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

Optimal. Leaf size=31 \[ \frac {2}{a c (1-a x)}+\frac {\log (1-a x)}{a c} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6129, 43} \[ \frac {2}{a c (1-a x)}+\frac {\log (1-a x)}{a c} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])/(c - a*c*x),x]

[Out]

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

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])

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

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Mathematica [A]  time = 0.02, size = 25, normalized size = 0.81 \[ \frac {\frac {2}{1-a x}+\log (1-a x)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2/(1 - a*x) + Log[1 - a*x])/(a*c)

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fricas [A]  time = 0.46, size = 28, normalized size = 0.90 \[ \frac {{\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{a^{2} c x - a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 30, normalized size = 0.97 \[ \frac {\log \left ({\left | a x - 1 \right |}\right )}{a c} - \frac {2}{{\left (a x - 1\right )} a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 30, normalized size = 0.97 \[ \frac {\ln \left (a x -1\right )}{c a}-\frac {2}{c a \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 29, normalized size = 0.94 \[ -\frac {2}{a^{2} c x - a c} + \frac {\log \left (a x - 1\right )}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.80, size = 28, normalized size = 0.90 \[ \frac {2}{a\,\left (c-a\,c\,x\right )}+\frac {\ln \left (a\,x-1\right )}{a\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.14, size = 20, normalized size = 0.65 \[ - \frac {2}{a^{2} c x - a c} + \frac {\log {\left (a x - 1 \right )}}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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