\(\int \frac {\log (\frac {a+b x^2}{x^2})}{x} \, dx\) [394]

Optimal result

Integrand size = 16, antiderivative size = 39 \[ \int \frac {\log \left (\frac {a+b x^2}{x^2}\right )}{x} \, dx=-\frac {1}{2} \log \left (b+\frac {a}{x^2}\right ) \log \left (-\frac {a}{b x^2}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,1+\frac {a}{b x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2511, 2504, 2441, 2352} \[ \int \frac {\log \left (\frac {a+b x^2}{x^2}\right )}{x} \, dx=-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {a}{b x^2}+1\right )-\frac {1}{2} \log \left (\frac {a}{x^2}+b\right ) \log \left (-\frac {a}{b x^2}\right ) \]

[In]

Int[Log[(a + b*x^2)/x^2]/x,x]

[Out]

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

Rule 2352

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

Rule 2441

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 2504

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 2511

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*} \text {integral}& = \int \frac {\log \left (b+\frac {a}{x^2}\right )}{x} \, dx \\ & = -\left (\frac {1}{2} \text {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 \text {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] (verified)

Time = 0.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03 \[ \int \frac {\log \left (\frac {a+b x^2}{x^2}\right )}{x} \, dx=-\frac {1}{2} \log \left (b+\frac {a}{x^2}\right ) \log \left (-\frac {a}{b x^2}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {b+\frac {a}{x^2}}{b}\right ) \]

[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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(107\) vs. \(2(35)=70\).

Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.77

method	result	size
risch	\(-\ln \left (\frac {1}{x}\right ) \ln \left (b +\frac {a}{x^{2}}\right )+\ln \left (\frac {1}{x}\right ) \ln \left (\frac {-\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )+\ln \left (\frac {1}{x}\right ) \ln \left (\frac {\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )+\operatorname {dilog}\left (\frac {-\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )+\operatorname {dilog}\left (\frac {\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )\)	\(108\)
parts	\(\ln \left (\frac {b \,x^{2}+a}{x^{2}}\right ) \ln \left (x \right )+\ln \left (x \right )^{2}-2 b \left (\frac {\ln \left (x \right ) \left (\ln \left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\ln \left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )\right )}{2 b}+\frac {\operatorname {dilog}\left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\operatorname {dilog}\left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 b}\right )\)	\(113\)
derivativedivides	\(-\ln \left (\frac {1}{x}\right ) \ln \left (b +\frac {a}{x^{2}}\right )+2 a \left (\frac {\ln \left (\frac {1}{x}\right ) \left (\ln \left (\frac {-\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )+\ln \left (\frac {\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )\right )}{2 a}+\frac {\operatorname {dilog}\left (\frac {-\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )+\operatorname {dilog}\left (\frac {\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 a}\right )\)	\(118\)
default	\(-\ln \left (\frac {1}{x}\right ) \ln \left (b +\frac {a}{x^{2}}\right )+2 a \left (\frac {\ln \left (\frac {1}{x}\right ) \left (\ln \left (\frac {-\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )+\ln \left (\frac {\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )\right )}{2 a}+\frac {\operatorname {dilog}\left (\frac {-\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )+\operatorname {dilog}\left (\frac {\frac {a}{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 a}\right )\)	\(118\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\log \left (\frac {a+b x^2}{x^2}\right )}{x} \, dx=\int { \frac {\log \left (\frac {b x^{2} + a}{x^{2}}\right )}{x} \,d x } \]

[In]

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

[Out]

integral(log((b*x^2 + a)/x^2)/x, x)

Sympy [F]

\[ \int \frac {\log \left (\frac {a+b x^2}{x^2}\right )}{x} \, dx=\int \frac {\log {\left (\frac {a}{x^{2}} + b \right )}}{x}\, dx \]

[In]

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

[Out]

Integral(log(a/x**2 + b)/x, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (34) = 68\).

Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.97 \[ \int \frac {\log \left (\frac {a+b x^2}{x^2}\right )}{x} \, dx=-{\left (\log \left (b x^{2} + a\right ) - 2 \, \log \left (x\right )\right )} \log \left (x\right ) + \log \left (b x^{2} + a\right ) \log \left (x\right ) - \log \left (\frac {b x^{2}}{a} + 1\right ) \log \left (x\right ) - \log \left (x\right )^{2} + \log \left (x\right ) \log \left (\frac {b x^{2} + a}{x^{2}}\right ) - \frac {1}{2} \, {\rm Li}_2\left (-\frac {b x^{2}}{a}\right ) \]

[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)

Giac [F]

\[ \int \frac {\log \left (\frac {a+b x^2}{x^2}\right )}{x} \, dx=\int { \frac {\log \left (\frac {b x^{2} + a}{x^{2}}\right )}{x} \,d x } \]

[In]

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

[Out]

integrate(log((b*x^2 + a)/x^2)/x, x)

Mupad [B] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {\log \left (\frac {a+b x^2}{x^2}\right )}{x} \, dx=-\frac {{\mathrm {Li}}_{\mathrm {2}}\left (-\frac {a}{b\,x^2}\right )}{2}-\frac {\ln \left (b+\frac {a}{x^2}\right )\,\ln \left (-\frac {a}{b\,x^2}\right )}{2} \]

[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