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

Optimal. Leaf size=11 \[ \log (x)-2 \log (a x+1) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 11, normalized size of antiderivative = 1.00, 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, 72} \[ \log (x)-2 \log (a x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - 2*Log[1 + a*x]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 11, normalized size = 1.00 \[ \log (x)-2 \log (a x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - 2*Log[1 + a*x]

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fricas [A]  time = 0.42, size = 11, normalized size = 1.00 \[ -2 \, \log \left (a x + 1\right ) + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(a*x + 1) + log(x)

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giac [B]  time = 0.36, size = 43, normalized size = 3.91 \[ a {\left (\frac {\log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a} + \frac {\log \left ({\left | -\frac {1}{a x + 1} + 1 \right |}\right )}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 12, normalized size = 1.09 \[ \ln \relax (x )-2 \ln \left (a x +1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 11, normalized size = 1.00 \[ -2 \, \log \left (a x + 1\right ) + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(a*x + 1) + log(x)

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mupad [B]  time = 0.79, size = 12, normalized size = 1.09 \[ \ln \relax (x)-2\,\ln \left (3\,a\,x+3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^2*x^2 - 1)/(x*(a*x + 1)^2),x)

[Out]

log(x) - 2*log(3*a*x + 3)

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sympy [A]  time = 0.12, size = 10, normalized size = 0.91 \[ \log {\relax (x )} - 2 \log {\left (x + \frac {1}{a} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - 2*log(x + 1/a)

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