3.1.48 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{x^2} \, dx\) [48]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 20, 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*} -2 a \log (x)+2 a \log (a x+1)-\frac {1}{x} \end {gather*}

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 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^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*}

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Mathematica [A]
time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} -\frac {1}{x}-2 a \log (x)+2 a \log (1+a x) \end {gather*}

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 = 0.58, size = 21, normalized size = 1.05

method	result	size
default	\(-\frac {1}{x}-2 a \ln \left (x \right )+2 a \ln \left (a x +1\right )\)	\(21\)
risch	\(-\frac {1}{x}-2 a \ln \left (x \right )+2 a \ln \left (-a x -1\right )\)	\(22\)
norman	\(\frac {a^{2} x^{2}-1}{\left (a x +1\right ) x}-2 a \ln \left (x \right )+2 a \ln \left (a x +1\right )\)	\(36\)
meijerg	\(-\frac {a^{2} x}{a x +1}+a \left (\frac {3 a x}{3 a x +3}+2 \ln \left (a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (a \right )-\frac {1}{a x}\right )\)	\(55\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 20, normalized size = 1.00 \begin {gather*} 2 \, a \log \left (a x + 1\right ) - 2 \, a \log \left (x\right ) - \frac {1}{x} \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^2,x, algorithm="maxima")

[Out]

2*a*log(a*x + 1) - 2*a*log(x) - 1/x

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Fricas [A]
time = 0.36, size = 22, normalized size = 1.10 \begin {gather*} \frac {2 \, a x \log \left (a x + 1\right ) - 2 \, a x \log \left (x\right ) - 1}{x} \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^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 = 0.07, size = 17, normalized size = 0.85 \begin {gather*} - 2 a \left (\log {\left (x \right )} - \log {\left (x + \frac {1}{a} \right )}\right ) - \frac {1}{x} \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**2,x)

[Out]

-2*a*(log(x) - log(x + 1/a)) - 1/x

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Giac [A]
time = 0.42, size = 30, normalized size = 1.50 \begin {gather*} -2 \, a \log \left ({\left | -\frac {1}{a x + 1} + 1 \right |}\right ) + \frac {a}{\frac {1}{a x + 1} - 1} \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^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.05, size = 16, normalized size = 0.80 \begin {gather*} 4\,a\,\mathrm {atanh}\left (2\,a\,x+1\right )-\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a*atanh(2*a*x + 1) - 1/x

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