Integrand size = 19, antiderivative size = 46 \[ \int \frac {\log \left (\frac {c x}{a+b x}\right )}{a+b x} \, dx=-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,1-\frac {a}{a+b x}\right )}{b} \]
Time = 0.01 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.83 \[ \int \frac {\log \left (\frac {c x}{a+b x}\right )}{a+b x} \, dx=\frac {\log \left (-\frac {b x}{a}\right ) \log \left (\frac {a}{a+b x}\right )}{b}+\frac {\log ^2\left (\frac {a}{a+b x}\right )}{2 b}-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,\frac {a+b x}{a}\right )}{b} \]
(Log[-((b*x)/a)]*Log[a/(a + b*x)])/b + Log[a/(a + b*x)]^2/(2*b) - (Log[a/( a + b*x)]*Log[(c*x)/(a + b*x)])/b - PolyLog[2, (a + b*x)/a]/b
Time = 0.39 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2944, 2858, 25, 27, 2778, 2005, 2752}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log \left (\frac {c x}{a+b x}\right )}{a+b x} \, dx\) |
\(\Big \downarrow \) 2944 |
\(\displaystyle \frac {a \int \frac {\log \left (\frac {a}{a+b x}\right )}{x (a+b x)}dx}{b}-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 2858 |
\(\displaystyle \frac {a \int \frac {\log \left (\frac {a}{a+b x}\right )}{x (a+b x)}d(a+b x)}{b^2}-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a \int -\frac {\log \left (\frac {a}{a+b x}\right )}{x (a+b x)}d(a+b x)}{b^2}-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \int -\frac {\log \left (\frac {a}{a+b x}\right )}{b x (a+b x)}d(a+b x)}{b}-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 2778 |
\(\displaystyle \frac {a \int -\frac {(a+b x) \log \left (\frac {a}{a+b x}\right )}{b x}d\frac {1}{a+b x}}{b}-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \frac {a \int \frac {\log \left (\frac {a}{a+b x}\right )}{\frac {a}{a+b x}-1}d\frac {1}{a+b x}}{b}-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle -\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,1-\frac {a}{a+b x}\right )}{b}\) |
3.1.88.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[((a_.) + Log[(c_.)*(x_)^(n_)]*(b_.))/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[1/n Subst[Int[(a + b*Log[c*x])/(x*(d + e*x^(r/n))), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IntegerQ[r/n]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_ .)*(x_))^(q_.)*((h_.) + (i_.)*(x_))^(r_.), x_Symbol] :> Simp[1/e Subst[In t[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[r, 0]) && IntegerQ[2*r]
Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ )]*(B_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(-Log[(b*c - a*d)/(b*(c + d*x))])*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])/g), x] + Simp[B*n*((b*c - a*d)/g) Int[Log[(b*c - a*d)/(b*(c + d*x))]/((a + b*x)*(c + d*x)), x], x ] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && EqQ[d*f - c*g, 0]
Time = 1.90 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.50
method | result | size |
parts | \(\frac {\ln \left (\frac {c x}{b x +a}\right ) \ln \left (b x +a \right )}{b}-\frac {-\frac {c \ln \left (b x +a \right )^{2}}{2}+c \left (\operatorname {dilog}\left (-\frac {x b}{a}\right )+\ln \left (b x +a \right ) \ln \left (-\frac {x b}{a}\right )\right )}{b c}\) | \(69\) |
derivativedivides | \(-\frac {\operatorname {dilog}\left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}-\frac {\ln \left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) \ln \left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}\) | \(97\) |
default | \(-\frac {\operatorname {dilog}\left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}-\frac {\ln \left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) \ln \left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}\) | \(97\) |
risch | \(-\frac {\operatorname {dilog}\left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}-\frac {\ln \left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) \ln \left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}\) | \(97\) |
ln(c*x/(b*x+a))*ln(b*x+a)/b-1/b/c*(-1/2*c*ln(b*x+a)^2+c*(dilog(-x/a*b)+ln( b*x+a)*ln(-x/a*b)))
\[ \int \frac {\log \left (\frac {c x}{a+b x}\right )}{a+b x} \, dx=\int { \frac {\log \left (\frac {c x}{b x + a}\right )}{b x + a} \,d x } \]
\[ \int \frac {\log \left (\frac {c x}{a+b x}\right )}{a+b x} \, dx=\int \frac {\log {\left (\frac {c x}{a + b x} \right )}}{a + b x}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (45) = 90\).
Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.07 \[ \int \frac {\log \left (\frac {c x}{a+b x}\right )}{a+b x} \, dx=\frac {\log \left (b x + a\right ) \log \left (\frac {c x}{b x + a}\right )}{b} - \frac {\frac {c \log \left (b x + a\right )^{2}}{b} - \frac {2 \, {\left (\log \left (\frac {b x}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a}\right )\right )} c}{b}}{2 \, c} + \frac {{\left (c \log \left (b x + a\right ) - c \log \left (x\right )\right )} \log \left (b x + a\right )}{b c} \]
log(b*x + a)*log(c*x/(b*x + a))/b - 1/2*(c*log(b*x + a)^2/b - 2*(log(b*x/a + 1)*log(x) + dilog(-b*x/a))*c/b)/c + (c*log(b*x + a) - c*log(x))*log(b*x + a)/(b*c)
\[ \int \frac {\log \left (\frac {c x}{a+b x}\right )}{a+b x} \, dx=\int { \frac {\log \left (\frac {c x}{b x + a}\right )}{b x + a} \,d x } \]
Timed out. \[ \int \frac {\log \left (\frac {c x}{a+b x}\right )}{a+b x} \, dx=\int \frac {\ln \left (\frac {c\,x}{a+b\,x}\right )}{a+b\,x} \,d x \]