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

Optimal. Leaf size=39 \[ -\frac {1}{2} \text {Li}_2\left (\frac {a}{b x^2}+1\right )-\frac {1}{2} \log \left (\frac {a}{x^2}+b\right ) \log \left (-\frac {a}{b x^2}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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 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 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]

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 = 0.00, size = 40, normalized size = 1.03 \[ -\frac {1}{2} \text {Li}_2\left (\frac {\frac {a}{x^2}+b}{b}\right )-\frac {1}{2} \log \left (\frac {a}{x^2}+b\right ) \log \left (-\frac {a}{b x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(a + b*x^2)/x^2]/x,x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (\frac {b x^{2} + a}{x^{2}}\right )}{x}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (\frac {b x^{2} + a}{x^{2}}\right )}{x}\,{d x} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.09, size = 108, normalized size = 2.77 \[ \ln \left (\frac {1}{x}\right ) \ln \left (\frac {-\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )+\ln \left (\frac {1}{x}\right ) \ln \left (\frac {\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )-\ln \left (\frac {1}{x}\right ) \ln \left (b +\frac {a}{x^{2}}\right )+\dilog \left (\frac {-\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )+\dilog \left (\frac {\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right ) \]

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))

________________________________________________________________________________________

maxima [B]  time = 0.47, size = 77, normalized size = 1.97 \[ -{\left (\log \left (b x^{2} + a\right ) - 2 \, \log \relax (x)\right )} \log \relax (x) + \log \left (b x^{2} + a\right ) \log \relax (x) - \log \left (\frac {b x^{2}}{a} + 1\right ) \log \relax (x) - \log \relax (x)^{2} + \log \relax (x) \log \left (\frac {b x^{2} + a}{x^{2}}\right ) - \frac {1}{2} \, {\rm Li}_2\left (-\frac {b x^{2}}{a}\right ) \]

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)

________________________________________________________________________________________

mupad [B]  time = 0.44, size = 33, normalized size = 0.85 \[ -\frac {{\mathrm {Li}}_{\mathrm {2}}\left (-\frac {a}{b\,x^2}\right )}{2}-\frac {\ln \left (b+\frac {a}{x^2}\right )\,\ln \left (-\frac {a}{b\,x^2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((a + b*x^2)/x^2)/x,x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (\frac {a}{x^{2}} + b \right )}}{x}\, dx \]

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)

________________________________________________________________________________________