Integrand size = 35, antiderivative size = 27 \[ \int \frac {\log \left (1-\frac {c (a-b x)}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {\operatorname {PolyLog}\left (2,\frac {c (a-b x)}{a+b x}\right )}{2 a b} \]
Leaf count is larger than twice the leaf count of optimal. \(259\) vs. \(2(27)=54\).
Time = 0.13 (sec) , antiderivative size = 259, normalized size of antiderivative = 9.59 \[ \int \frac {\log \left (1-\frac {c (a-b x)}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {4 \text {arctanh}\left (\frac {b x}{a}\right ) \log \left (\frac {a}{b}+x\right )-\log ^2\left (\frac {a}{b}+x\right )-4 \text {arctanh}\left (\frac {b x}{a}\right ) \log \left (\frac {a-a c}{b+b c}+x\right )+2 \log \left (\frac {a}{b}+x\right ) \log \left (\frac {a-b x}{2 a}\right )-2 \log \left (\frac {a-a c}{b+b c}+x\right ) \log \left (\frac {(1+c) (a-b x)}{2 a}\right )+2 \log \left (\frac {a-a c}{b+b c}+x\right ) \log \left (\frac {(1+c) (a+b x)}{2 a c}\right )+4 \text {arctanh}\left (\frac {b x}{a}\right ) \log \left (\frac {a-a c+b (1+c) x}{a+b x}\right )+2 \operatorname {PolyLog}\left (2,\frac {a+b x}{2 a}\right )-2 \operatorname {PolyLog}\left (2,\frac {a-a c+b (1+c) x}{2 a}\right )+2 \operatorname {PolyLog}\left (2,-\frac {a-a c+b (1+c) x}{2 a c}\right )}{4 a b} \]
(4*ArcTanh[(b*x)/a]*Log[a/b + x] - Log[a/b + x]^2 - 4*ArcTanh[(b*x)/a]*Log [(a - a*c)/(b + b*c) + x] + 2*Log[a/b + x]*Log[(a - b*x)/(2*a)] - 2*Log[(a - a*c)/(b + b*c) + x]*Log[((1 + c)*(a - b*x))/(2*a)] + 2*Log[(a - a*c)/(b + b*c) + x]*Log[((1 + c)*(a + b*x))/(2*a*c)] + 4*ArcTanh[(b*x)/a]*Log[(a - a*c + b*(1 + c)*x)/(a + b*x)] + 2*PolyLog[2, (a + b*x)/(2*a)] - 2*PolyLo g[2, (a - a*c + b*(1 + c)*x)/(2*a)] + 2*PolyLog[2, -1/2*(a - a*c + b*(1 + c)*x)/(a*c)])/(4*a*b)
Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {2997, 2965, 27, 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.
\(\displaystyle \int \frac {\log \left (1-\frac {c (a-b x)}{a+b x}\right )}{(a-b x) (a+b x)} \, dx\) |
\(\Big \downarrow \) 2997 |
\(\displaystyle \int \frac {\log \left (\frac {a (1-c)+b (c+1) x}{a+b x}\right )}{(a-b x) (a+b x)}dx\) |
\(\Big \downarrow \) 2965 |
\(\displaystyle \int \frac {\log \left (\frac {a (1-c)+b (c+1) x}{a+b x}\right )}{2 a b \left (1-\frac {a (1-c)+b (c+1) x}{a+b x}\right )}d\frac {a (1-c)+b (c+1) x}{a+b x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\log \left (\frac {a (1-c)+b (c+1) x}{a+b x}\right )}{1-\frac {a (1-c)+b (c+1) x}{a+b x}}d\frac {a (1-c)+b (c+1) x}{a+b x}}{2 a b}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {\operatorname {PolyLog}\left (2,1-\frac {a (1-c)+b (c+1) x}{a+b x}\right )}{2 a b}\) |
3.4.42.3.1 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[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[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol ] :> Simp[(b*c - a*d)^(q + 1)*(i/d)^q Subst[Int[(b*f - a*g - (d*f - c*g)* x)^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d* x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, A, B, n}, x] && NeQ[b*c - a*d, 0] && IntegersQ[m, q] && IGtQ[p, 0] && EqQ[d*h - c*i, 0]
Int[Log[(e_.)*((f_.)*((g_) + (v_.)/(w_)))^(r_.)]^(s_.)*(u_.), x_Symbol] :> Int[u*Log[e*(f*(ExpandToSum[v + g*w, x]/ExpandToSum[w, x]))^r]^s, x] /; Fre eQ[{e, f, g, r, s}, x] && LinearQ[w, x] && (FreeQ[v, x] || LinearQ[v, x]) & & AlgebraicFunctionQ[u, x]
Time = 0.96 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\operatorname {dilog}\left (1+c -\frac {2 c a}{b x +a}\right )}{2 b a}\) | \(24\) |
default | \(\frac {\operatorname {dilog}\left (1+c -\frac {2 c a}{b x +a}\right )}{2 b a}\) | \(24\) |
risch | \(\frac {\operatorname {dilog}\left (1+c -\frac {2 c a}{b x +a}\right )}{2 b a}\) | \(24\) |
parts | \(\frac {\ln \left (1-\frac {c \left (-b x +a \right )}{b x +a}\right ) \ln \left (b x +a \right )}{2 a b}-\frac {\ln \left (1-\frac {c \left (-b x +a \right )}{b x +a}\right ) \ln \left (-b x +a \right )}{2 a b}+\frac {c \left (\frac {\left (1+c \right ) \left (\frac {\operatorname {dilog}\left (-\frac {\left (1+c \right ) \left (-b x +a \right )-2 a}{2 a}\right )}{1+c}+\frac {\ln \left (-b x +a \right ) \ln \left (-\frac {\left (1+c \right ) \left (-b x +a \right )-2 a}{2 a}\right )}{1+c}\right )}{2 c a}-\frac {\operatorname {dilog}\left (-\frac {-b x -a}{2 a}\right )+\ln \left (-b x +a \right ) \ln \left (-\frac {-b x -a}{2 a}\right )}{2 c a}\right )}{b}+\frac {c \left (\frac {\ln \left (b x +a \right )^{2}}{4 c a}+\frac {\left (-1-c \right ) \left (\frac {\operatorname {dilog}\left (-\frac {\left (1+c \right ) \left (b x +a \right )-2 c a}{2 c a}\right )}{1+c}+\frac {\ln \left (b x +a \right ) \ln \left (-\frac {\left (1+c \right ) \left (b x +a \right )-2 c a}{2 c a}\right )}{1+c}\right )}{2 c a}\right )}{b}\) | \(291\) |
Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\log \left (1-\frac {c (a-b x)}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {{\rm Li}_2\left (\frac {a c - {\left (b c + b\right )} x - a}{b x + a} + 1\right )}{2 \, a b} \]
Timed out. \[ \int \frac {\log \left (1-\frac {c (a-b x)}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (26) = 52\).
Time = 0.21 (sec) , antiderivative size = 243, normalized size of antiderivative = 9.00 \[ \int \frac {\log \left (1-\frac {c (a-b x)}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {1}{2} \, {\left (\frac {\log \left (b x + a\right )}{a b} - \frac {\log \left (b x - a\right )}{a b}\right )} \log \left (\frac {{\left (b x - a\right )} c}{b x + a} + 1\right ) + \frac {\log \left (b x + a\right )^{2} - 2 \, \log \left (b x + a\right ) \log \left (b x - a\right )}{4 \, a b} + \frac {\log \left (b x - a\right ) \log \left (\frac {b {\left (c + 1\right )} x - a {\left (c + 1\right )}}{2 \, a} + 1\right ) + {\rm Li}_2\left (-\frac {b {\left (c + 1\right )} x - a {\left (c + 1\right )}}{2 \, a}\right )}{2 \, a b} + \frac {\log \left (b x + a\right ) \log \left (-\frac {b x + a}{2 \, a} + 1\right ) + {\rm Li}_2\left (\frac {b x + a}{2 \, a}\right )}{2 \, a b} - \frac {\log \left (b x + a\right ) \log \left (-\frac {b {\left (c + 1\right )} x + a {\left (c + 1\right )}}{2 \, a c} + 1\right ) + {\rm Li}_2\left (\frac {b {\left (c + 1\right )} x + a {\left (c + 1\right )}}{2 \, a c}\right )}{2 \, a b} \]
1/2*(log(b*x + a)/(a*b) - log(b*x - a)/(a*b))*log((b*x - a)*c/(b*x + a) + 1) + 1/4*(log(b*x + a)^2 - 2*log(b*x + a)*log(b*x - a))/(a*b) + 1/2*(log(b *x - a)*log(1/2*(b*(c + 1)*x - a*(c + 1))/a + 1) + dilog(-1/2*(b*(c + 1)*x - a*(c + 1))/a))/(a*b) + 1/2*(log(b*x + a)*log(-1/2*(b*x + a)/a + 1) + di log(1/2*(b*x + a)/a))/(a*b) - 1/2*(log(b*x + a)*log(-1/2*(b*(c + 1)*x + a* (c + 1))/(a*c) + 1) + dilog(1/2*(b*(c + 1)*x + a*(c + 1))/(a*c)))/(a*b)
\[ \int \frac {\log \left (1-\frac {c (a-b x)}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\int { -\frac {\log \left (\frac {{\left (b x - a\right )} c}{b x + a} + 1\right )}{{\left (b x + a\right )} {\left (b x - a\right )}} \,d x } \]
Timed out. \[ \int \frac {\log \left (1-\frac {c (a-b x)}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\int \frac {\ln \left (1-\frac {c\,\left (a-b\,x\right )}{a+b\,x}\right )}{\left (a+b\,x\right )\,\left (a-b\,x\right )} \,d x \]