3.391 \(\int \frac{\log (\frac{a+x^2}{x^2})}{x} \, dx\)

Optimal. Leaf size=12 \[ \frac{1}{2} \text{PolyLog}\left (2,-\frac{a}{x^2}\right ) \]

[Out]

PolyLog[2, -(a/x^2)]/2

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Rubi [A]  time = 0.0184693, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2461, 2391} \[ \frac{1}{2} \text{PolyLog}\left (2,-\frac{a}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[(a + x^2)/x^2]/x,x]

[Out]

PolyLog[2, -(a/x^2)]/2

Rule 2461

Int[((a_.) + Log[(c_.)*(v_)^(p_.)]*(b_.))^(q_.)*((f_.)*(x_))^(m_.), x_Symbol] :> Int[(f*x)^m*(a + b*Log[c*Expa
ndToSum[v, x]^p])^q, x] /; FreeQ[{a, b, c, f, m, p, q}, x] && BinomialQ[v, x] &&  !BinomialMatchQ[v, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\log \left (\frac{a+x^2}{x^2}\right )}{x} \, dx &=\int \frac{\log \left (1+\frac{a}{x^2}\right )}{x} \, dx\\ &=\frac{1}{2} \text{Li}_2\left (-\frac{a}{x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.002811, size = 12, normalized size = 1. \[ \frac{1}{2} \text{PolyLog}\left (2,-\frac{a}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(a + x^2)/x^2]/x,x]

[Out]

PolyLog[2, -(a/x^2)]/2

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Maple [B]  time = 0.139, size = 76, normalized size = 6.3 \begin{align*} -\ln \left ({x}^{-1} \right ) \ln \left ( 1+{\frac{a}{{x}^{2}}} \right ) +\ln \left ({x}^{-1} \right ) \ln \left ( 1-{\frac{1}{x}\sqrt{-a}} \right ) +\ln \left ({x}^{-1} \right ) \ln \left ( 1+{\frac{1}{x}\sqrt{-a}} \right ) +{\it dilog} \left ( 1+{\frac{1}{x}\sqrt{-a}} \right ) +{\it dilog} \left ( 1-{\frac{1}{x}\sqrt{-a}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln((x^2+a)/x^2)/x,x)

[Out]

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

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Maxima [B]  time = 1.06037, size = 93, normalized size = 7.75 \begin{align*} -{\left (\log \left (x^{2} + a\right ) - 2 \, \log \left (x\right )\right )} \log \left (x\right ) + \log \left (x^{2} + a\right ) \log \left (x\right ) - \log \left (x\right )^{2} - \log \left (x\right ) \log \left (\frac{x^{2}}{a} + 1\right ) + \log \left (x\right ) \log \left (\frac{x^{2} + a}{x^{2}}\right ) - \frac{1}{2} \,{\rm Li}_2\left (-\frac{x^{2}}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log((x^2+a)/x^2)/x,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.59569, size = 42, normalized size = 3.5 \begin{align*} \frac{1}{2} \,{\rm Li}_2\left (-\frac{x^{2} + a}{x^{2}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log((x^2+a)/x^2)/x,x, algorithm="fricas")

[Out]

1/2*dilog(-(x^2 + a)/x^2 + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log{\left (\frac{a}{x^{2}} + 1 \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln((x**2+a)/x**2)/x,x)

[Out]

Integral(log(a/x**2 + 1)/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (\frac{x^{2} + a}{x^{2}}\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log((x^2+a)/x^2)/x,x, algorithm="giac")

[Out]

integrate(log((x^2 + a)/x^2)/x, x)