3.49 \(\int \frac{e^{-2 \tanh ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=32 \[ 2 a^2 \log (x)-2 a^2 \log (a x+1)+\frac{2 a}{x}-\frac{1}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0297747, antiderivative size = 32, 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} \[ 2 a^2 \log (x)-2 a^2 \log (a x+1)+\frac{2 a}{x}-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*x^3),x]

[Out]

-1/(2*x^2) + (2*a)/x + 2*a^2*Log[x] - 2*a^2*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^3} \, dx &=\int \frac{1-a x}{x^3 (1+a x)} \, dx\\ &=\int \left (\frac{1}{x^3}-\frac{2 a}{x^2}+\frac{2 a^2}{x}-\frac{2 a^3}{1+a x}\right ) \, dx\\ &=-\frac{1}{2 x^2}+\frac{2 a}{x}+2 a^2 \log (x)-2 a^2 \log (1+a x)\\ \end{align*}

Mathematica [A]  time = 0.0101006, size = 32, normalized size = 1. \[ 2 a^2 \log (x)-2 a^2 \log (a x+1)+\frac{2 a}{x}-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*x^3),x]

[Out]

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

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Maple [A]  time = 0.035, size = 31, normalized size = 1. \begin{align*} -{\frac{1}{2\,{x}^{2}}}+2\,{\frac{a}{x}}+2\,{a}^{2}\ln \left ( x \right ) -2\,{a}^{2}\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^3,x)

[Out]

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

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Maxima [A]  time = 0.991817, size = 41, normalized size = 1.28 \begin{align*} -2 \, a^{2} \log \left (a x + 1\right ) + 2 \, a^{2} \log \left (x\right ) + \frac{4 \, a x - 1}{2 \, x^{2}} \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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.8204, size = 89, normalized size = 2.78 \begin{align*} -\frac{4 \, a^{2} x^{2} \log \left (a x + 1\right ) - 4 \, a^{2} x^{2} \log \left (x\right ) - 4 \, a x + 1}{2 \, x^{2}} \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^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.669226, size = 26, normalized size = 0.81 \begin{align*} - 2 a^{2} \left (- \log{\left (x \right )} + \log{\left (x + \frac{1}{a} \right )}\right ) + \frac{4 a x - 1}{2 x^{2}} \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**3,x)

[Out]

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

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Giac [A]  time = 1.15121, size = 68, normalized size = 2.12 \begin{align*} 2 \, a^{2} \log \left ({\left | -\frac{1}{a x + 1} + 1 \right |}\right ) + \frac{5 \, a^{2} - \frac{6 \, a^{2}}{a x + 1}}{2 \,{\left (\frac{1}{a x + 1} - 1\right )}^{2}} \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^3,x, algorithm="giac")

[Out]

2*a^2*log(abs(-1/(a*x + 1) + 1)) + 1/2*(5*a^2 - 6*a^2/(a*x + 1))/(1/(a*x + 1) - 1)^2