3.48 \(\int \frac{e^{-2 \tanh ^{-1}(a x)}}{x^2} \, dx\)

Optimal. Leaf size=20 \[ -2 a \log (x)+2 a \log (a x+1)-\frac{1}{x} \]

[Out]

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

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Rubi [A]  time = 0.0270994, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6126, 77} \[ -2 a \log (x)+2 a \log (a x+1)-\frac{1}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 \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 = 0.0086829, size = 20, 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*ArcTanh[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 = 0.034, size = 21, 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)^2*(-a^2*x^2+1)/x^2,x)

[Out]

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

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Maxima [A]  time = 0.946294, size = 27, normalized size = 1.35 \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(1/(a*x+1)^2*(-a^2*x^2+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 = 1.96922, size = 58, normalized size = 2.9 \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(1/(a*x+1)^2*(-a^2*x^2+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 = 1.07266, size = 17, normalized size = 0.85 \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(1/(a*x+1)**2*(-a**2*x**2+1)/x**2,x)

[Out]

-2*a*(log(x) - log(x + 1/a)) - 1/x

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Giac [A]  time = 1.19509, size = 41, normalized size = 2.05 \begin{align*} -2 \, a \log \left ({\left | -\frac{1}{a x + 1} + 1 \right |}\right ) + \frac{a}{\frac{1}{a x + 1} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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