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

Optimal. Leaf size=32 \[ -\frac {2}{a c (1-a x)}-\frac {\log (1-a x)}{a c} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6167, 6129, 43} \[ -\frac {2}{a c (1-a x)}-\frac {\log (1-a x)}{a c} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[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])

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]

Rubi steps

\begin {align*} \int \frac {e^{2 \coth ^{-1}(a x)}}{c-a c x} \, dx &=-\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 = 30, normalized size = 0.94 \[ -\frac {\frac {2}{a (1-a x)}+\frac {\log (1-a x)}{a}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 29, normalized size = 0.91 \[ -\frac {{\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{a^{2} c x - a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+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 = 0.13, size = 31, 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(1/(a*x-1)*(a*x+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 = 0.04, size = 31, 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)/(a*x-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 = 0.31, size = 30, 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(1/(a*x-1)*(a*x+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 = 1.20, size = 29, normalized size = 0.91 \[ -\frac {2}{a\,\left (c-a\,c\,x\right )}-\frac {\ln \left (a\,x-1\right )}{a\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 20, normalized size = 0.62 \[ \frac {2}{a^{2} c x - a c} - \frac {\log {\left (a x - 1 \right )}}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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