Integrand size = 13, antiderivative size = 67 \[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\frac {(b+a x) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \log \left (c \left (a+\frac {b}{x}\right )\right ) \log \left (-\frac {b}{a x}\right )}{a}-\frac {2 b \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )}{a} \]
Time = 0.01 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94 \[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\frac {\log \left (\frac {c (b+a x)}{x}\right ) \left (-2 b \log \left (-\frac {b}{a x}\right )+(b+a x) \log \left (\frac {c (b+a x)}{x}\right )\right )-2 b \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )}{a} \]
(Log[(c*(b + a*x))/x]*(-2*b*Log[-(b/(a*x))] + (b + a*x)*Log[(c*(b + a*x))/ x]) - 2*b*PolyLog[2, 1 + b/(a*x)])/a
Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2903, 2899, 2904, 2841, 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.
\(\displaystyle \int \log ^2\left (\frac {c (a x+b)}{x}\right ) \, dx\) |
\(\Big \downarrow \) 2903 |
\(\displaystyle \int \log ^2\left (a c+\frac {b c}{x}\right )dx\) |
\(\Big \downarrow \) 2899 |
\(\displaystyle \frac {2 b \int \frac {\log \left (a c+\frac {b c}{x}\right )}{x}dx}{a}+\frac {(a x+b) \log ^2\left (a c+\frac {b c}{x}\right )}{a}\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {(a x+b) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \int x \log \left (a c+\frac {b c}{x}\right )d\frac {1}{x}}{a}\) |
\(\Big \downarrow \) 2841 |
\(\displaystyle \frac {(a x+b) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \left (\log \left (-\frac {b}{a x}\right ) \log \left (a c+\frac {b c}{x}\right )-b c \int \frac {\log \left (-\frac {b}{a x}\right )}{c \left (a+\frac {b}{x}\right )}d\frac {1}{x}\right )}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a x+b) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \left (\log \left (-\frac {b}{a x}\right ) \log \left (a c+\frac {b c}{x}\right )-b \int \frac {\log \left (-\frac {b}{a x}\right )}{a+\frac {b}{x}}d\frac {1}{x}\right )}{a}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {(a x+b) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \left (\log \left (-\frac {b}{a x}\right ) \log \left (a c+\frac {b c}{x}\right )+\operatorname {PolyLog}\left (2,\frac {b}{a x}+1\right )\right )}{a}\) |
((b + a*x)*Log[a*c + (b*c)/x]^2)/a - (2*b*(Log[a*c + (b*c)/x]*Log[-(b/(a*x ))] + PolyLog[2, 1 + b/(a*x)]))/a
3.1.94.3.1 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[(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] - Simp[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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)/(x_))^(p_.)]*(b_.))^(q_), x_Symbol] :> Simp[(e + d*x)*((a + b*Log[c*(d + e/x)^p])^q/d), x] + Simp[b*e*p*(q/d) I nt[(a + b*Log[c*(d + e/x)^p])^(q - 1)/x, x], x] /; FreeQ[{a, b, c, d, e, p} , x] && IGtQ[q, 0]
Int[((a_.) + Log[(c_.)*(v_)^(p_.)]*(b_.))^(q_.), x_Symbol] :> Int[(a + b*Lo g[c*ExpandToSum[v, x]^p])^q, x] /; FreeQ[{a, b, c, p, q}, x] && BinomialQ[v , x] && !BinomialMatchQ[v, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[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])
\[\int \ln \left (\frac {c \left (a x +b \right )}{x}\right )^{2}d x\]
\[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\int { \log \left (\frac {{\left (a x + b\right )} c}{x}\right )^{2} \,d x } \]
\[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=2 b \int \frac {\log {\left (a c + \frac {b c}{x} \right )}}{a x + b}\, dx + x \log {\left (\frac {c \left (a x + b\right )}{x} \right )}^{2} \]
Time = 0.22 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.69 \[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=x \log \left (\frac {{\left (a x + b\right )} c}{x}\right )^{2} + \frac {2 \, b \log \left (a x + b\right ) \log \left (\frac {{\left (a x + b\right )} c}{x}\right )}{a} + \frac {{\left (\frac {c \log \left (a x + b\right )^{2}}{a} - \frac {2 \, {\left (\log \left (\frac {a x}{b} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {a x}{b}\right )\right )} c}{a}\right )} b - \frac {2 \, {\left (c \log \left (a x + b\right ) - c \log \left (x\right )\right )} b \log \left (a x + b\right )}{a}}{c} \]
x*log((a*x + b)*c/x)^2 + 2*b*log(a*x + b)*log((a*x + b)*c/x)/a + ((c*log(a *x + b)^2/a - 2*(log(a*x/b + 1)*log(x) + dilog(-a*x/b))*c/a)*b - 2*(c*log( a*x + b) - c*log(x))*b*log(a*x + b)/a)/c
\[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\int { \log \left (\frac {{\left (a x + b\right )} c}{x}\right )^{2} \,d x } \]
Timed out. \[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\int {\ln \left (\frac {c\,\left (b+a\,x\right )}{x}\right )}^2 \,d x \]