3.64 \(\int \frac {\log (x) \log (\frac {a+b x}{(b c-a d) x})}{x} \, dx\)

Optimal. Leaf size=82 \[ \frac {1}{2} \log ^2(x) \log \left (\frac {a}{x (b c-a d)}+\frac {b}{b c-a d}\right )+\text {Li}_3\left (-\frac {a}{b x}\right )+\log (x) \text {Li}_2\left (-\frac {a}{b x}\right )-\frac {1}{2} \log ^2(x) \log \left (\frac {a}{b x}+1\right ) \]

[Out]

-1/2*ln(1+a/b/x)*ln(x)^2+1/2*ln(b/(-a*d+b*c)+a/(-a*d+b*c)/x)*ln(x)^2+ln(x)*polylog(2,-a/b/x)+polylog(3,-a/b/x)

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Rubi [A]  time = 0.18, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2380, 2375, 2337, 2374, 6589} \[ \text {PolyLog}\left (3,-\frac {a}{b x}\right )+\log (x) \text {PolyLog}\left (2,-\frac {a}{b x}\right )+\frac {1}{2} \log ^2(x) \log \left (\frac {a}{x (b c-a d)}+\frac {b}{b c-a d}\right )-\frac {1}{2} \log ^2(x) \log \left (\frac {a}{b x}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(Log[x]*Log[(a + b*x)/((b*c - a*d)*x)])/x,x]

[Out]

-(Log[1 + a/(b*x)]*Log[x]^2)/2 + (Log[b/(b*c - a*d) + a/((b*c - a*d)*x)]*Log[x]^2)/2 + Log[x]*PolyLog[2, -(a/(
b*x))] + PolyLog[3, -(a/(b*x))]

Rule 2337

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[(f^m*Log[1 + (e*x^r)/d]*(a + b*Log[c*x^n])^p)/(e*r), x] - Dist[(b*f^m*n*p)/(e*r), Int[(Log[1 + (e*x^r)/d]*(
a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2375

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :
> Simp[(Log[d*(e + f*x^m)^r]*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] - Dist[(f*m*r)/(b*n*(p + 1)), Int[(
x^(m - 1)*(a + b*Log[c*x^n])^(p + 1))/(e + f*x^m), x], x] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p,
0] && NeQ[d*e, 1]

Rule 2380

Int[Log[(d_.)*(u_)^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((g_.)*(x_))^(q_.), x_Symbol] :> Int[(g*
x)^q*Log[d*ExpandToSum[u, x]^r]*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, g, r, n, p, q}, x] && BinomialQ
[u, x] &&  !BinomialMatchQ[u, x]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\log (x) \log \left (\frac {a+b x}{(b c-a d) x}\right )}{x} \, dx &=\int \frac {\log \left (\frac {b}{b c-a d}+\frac {a}{(b c-a d) x}\right ) \log (x)}{x} \, dx\\ &=\frac {1}{2} \log \left (\frac {b}{b c-a d}+\frac {a}{(b c-a d) x}\right ) \log ^2(x)+\frac {a \int \frac {\log ^2(x)}{\left (\frac {b}{b c-a d}+\frac {a}{(b c-a d) x}\right ) x^2} \, dx}{2 (b c-a d)}\\ &=-\frac {1}{2} \log \left (1+\frac {a}{b x}\right ) \log ^2(x)+\frac {1}{2} \log \left (\frac {b}{b c-a d}+\frac {a}{(b c-a d) x}\right ) \log ^2(x)+\int \frac {\log \left (1+\frac {a}{b x}\right ) \log (x)}{x} \, dx\\ &=-\frac {1}{2} \log \left (1+\frac {a}{b x}\right ) \log ^2(x)+\frac {1}{2} \log \left (\frac {b}{b c-a d}+\frac {a}{(b c-a d) x}\right ) \log ^2(x)+\log (x) \text {Li}_2\left (-\frac {a}{b x}\right )-\int \frac {\text {Li}_2\left (-\frac {a}{b x}\right )}{x} \, dx\\ &=-\frac {1}{2} \log \left (1+\frac {a}{b x}\right ) \log ^2(x)+\frac {1}{2} \log \left (\frac {b}{b c-a d}+\frac {a}{(b c-a d) x}\right ) \log ^2(x)+\log (x) \text {Li}_2\left (-\frac {a}{b x}\right )+\text {Li}_3\left (-\frac {a}{b x}\right )\\ \end {align*}

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Mathematica [A]  time = 5.03, size = 66, normalized size = 0.80 \[ \frac {1}{6} \log ^2(x) \left (3 \log \left (\frac {a+b x}{b c x-a d x}\right )-3 \log \left (\frac {b x}{a}+1\right )+\log (x)\right )+\text {Li}_3\left (-\frac {b x}{a}\right )-\log (x) \text {Li}_2\left (-\frac {b x}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Log[x]*Log[(a + b*x)/((b*c - a*d)*x)])/x,x]

[Out]

(Log[x]^2*(Log[x] - 3*Log[1 + (b*x)/a] + 3*Log[(a + b*x)/(b*c*x - a*d*x)]))/6 - Log[x]*PolyLog[2, -((b*x)/a)]
+ PolyLog[3, -((b*x)/a)]

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \relax (x) \log \left (\frac {b x + a}{{\left (b c - a d\right )} x}\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(x)*log((b*x + a)/((b*c - a*d)*x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(x)*log((b*x + a)/((b*c - a*d)*x))/x, x)

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maple [C]  time = 0.18, size = 450, normalized size = 5.49 \[ -\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{a d -b c}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{\left (a d -b c \right ) x}\right ) \ln \relax (x )^{2}}{4}+\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{\left (a d -b c \right ) x}\right )^{2} \ln \relax (x )^{2}}{4}-\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )\right ) \mathrm {csgn}\left (\frac {i}{a d -b c}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{a d -b c}\right ) \ln \relax (x )^{2}}{4}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{a d -b c}\right )^{2} \ln \relax (x )^{2}}{4}+\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{a d -b c}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{a d -b c}\right )^{2} \ln \relax (x )^{2}}{4}-\frac {i \pi \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{a d -b c}\right )^{3} \ln \relax (x )^{2}}{4}+\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (b x +a \right )}{a d -b c}\right ) \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{\left (a d -b c \right ) x}\right )^{2} \ln \relax (x )^{2}}{4}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{\left (a d -b c \right ) x}\right )^{3} \ln \relax (x )^{2}}{4}-\frac {i \pi \mathrm {csgn}\left (\frac {i \left (b x +a \right )}{\left (a d -b c \right ) x}\right )^{2} \ln \relax (x )^{2}}{2}-\frac {\ln \relax (x )^{3}}{3}-\frac {\ln \relax (x )^{2} \ln \left (\frac {b x}{a}+1\right )}{2}+\frac {\ln \relax (x )^{2} \ln \left (b x +a \right )}{2}-\frac {\ln \relax (x )^{2} \ln \left (a d -b c \right )}{2}-\polylog \left (2, -\frac {b x}{a}\right ) \ln \relax (x )+\frac {i \pi \ln \relax (x )^{2}}{2}+\polylog \left (3, -\frac {b x}{a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)*ln((b*x+a)/(-a*d+b*c)/x)/x,x)

[Out]

1/2*ln(x)^2*ln(b*x+a)-1/3*ln(x)^3+1/4*I*ln(x)^2*Pi*csgn(I/x)*csgn(I/x*(b*x+a)/(a*d-b*c))^2+1/2*I*ln(x)^2*Pi-1/
4*I*ln(x)^2*Pi*csgn(I*(b*x+a)/(a*d-b*c))^3-1/4*I*ln(x)^2*Pi*csgn(I/x)*csgn(I*(b*x+a)/(a*d-b*c))*csgn(I/x*(b*x+
a)/(a*d-b*c))+1/4*I*ln(x)^2*Pi*csgn(I*(b*x+a))*csgn(I*(b*x+a)/(a*d-b*c))^2+1/4*I*ln(x)^2*Pi*csgn(I/x*(b*x+a)/(
a*d-b*c))^3+1/4*I*ln(x)^2*Pi*csgn(I/(a*d-b*c))*csgn(I*(b*x+a)/(a*d-b*c))^2-1/4*I*ln(x)^2*Pi*csgn(I*(b*x+a))*cs
gn(I/(a*d-b*c))*csgn(I*(b*x+a)/(a*d-b*c))-1/2*I*ln(x)^2*Pi*csgn(I/x*(b*x+a)/(a*d-b*c))^2+1/4*I*ln(x)^2*Pi*csgn
(I*(b*x+a)/(a*d-b*c))*csgn(I/x*(b*x+a)/(a*d-b*c))^2-1/2*ln(x)^2*ln(a*d-b*c)-1/2*ln(x)^2*ln(1+b*x/a)-ln(x)*poly
log(2,-b*x/a)+polylog(3,-b*x/a)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> unable to make sense of Maxima expression 'li[2]' in Sage

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (-\frac {a+b\,x}{x\,\left (a\,d-b\,c\right )}\right )\,\ln \relax (x)}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-(a + b*x)/(x*(a*d - b*c)))*log(x))/x,x)

[Out]

int((log(-(a + b*x)/(x*(a*d - b*c)))*log(x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(x)*ln((b*x+a)/(-a*d+b*c)/x)/x,x)

[Out]

a*Integral(log(x)**2/(a*x + b*x**2), x)/2 + log(x)**2*log((a + b*x)/(x*(-a*d + b*c)))/2

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