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

Optimal. Leaf size=39 \[ -\frac{1}{2} \text{PolyLog}\left (2,\frac{a}{b x^2}+1\right )-\frac{1}{2} \log \left (\frac{a}{x^2}+b\right ) \log \left (-\frac{a}{b x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0470272, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2461, 2454, 2394, 2315} \[ -\frac{1}{2} \text{PolyLog}\left (2,\frac{a}{b x^2}+1\right )-\frac{1}{2} \log \left (\frac{a}{x^2}+b\right ) \log \left (-\frac{a}{b x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[(a + b*x^2)/x^2]/x,x]

[Out]

-(Log[b + a/x^2]*Log[-(a/(b*x^2))])/2 - PolyLog[2, 1 + a/(b*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 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.0040833, size = 40, normalized size = 1.03 \[ -\frac{1}{2} \text{PolyLog}\left (2,\frac{\frac{a}{x^2}+b}{b}\right )-\frac{1}{2} \log \left (\frac{a}{x^2}+b\right ) \log \left (-\frac{a}{b x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(a + b*x^2)/x^2]/x,x]

[Out]

-(Log[b + a/x^2]*Log[-(a/(b*x^2))])/2 - PolyLog[2, (b + a/x^2)/b]/2

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Maple [B]  time = 0.093, size = 108, normalized size = 2.8 \begin{align*} -\ln \left ({x}^{-1} \right ) \ln \left ( b+{\frac{a}{{x}^{2}}} \right ) +\ln \left ({x}^{-1} \right ) \ln \left ({ \left ( -{\frac{a}{x}}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ) +\ln \left ({x}^{-1} \right ) \ln \left ({ \left ({\frac{a}{x}}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ) +{\it dilog} \left ({ \left ( -{\frac{a}{x}}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ) +{\it dilog} \left ({ \left ({\frac{a}{x}}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.02814, size = 104, normalized size = 2.67 \begin{align*} -{\left (\log \left (b x^{2} + a\right ) - 2 \, \log \left (x\right )\right )} \log \left (x\right ) + \log \left (b x^{2} + a\right ) \log \left (x\right ) - \log \left (\frac{b x^{2}}{a} + 1\right ) \log \left (x\right ) - \log \left (x\right )^{2} + \log \left (x\right ) \log \left (\frac{b x^{2} + a}{x^{2}}\right ) - \frac{1}{2} \,{\rm Li}_2\left (-\frac{b x^{2}}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((b*x^2 + a)/x^2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a/x**2 + b)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((b*x^2 + a)/x^2)/x, x)