c} \]

[Out]

x/c - Log[1 + a*x]/(a*c)

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Rubi [A]  time = 0.11509, antiderivative size = 20, 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, 43} \[ \frac{x}{c}-\frac{\log (a x+1)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/c - 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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A]  time = 0.0175695, size = 22, normalized size = 1.1 \[ \frac{a \left (\frac{x}{a}-\frac{\log (a x+1)}{a^2}\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(x/a - Log[1 + a*x]/a^2))/c

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Maple [A]  time = 0.037, size = 21, normalized size = 1.1 \begin{align*}{\frac{x}{c}}-{\frac{\ln \left ( ax+1 \right ) }{ac}} \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),x)

[Out]

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

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Maxima [A]  time = 1.02752, size = 27, normalized size = 1.35 \begin{align*} \frac{x}{c} - \frac{\log \left (a x + 1\right )}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c - log(a*x + 1)/(a*c)

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Fricas [A]  time = 1.76688, size = 38, normalized size = 1.9 \begin{align*} \frac{a x - \log \left (a x + 1\right )}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x - log(a*x + 1))/(a*c)

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Sympy [A]  time = 0.269226, size = 17, normalized size = 0.85 \begin{align*} a \left (\frac{x}{a c} - \frac{\log{\left (a x + 1 \right )}}{a^{2} c}\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),x)

[Out]

a*(x/(a*c) - log(a*x + 1)/(a**2*c))

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Giac [A]  time = 1.13968, size = 28, normalized size = 1.4 \begin{align*} \frac{x}{c} - \frac{\log \left ({\left | a x + 1 \right |}\right )}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c - log(abs(a*x + 1))/(a*c)