### 3.393 $$\int \frac{e^{2 \coth ^{-1}(a x)}}{(c-\frac{c}{a x})^2} \, dx$$

Optimal. Leaf size=53 $\frac{5}{a c^2 (1-a x)}-\frac{1}{a c^2 (1-a x)^2}+\frac{4 \log (1-a x)}{a c^2}+\frac{x}{c^2}$

[Out]

x/c^2 - 1/(a*c^2*(1 - a*x)^2) + 5/(a*c^2*(1 - a*x)) + (4*Log[1 - a*x])/(a*c^2)

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Rubi [A]  time = 0.15467, 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} $\frac{5}{a c^2 (1-a x)}-\frac{1}{a c^2 (1-a x)^2}+\frac{4 \log (1-a x)}{a c^2}+\frac{x}{c^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/c^2 - 1/(a*c^2*(1 - a*x)^2) + 5/(a*c^2*(1 - a*x)) + (4*Log[1 - a*x])/(a*c^2)

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

Mathematica [A]  time = 0.09914, size = 51, normalized size = 0.96 $-\frac{5}{a c^2 (a x-1)}-\frac{1}{a c^2 (a x-1)^2}+\frac{4 \log (1-a x)}{a c^2}+\frac{x}{c^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/c^2 - 1/(a*c^2*(-1 + a*x)^2) - 5/(a*c^2*(-1 + a*x)) + (4*Log[1 - a*x])/(a*c^2)

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Maple [A]  time = 0.043, size = 51, normalized size = 1. \begin{align*}{\frac{x}{{c}^{2}}}-{\frac{1}{a{c}^{2} \left ( ax-1 \right ) ^{2}}}+4\,{\frac{\ln \left ( ax-1 \right ) }{a{c}^{2}}}-5\,{\frac{1}{a{c}^{2} \left ( ax-1 \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04903, size = 74, normalized size = 1.4 \begin{align*} -\frac{5 \, a x - 4}{a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}} + \frac{x}{c^{2}} + \frac{4 \, \log \left (a x - 1\right )}{a c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.5058, size = 149, normalized size = 2.81 \begin{align*} \frac{a^{3} x^{3} - 2 \, a^{2} x^{2} - 4 \, a x + 4 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 4}{a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.435774, size = 49, normalized size = 0.92 \begin{align*} - \frac{5 a x - 4}{a^{3} c^{2} x^{2} - 2 a^{2} c^{2} x + a c^{2}} + \frac{x}{c^{2}} + \frac{4 \log{\left (a x - 1 \right )}}{a c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14232, size = 57, normalized size = 1.08 \begin{align*} \frac{x}{c^{2}} + \frac{4 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{2}} - \frac{5 \, a x - 4}{{\left (a x - 1\right )}^{2} a c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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