\(\int \log ^2(\frac {c (b+a x)}{x}) \, dx\) [94]

Optimal result

Integrand size = 13, antiderivative size = 68 \[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\frac {(b+a x) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \log \left (a c+\frac {b c}{x}\right ) \log \left (-\frac {b}{a x}\right )}{a}-\frac {2 b \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )}{a} \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93 \[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\frac {\log \left (\frac {c (b+a x)}{x}\right ) \left (-2 b \log \left (-\frac {b}{a x}\right )+(b+a x) \log \left (\frac {c (b+a x)}{x}\right )\right )-2 b \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )}{a} \] Input:

Integrate[Log[(c*(b + a*x))/x]^2,x]
 

Output:

(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
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94, 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

\(\displaystyle \frac {(a x+b) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \left (\log \left (-\frac {b}{a x}\right ) \log \left (a c+\frac {b c}{x}\right )+\operatorname {PolyLog}\left (2,\frac {b}{a x}+1\right )\right )}{a}\)

Input:

Int[Log[(c*(b + a*x))/x]^2,x]
 

Output:

((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
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2752

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]
 

rule 2841

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]
 

rule 2899

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]
 

rule 2903

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]
 

rule 2904

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])
 

Maple [F]

\[\int \ln \left (\frac {c \left (a x +b \right )}{x}\right )^{2}d x\]

Input:

int(ln(c*(a*x+b)/x)^2,x)
 

Output:

int(ln(c*(a*x+b)/x)^2,x)
 

Fricas [F]

\[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\int { \log \left (\frac {{\left (a x + b\right )} c}{x}\right )^{2} \,d x } \] Input:

integrate(log(c*(a*x+b)/x)^2,x, algorithm="fricas")
 

Output:

integral(log((a*c*x + b*c)/x)^2, x)
 

Sympy [F]

\[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=2 b \int \frac {\log {\left (a c + \frac {b c}{x} \right )}}{a x + b}\, dx + x \log {\left (\frac {c \left (a x + b\right )}{x} \right )}^{2} \] Input:

integrate(ln(c*(a*x+b)/x)**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.66 \[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=x \log \left (\frac {{\left (a x + b\right )} c}{x}\right )^{2} + \frac {2 \, b \log \left (a x + b\right ) \log \left (\frac {{\left (a x + b\right )} c}{x}\right )}{a} + \frac {{\left (\frac {c \log \left (a x + b\right )^{2}}{a} - \frac {2 \, {\left (\log \left (\frac {a x}{b} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {a x}{b}\right )\right )} c}{a}\right )} b - \frac {2 \, {\left (c \log \left (a x + b\right ) - c \log \left (x\right )\right )} b \log \left (a x + b\right )}{a}}{c} \] Input:

integrate(log(c*(a*x+b)/x)^2,x, algorithm="maxima")
 

Output:

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
 

Giac [F]

\[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\int { \log \left (\frac {{\left (a x + b\right )} c}{x}\right )^{2} \,d x } \] Input:

integrate(log(c*(a*x+b)/x)^2,x, algorithm="giac")
 

Output:

integrate(log((a*x + b)*c/x)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\int {\ln \left (\frac {c\,\left (b+a\,x\right )}{x}\right )}^2 \,d x \] Input:

int(log((c*(b + a*x))/x)^2,x)
 

Output:

int(log((c*(b + a*x))/x)^2, x)
 

Reduce [F]

\[ \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx=\int \mathrm {log}\left (\frac {a c x +b c}{x}\right )^{2}d x \] Input:

int(log(c*(a*x+b)/x)^2,x)
                                                                                    
                                                                                    
 

Output:

int(log((a*c*x + b*c)/x)**2,x)