3.393 \(\int \frac {\log (\frac {a+b x}{x})}{x} \, dx\)

Optimal. Leaf size=35 \[ -\text {Li}_2\left (\frac {a}{b x}+1\right )-\log \left (\frac {a}{x}+b\right ) \log \left (-\frac {a}{b x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2461, 2454, 2394, 2315} \[ -\text {PolyLog}\left (2,\frac {a}{b x}+1\right )-\log \left (\frac {a}{x}+b\right ) \log \left (-\frac {a}{b x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[(a + b*x)/x]/x,x]

[Out]

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

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

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Mathematica [A]  time = 0.00, size = 36, normalized size = 1.03 \[ -\text {Li}_2\left (\frac {\frac {a}{x}+b}{b}\right )-\log \left (\frac {a}{x}+b\right ) \log \left (-\frac {a}{b x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(a + b*x)/x]/x,x]

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (\frac {b x + a}{x}\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((b*x + a)/x)/x, x)

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giac [B]  time = 0.66, size = 204, normalized size = 5.83 \[ \frac {a^{3} {\left (\frac {\log \left (\frac {{\left | b x + a \right |}}{{\left | x \right |}}\right )}{b^{2}} - \frac {\log \left ({\left | -b + \frac {b x + a}{x} \right |}\right )}{b^{2}} + \frac {1}{{\left (b - \frac {b x + a}{x}\right )} b}\right )} - \frac {a^{3} \log \left (-{\left (a - \frac {b}{\frac {{\left (a - \frac {b}{\frac {b}{a} - \frac {b x + a}{a x}}\right )} {\left (\frac {b}{a} - \frac {b x + a}{a x}\right )}}{a} + \frac {b}{a}}\right )} {\left (\frac {{\left (a - \frac {b}{\frac {b}{a} - \frac {b x + a}{a x}}\right )} {\left (\frac {b}{a} - \frac {b x + a}{a x}\right )}}{a} + \frac {b}{a}\right )}\right )}{{\left (b - \frac {b x + a}{x}\right )}^{2}}}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^3*(log(abs(b*x + a)/abs(x))/b^2 - log(abs(-b + (b*x + a)/x))/b^2 + 1/((b - (b*x + a)/x)*b)) - a^3*log(-
(a - b/((a - b/(b/a - (b*x + a)/(a*x)))*(b/a - (b*x + a)/(a*x))/a + b/a))*((a - b/(b/a - (b*x + a)/(a*x)))*(b/
a - (b*x + a)/(a*x))/a + b/a))/(b - (b*x + a)/x)^2)/a^2

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maple [A]  time = 0.08, size = 34, normalized size = 0.97 \[ -\ln \left (-\frac {a}{b x}\right ) \ln \left (b +\frac {a}{x}\right )-\dilog \left (-\frac {a}{b x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-dilog(-a/b/x)-ln(b+a/x)*ln(-a/b/x)

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maxima [A]  time = 0.48, size = 67, normalized size = 1.91 \[ -{\left (\log \left (b x + a\right ) - \log \relax (x)\right )} \log \relax (x) + \log \left (b x + a\right ) \log \relax (x) - \log \left (\frac {b x}{a} + 1\right ) \log \relax (x) - \frac {1}{2} \, \log \relax (x)^{2} + \log \relax (x) \log \left (\frac {b x + a}{x}\right ) - {\rm Li}_2\left (-\frac {b x}{a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.41, size = 37, normalized size = 1.06 \[ -\mathrm {polylog}\left (2,\frac {a}{b\,x}+1\right )-\ln \left (\frac {a+b\,x}{x}\right )\,\ln \left (-\frac {a}{b\,x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((a + b*x)/x)/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a/x + b)/x, x)

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