Integrand size = 55, antiderivative size = 150 \[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=-\frac {\log ^2\left (\frac {e (a+b x)}{c+d x}\right ) \operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac {2 \log \left (\frac {e (a+b x)}{c+d x}\right ) \operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac {2 \operatorname {PolyLog}\left (4,1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g} \]
-ln(e*(b*x+a)/(d*x+c))^2*polylog(2,1+(a*d-b*c)/b/(d*x+c))/(-a*d+b*c)/g+2*l n(e*(b*x+a)/(d*x+c))*polylog(3,1+(a*d-b*c)/b/(d*x+c))/(-a*d+b*c)/g-2*polyl og(4,1+(a*d-b*c)/b/(d*x+c))/(-a*d+b*c)/g
Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.73 \[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\frac {-\log ^2\left (\frac {e (a+b x)}{c+d x}\right ) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )+2 \log \left (\frac {e (a+b x)}{c+d x}\right ) \operatorname {PolyLog}\left (3,\frac {d (a+b x)}{b (c+d x)}\right )-2 \operatorname {PolyLog}\left (4,\frac {d (a+b x)}{b (c+d x)}\right )}{(b c-a d) g} \]
Integrate[(Log[(b*c - a*d)/(b*(c + d*x))]*Log[(e*(a + b*x))/(c + d*x)]^2)/ ((c + d*x)*(a*g + b*g*x)),x]
(-(Log[(e*(a + b*x))/(c + d*x)]^2*PolyLog[2, (d*(a + b*x))/(b*(c + d*x))]) + 2*Log[(e*(a + b*x))/(c + d*x)]*PolyLog[3, (d*(a + b*x))/(b*(c + d*x))] - 2*PolyLog[4, (d*(a + b*x))/(b*(c + d*x))])/((b*c - a*d)*g)
Time = 0.62 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {2988, 2990, 7164}
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.
| \(\displaystyle \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx\) |
| \(\Big \downarrow \) 2988 |
| \(\displaystyle \frac {2 \int \frac {\log \left (\frac {e (a+b x)}{c+d x}\right ) \operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)}dx}{g}-\frac {\operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{g (b c-a d)}\) |
| \(\Big \downarrow \) 2990 |
| \(\displaystyle \frac {2 \left (\frac {\operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right ) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b c-a d}-\int \frac {\operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)}dx\right )}{g}-\frac {\operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{g (b c-a d)}\) |
| \(\Big \downarrow \) 7164 |
| \(\displaystyle \frac {2 \left (\frac {\operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right ) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b c-a d}-\frac {\operatorname {PolyLog}\left (4,1-\frac {b c-a d}{b (c+d x)}\right )}{b c-a d}\right )}{g}-\frac {\operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{g (b c-a d)}\) |
Int[(Log[(b*c - a*d)/(b*(c + d*x))]*Log[(e*(a + b*x))/(c + d*x)]^2)/((c + d*x)*(a*g + b*g*x)),x]
-((Log[(e*(a + b*x))/(c + d*x)]^2*PolyLog[2, 1 - (b*c - a*d)/(b*(c + d*x)) ])/((b*c - a*d)*g)) + (2*((Log[(e*(a + b*x))/(c + d*x)]*PolyLog[3, 1 - (b* c - a*d)/(b*(c + d*x))])/(b*c - a*d) - PolyLog[4, 1 - (b*c - a*d)/(b*(c + d*x))]/(b*c - a*d)))/g
3.1.91.3.1 Defintions of rubi rules used
Int[Log[v_]*Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_) )^(q_.))^(r_.)]^(s_.)*(u_), x_Symbol] :> With[{g = Simplify[(v - 1)*((c + d *x)/(a + b*x))], h = Simplify[u*(a + b*x)*(c + d*x)]}, Simp[(-h)*PolyLog[2, 1 - v]*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s/(b*c - a*d)), x] + Simp[h*p *r*s Int[PolyLog[2, 1 - v]*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1)/ ((a + b*x)*(c + d*x))), x], x] /; FreeQ[{g, h}, x]] /; FreeQ[{a, b, c, d, e , f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && IGtQ[s, 0] && EqQ[p + q, 0]
Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.)) ^(r_.)]^(s_.)*(u_)*PolyLog[n_, v_], x_Symbol] :> With[{g = Simplify[v*((c + d*x)/(a + b*x))], h = Simplify[u*(a + b*x)*(c + d*x)]}, Simp[h*PolyLog[n + 1, v]*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s/(b*c - a*d)), x] - Simp[h*p* r*s Int[PolyLog[n + 1, v]*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1)/( (a + b*x)*(c + d*x))), x], x] /; FreeQ[{g, h}, x]] /; FreeQ[{a, b, c, d, e, f, n, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && IGtQ[s, 0] && EqQ[p + q, 0]
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /; !FalseQ[w]] /; FreeQ[n, x]
Time = 4.11 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.33
| method | result | size |
| default | \(-\frac {\frac {\ln \left (-\frac {\frac {e \left (b x +a \right ) d}{d x +c}-b e}{b e}\right ) \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{3}}{3}-\frac {\ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{3} \ln \left (1-\frac {d \left (b x +a \right )}{b \left (d x +c \right )}\right )}{3}-\ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{2} \operatorname {Li}_{2}\left (\frac {d \left (b x +a \right )}{b \left (d x +c \right )}\right )+2 \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) \operatorname {Li}_{3}\left (\frac {d \left (b x +a \right )}{b \left (d x +c \right )}\right )-2 \,\operatorname {Li}_{4}\left (\frac {d \left (b x +a \right )}{b \left (d x +c \right )}\right )}{g \left (a d -c b \right )}\) | \(200\) |
int(ln((-a*d+b*c)/b/(d*x+c))*ln(e*(b*x+a)/(d*x+c))^2/(d*x+c)/(b*g*x+a*g),x ,method=_RETURNVERBOSE)
-1/g/(a*d-b*c)*(1/3*ln(-(e*(b*x+a)/(d*x+c)*d-b*e)/b/e)*ln(e*(b*x+a)/(d*x+c ))^3-1/3*ln(e*(b*x+a)/(d*x+c))^3*ln(1-d*(b*x+a)/b/(d*x+c))-ln(e*(b*x+a)/(d *x+c))^2*polylog(2,d*(b*x+a)/b/(d*x+c))+2*ln(e*(b*x+a)/(d*x+c))*polylog(3, d*(b*x+a)/b/(d*x+c))-2*polylog(4,d*(b*x+a)/b/(d*x+c)))
\[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\int { \frac {\log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}} \,d x } \]
integrate(log((-a*d+b*c)/b/(d*x+c))*log(e*(b*x+a)/(d*x+c))^2/(d*x+c)/(b*g* x+a*g),x, algorithm="fricas")
integral(log((b*c - a*d)/(b*d*x + b*c))*log((b*e*x + a*e)/(d*x + c))^2/(b* d*g*x^2 + a*c*g + (b*c + a*d)*g*x), x)
\[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=- \frac {d \int \frac {\log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}^{3}}{c + d x}\, dx}{3 g \left (a d - b c\right )} - \frac {\log {\left (\frac {- a d + b c}{b \left (c + d x\right )} \right )} \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}^{3}}{3 a d g - 3 b c g} \]
-d*Integral(log(a*e/(c + d*x) + b*e*x/(c + d*x))**3/(c + d*x), x)/(3*g*(a* d - b*c)) - log((-a*d + b*c)/(b*(c + d*x)))*log(e*(a + b*x)/(c + d*x))**3/ (3*a*d*g - 3*b*c*g)
\[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\int { \frac {\log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}} \,d x } \]
integrate(log((-a*d+b*c)/b/(d*x+c))*log(e*(b*x+a)/(d*x+c))^2/(d*x+c)/(b*g* x+a*g),x, algorithm="maxima")
-1/4*(4*log(b*x + a)*log(d*x + c)^3 - log(d*x + c)^4)/(b*c*g - a*d*g) - in tegrate((((d*log(b*c - a*d) - d*log(b))*a - (c*log(b*c - a*d) - c*log(b))* b)*log(b*x + a)^2 + ((d*log(b*c - a*d) - d*log(b) + 2*d*log(e))*a - (c*(lo g(b*c - a*d) + 2*log(e)) - c*log(b))*b - (3*b*d*x + 2*b*c + a*d)*log(b*x + a))*log(d*x + c)^2 + (d*log(b*c - a*d)*log(e)^2 - d*log(b)*log(e)^2)*a - (c*log(b*c - a*d)*log(e)^2 - c*log(b)*log(e)^2)*b + 2*((d*log(b*c - a*d)*l og(e) - d*log(b)*log(e))*a - (c*log(b*c - a*d)*log(e) - c*log(b)*log(e))*b )*log(b*x + a) + ((b*c - a*d)*log(b*x + a)^2 - (2*d*log(b*c - a*d)*log(e) - 2*d*log(b)*log(e) + d*log(e)^2)*a - (2*c*log(b)*log(e) - (2*log(b*c - a* d)*log(e) + log(e)^2)*c)*b - 2*((d*log(b*c - a*d) - d*log(b) + d*log(e))*a - (c*(log(b*c - a*d) + log(e)) - c*log(b))*b)*log(b*x + a))*log(d*x + c)) /(a*b*c^2*g - a^2*c*d*g + (b^2*c*d*g - a*b*d^2*g)*x^2 + (b^2*c^2*g - a^2*d ^2*g)*x), x)
\[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\int { \frac {\log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}} \,d x } \]
integrate(log((-a*d+b*c)/b/(d*x+c))*log(e*(b*x+a)/(d*x+c))^2/(d*x+c)/(b*g* x+a*g),x, algorithm="giac")
integrate(log((b*x + a)*e/(d*x + c))^2*log((b*c - a*d)/((d*x + c)*b))/((b* g*x + a*g)*(d*x + c)), x)
Timed out. \[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\int \frac {{\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}^2\,\ln \left (-\frac {a\,d-b\,c}{b\,\left (c+d\,x\right )}\right )}{\left (a\,g+b\,g\,x\right )\,\left (c+d\,x\right )} \,d x \]
int((log((e*(a + b*x))/(c + d*x))^2*log(-(a*d - b*c)/(b*(c + d*x))))/((a*g + b*g*x)*(c + d*x)),x)