3.193 \(\int \frac{e^{4 \coth ^{-1}(a x)}}{c-a c x} \, dx\)

Optimal. Leaf size=48 \[ -\frac{4}{a c (1-a x)}+\frac{2}{a c (1-a x)^2}-\frac{\log (1-a x)}{a c} \]

[Out]

2/(a*c*(1 - a*x)^2) - 4/(a*c*(1 - a*x)) - Log[1 - a*x]/(a*c)

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Rubi [A]  time = 0.0677062, antiderivative size = 48, normalized size of antiderivative = 1., 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} \[ -\frac{4}{a c (1-a x)}+\frac{2}{a c (1-a x)^2}-\frac{\log (1-a x)}{a c} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcCoth[a*x])/(c - a*c*x),x]

[Out]

2/(a*c*(1 - a*x)^2) - 4/(a*c*(1 - a*x)) - 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 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^{4 \coth ^{-1}(a x)}}{c-a c x} \, dx &=\int \frac{e^{4 \tanh ^{-1}(a x)}}{c-a c x} \, dx\\ &=\frac{\int \frac{(1+a x)^2}{(1-a x)^3} \, dx}{c}\\ &=\frac{\int \left (\frac{1}{1-a x}-\frac{4}{(-1+a x)^3}-\frac{4}{(-1+a x)^2}\right ) \, dx}{c}\\ &=\frac{2}{a c (1-a x)^2}-\frac{4}{a c (1-a x)}-\frac{\log (1-a x)}{a c}\\ \end{align*}

Mathematica [A]  time = 0.0206116, size = 36, normalized size = 0.75 \[ \frac{4 a x+(a x-1)^2 (-\log (1-a x))-2}{a c (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcCoth[a*x])/(c - a*c*x),x]

[Out]

(-2 + 4*a*x - (-1 + a*x)^2*Log[1 - a*x])/(a*c*(-1 + a*x)^2)

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Maple [A]  time = 0.047, size = 46, normalized size = 1. \begin{align*} 2\,{\frac{1}{ac \left ( ax-1 \right ) ^{2}}}+4\,{\frac{1}{ac \left ( ax-1 \right ) }}-{\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*x+1)^2/(-a*c*x+c),x)

[Out]

2/c/a/(a*x-1)^2+4/c/a/(a*x-1)-1/c/a*ln(a*x-1)

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Maxima [A]  time = 1.05803, size = 59, normalized size = 1.23 \begin{align*} \frac{2 \,{\left (2 \, a x - 1\right )}}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} - \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*x+1)^2/(-a*c*x+c),x, algorithm="maxima")

[Out]

2*(2*a*x - 1)/(a^3*c*x^2 - 2*a^2*c*x + a*c) - log(a*x - 1)/(a*c)

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Fricas [A]  time = 1.46157, size = 108, normalized size = 2.25 \begin{align*} \frac{4 \, a x -{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 2}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a*x - (a^2*x^2 - 2*a*x + 1)*log(a*x - 1) - 2)/(a^3*c*x^2 - 2*a^2*c*x + a*c)

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Sympy [A]  time = 0.729809, size = 36, normalized size = 0.75 \begin{align*} \frac{4 a x - 2}{a^{3} c x^{2} - 2 a^{2} c x + a c} - \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*x+1)**2/(-a*c*x+c),x)

[Out]

(4*a*x - 2)/(a**3*c*x**2 - 2*a**2*c*x + a*c) - log(a*x - 1)/(a*c)

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Giac [A]  time = 1.15814, size = 77, normalized size = 1.6 \begin{align*} \frac{\log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | a \right |}}\right )}{a c} + \frac{2 \,{\left (\frac{2 \, a c}{a x - 1} + \frac{a c}{{\left (a x - 1\right )}^{2}}\right )}}{a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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