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]

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

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Rubi [A]  time = 0.03, antiderivative size = 13, normalized size of antiderivative = 1.00, 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 31

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

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])

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*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \[ \frac {\log (a x+1)}{a c} \]

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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fricas [A]  time = 0.50, size = 13, normalized size = 1.00 \[ \frac {\log \left (a x + 1\right )}{a c} \]

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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giac [B]  time = 0.76, size = 27, normalized size = 2.08 \[ -\frac {\log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a c} \]

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)

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maple [A]  time = 0.03, size = 14, normalized size = 1.08 \[ \frac {\ln \left (a x +1\right )}{a c} \]

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 = 0.31, size = 13, normalized size = 1.00 \[ \frac {\log \left (a x + 1\right )}{a c} \]

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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mupad [B]  time = 0.03, size = 13, normalized size = 1.00 \[ \frac {\ln \left (a\,x+1\right )}{a\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*x + 1)/(a*c)

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sympy [A]  time = 0.07, size = 10, normalized size = 0.77 \[ \frac {\log {\left (a c x + c \right )}}{a c} \]

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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