3.1.47 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{x} \, dx\) [47]

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {6261, 78} \begin {gather*} \log (x)-2 \log (a x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - 2*Log[1 + a*x]

Rule 78

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]))))

Rule 6261

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 \begin {gather*} \log (x)-2 \log (1+a x) \end {gather*}

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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Maple [A]
time = 0.50, size = 12, normalized size = 1.09

method	result	size
default	\(\ln \left (x \right )-2 \ln \left (a x +1\right )\)	\(12\)
norman	\(\ln \left (x \right )-2 \ln \left (a x +1\right )\)	\(12\)
risch	\(\ln \left (-x \right )-2 \ln \left (a x +1\right )\)	\(14\)
meijerg	\(\frac {a x}{a x +1}-2 \ln \left (a x +1\right )-\frac {2 a x}{2 a x +2}+1+\ln \left (x \right )+\ln \left (a \right )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 11, normalized size = 1.00 \begin {gather*} -2 \, \log \left (a x + 1\right ) + \log \left (x\right ) \end {gather*}

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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Fricas [A]
time = 0.33, size = 11, normalized size = 1.00 \begin {gather*} -2 \, \log \left (a x + 1\right ) + \log \left (x\right ) \end {gather*}

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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Sympy [A]
time = 0.06, size = 10, normalized size = 0.91 \begin {gather*} \log {\left (x \right )} - 2 \log {\left (x + \frac {1}{a} \right )} \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (11) = 22\).
time = 0.41, size = 43, normalized size = 3.91 \begin {gather*} 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 )} \end {gather*}

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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Mupad [B]
time = 0.79, size = 12, normalized size = 1.09 \begin {gather*} \ln \left (x\right )-2\,\ln \left (3\,a\,x+3\right ) \end {gather*}

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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