### 3.785 $$\int \frac{e^{2 \coth ^{-1}(a x)}}{c-\frac{c}{a^2 x^2}} \, dx$$

Optimal. Leaf size=36 $\frac{1}{a c (1-a x)}+\frac{2 \log (1-a x)}{a c}+\frac{x}{c}$

[Out]

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

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Rubi [A]  time = 0.159508, antiderivative size = 36, 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, 6157, 6150, 43} $\frac{1}{a c (1-a x)}+\frac{2 \log (1-a x)}{a c}+\frac{x}{c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/c + 1/(a*c*(1 - a*x)) + (2*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 6157

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

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 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^{2 \coth ^{-1}(a x)}}{c-\frac{c}{a^2 x^2}} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)}}{c-\frac{c}{a^2 x^2}} \, dx\\ &=\frac{a^2 \int \frac{e^{2 \tanh ^{-1}(a x)} x^2}{1-a^2 x^2} \, dx}{c}\\ &=\frac{a^2 \int \frac{x^2}{(1-a x)^2} \, dx}{c}\\ &=\frac{a^2 \int \left (\frac{1}{a^2}+\frac{1}{a^2 (-1+a x)^2}+\frac{2}{a^2 (-1+a x)}\right ) \, dx}{c}\\ &=\frac{x}{c}+\frac{1}{a c (1-a x)}+\frac{2 \log (1-a x)}{a c}\\ \end{align*}

Mathematica [A]  time = 0.0290702, size = 28, normalized size = 0.78 $\frac{\frac{1}{a-a^2 x}+\frac{2 \log (1-a x)}{a}+x}{c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 36, normalized size = 1. \begin{align*}{\frac{x}{c}}+2\,{\frac{\ln \left ( ax-1 \right ) }{ac}}-{\frac{1}{ac \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^2/x^2),x)

[Out]

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

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Maxima [A]  time = 1.05927, size = 47, normalized size = 1.31 \begin{align*} \frac{x}{c} - \frac{1}{a^{2} c x - a c} + \frac{2 \, \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)*(a*x+1)/(c-c/a^2/x^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.52547, size = 86, normalized size = 2.39 \begin{align*} \frac{a^{2} x^{2} - a x + 2 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 1}{a^{2} c x - a c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.31619, size = 36, normalized size = 1. \begin{align*} a^{2} \left (- \frac{1}{a^{4} c x - a^{3} c} + \frac{x}{a^{2} c} + \frac{2 \log{\left (a x - 1 \right )}}{a^{3} c}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.08939, size = 49, normalized size = 1.36 \begin{align*} \frac{x}{c} + \frac{2 \, \log \left ({\left | a x - 1 \right |}\right )}{a c} - \frac{1}{{\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)*(a*x+1)/(c-c/a^2/x^2),x, algorithm="giac")

[Out]

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