### 3.47 $$\int \frac{e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx$$

Optimal. Leaf size=18 $2 a \log (x)-2 a \log (a x+1)+\frac{1}{x}$

[Out]

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

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Rubi [A]  time = 0.0432672, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {6167, 6126, 77} $2 a \log (x)-2 a \log (a x+1)+\frac{1}{x}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(E^(2*ArcCoth[a*x])*x^2),x]

[Out]

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

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 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

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

Mathematica [A]  time = 0.0092316, size = 18, normalized size = 1. $2 a \log (x)-2 a \log (a x+1)+\frac{1}{x}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*x^2),x]

[Out]

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

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Maple [A]  time = 0.045, size = 19, normalized size = 1.1 \begin{align*}{x}^{-1}+2\,a\ln \left ( x \right ) -2\,a\ln \left ( ax+1 \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00079, size = 24, normalized size = 1.33 \begin{align*} -2 \, a \log \left (a x + 1\right ) + 2 \, a \log \left (x\right ) + \frac{1}{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*log(a*x + 1) + 2*a*log(x) + 1/x

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Fricas [A]  time = 1.52237, size = 59, normalized size = 3.28 \begin{align*} -\frac{2 \, a x \log \left (a x + 1\right ) - 2 \, a x \log \left (x\right ) - 1}{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.318615, size = 15, normalized size = 0.83 \begin{align*} 2 a \left (\log{\left (x \right )} - \log{\left (x + \frac{1}{a} \right )}\right ) + \frac{1}{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*(log(x) - log(x + 1/a)) + 1/x

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Giac [A]  time = 1.14546, size = 27, normalized size = 1.5 \begin{align*} -2 \, a \log \left ({\left | a x + 1 \right |}\right ) + 2 \, a \log \left ({\left | x \right |}\right ) + \frac{1}{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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