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

Optimal. Leaf size=13 \[ \frac{\log (a x+1)}{a c} \]

[Out]

Log[1 + a*x]/(a*c)

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Rubi [A]  time = 0.0256917, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6129, 31} \[ \frac{\log (a x+1)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + a*x]/(a*c)

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0059627, size = 13, normalized size = 1. \[ \frac{\log (a x+1)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + a*x]/(a*c)

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Maple [A]  time = 0.03, size = 14, normalized size = 1.1 \begin{align*}{\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)^2*(-a^2*x^2+1)/(-a*c*x+c),x)

[Out]

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

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Maxima [A]  time = 0.943681, size = 18, normalized size = 1.38 \begin{align*} \frac{\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)^2*(-a^2*x^2+1)/(-a*c*x+c),x, algorithm="maxima")

[Out]

log(a*x + 1)/(a*c)

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Fricas [A]  time = 1.57094, size = 27, normalized size = 2.08 \begin{align*} \frac{\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)^2*(-a^2*x^2+1)/(-a*c*x+c),x, algorithm="fricas")

[Out]

log(a*x + 1)/(a*c)

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Sympy [A]  time = 0.084929, size = 10, normalized size = 0.77 \begin{align*} \frac{\log{\left (a c x + c \right )}}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*c*x + c)/(a*c)

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Giac [A]  time = 1.23518, size = 36, normalized size = 2.77 \begin{align*} -\frac{\log \left (\frac{{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2}{\left | a \right |}}\right )}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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