### 3.409 $$\int \frac{e^{4 \coth ^{-1}(a x)}}{c-\frac{c}{a x}} \, dx$$

Optimal. Leaf size=53 $\frac{8}{a c (1-a x)}-\frac{2}{a c (1-a x)^2}+\frac{5 \log (1-a x)}{a c}+\frac{x}{c}$

[Out]

x/c - 2/(a*c*(1 - a*x)^2) + 8/(a*c*(1 - a*x)) + (5*Log[1 - a*x])/(a*c)

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Rubi [A]  time = 0.133421, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {6167, 6131, 6129, 77} $\frac{8}{a c (1-a x)}-\frac{2}{a c (1-a x)^2}+\frac{5 \log (1-a x)}{a c}+\frac{x}{c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/c - 2/(a*c*(1 - a*x)^2) + 8/(a*c*(1 - a*x)) + (5*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 6131

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 + (c*x)/d)
^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{e^{4 \coth ^{-1}(a x)}}{c-\frac{c}{a x}} \, dx &=\int \frac{e^{4 \tanh ^{-1}(a x)}}{c-\frac{c}{a x}} \, dx\\ &=-\frac{a \int \frac{e^{4 \tanh ^{-1}(a x)} x}{1-a x} \, dx}{c}\\ &=-\frac{a \int \frac{x (1+a x)^2}{(1-a x)^3} \, dx}{c}\\ &=-\frac{a \int \left (-\frac{1}{a}-\frac{4}{a (-1+a x)^3}-\frac{8}{a (-1+a x)^2}-\frac{5}{a (-1+a x)}\right ) \, dx}{c}\\ &=\frac{x}{c}-\frac{2}{a c (1-a x)^2}+\frac{8}{a c (1-a x)}+\frac{5 \log (1-a x)}{a c}\\ \end{align*}

Mathematica [A]  time = 0.0304802, size = 51, normalized size = 0.96 $-\frac{a \left (-\frac{8}{a^2 (1-a x)}+\frac{2}{a^2 (1-a x)^2}-\frac{5 \log (1-a x)}{a^2}-\frac{x}{a}\right )}{c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((a*(-(x/a) + 2/(a^2*(1 - a*x)^2) - 8/(a^2*(1 - a*x)) - (5*Log[1 - a*x])/a^2))/c)

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Maple [A]  time = 0.045, size = 51, normalized size = 1. \begin{align*}{\frac{x}{c}}-2\,{\frac{1}{ac \left ( ax-1 \right ) ^{2}}}+5\,{\frac{\ln \left ( ax-1 \right ) }{ac}}-8\,{\frac{1}{ac \left ( ax-1 \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c-2/a/c/(a*x-1)^2+5/a/c*ln(a*x-1)-8/a/c/(a*x-1)

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Maxima [A]  time = 1.03053, size = 66, normalized size = 1.25 \begin{align*} -\frac{2 \,{\left (4 \, a x - 3\right )}}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} + \frac{x}{c} + \frac{5 \, \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/(c-c/a/x),x, algorithm="maxima")

[Out]

-2*(4*a*x - 3)/(a^3*c*x^2 - 2*a^2*c*x + a*c) + x/c + 5*log(a*x - 1)/(a*c)

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Fricas [A]  time = 1.43967, size = 140, normalized size = 2.64 \begin{align*} \frac{a^{3} x^{3} - 2 \, a^{2} x^{2} - 7 \, a x + 5 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 6}{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/(c-c/a/x),x, algorithm="fricas")

[Out]

(a^3*x^3 - 2*a^2*x^2 - 7*a*x + 5*(a^2*x^2 - 2*a*x + 1)*log(a*x - 1) + 6)/(a^3*c*x^2 - 2*a^2*c*x + a*c)

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Sympy [A]  time = 0.429769, size = 41, normalized size = 0.77 \begin{align*} - \frac{8 a x - 6}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac{x}{c} + \frac{5 \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/(c-c/a/x),x)

[Out]

-(8*a*x - 6)/(a**3*c*x**2 - 2*a**2*c*x + a*c) + x/c + 5*log(a*x - 1)/(a*c)

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Giac [A]  time = 1.11963, size = 100, normalized size = 1.89 \begin{align*} \frac{a x - 1}{a c} - \frac{5 \, \log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | a \right |}}\right )}{a c} - \frac{2 \,{\left (\frac{4 \, a^{3} c}{a x - 1} + \frac{a^{3} c}{{\left (a x - 1\right )}^{2}}\right )}}{a^{4} 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/(c-c/a/x),x, algorithm="giac")

[Out]

(a*x - 1)/(a*c) - 5*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/(a*c) - 2*(4*a^3*c/(a*x - 1) + a^3*c/(a*x - 1)^2)/(
a^4*c^2)