Integrand size = 58, antiderivative size = 160 \[ \int \frac {\log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx=-\frac {\log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac {2 n \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac {2 n^2 \operatorname {PolyLog}\left (4,1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g} \] Output:
-ln(e*((b*x+a)/(d*x+c))^n)^2*polylog(2,1-(-a*d+b*c)/b/(d*x+c))/(-a*d+b*c)/ g+2*n*ln(e*((b*x+a)/(d*x+c))^n)*polylog(3,1-(-a*d+b*c)/b/(d*x+c))/(-a*d+b* c)/g-2*n^2*polylog(4,1-(-a*d+b*c)/b/(d*x+c))/(-a*d+b*c)/g
Leaf count is larger than twice the leaf count of optimal. \(785\) vs. \(2(160)=320\).
Time = 0.53 (sec) , antiderivative size = 785, normalized size of antiderivative = 4.91 \[ \int \frac {\log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx =\text {Too large to display} \] Input:
Integrate[(Log[e*((a + b*x)/(c + d*x))^n]^2*Log[(b*c - a*d)/(b*(c + d*x))] )/((c + d*x)*(a*g + b*g*x)),x]
Output:
(Log[(a + b*x)/(c + d*x)]*(3*Log[e*((a + b*x)/(c + d*x))^n]^2 - 3*n*Log[e* ((a + b*x)/(c + d*x))^n]*Log[(a + b*x)/(c + d*x)] + n^2*Log[(a + b*x)/(c + d*x)]^2)*Log[(b*c - a*d)/(b*c + b*d*x)] + (3*(Log[e*((a + b*x)/(c + d*x)) ^n] - n*Log[(a + b*x)/(c + d*x)])^2*(-Log[c/d + x]^2 - 2*Log[a/b + x]*Log[ c + d*x] + 2*Log[c/d + x]*Log[c + d*x] + 2*Log[(a + b*x)/(c + d*x)]*Log[c + d*x] + 2*Log[a/b + x]*Log[(b*(c + d*x))/(b*c - a*d)] + 2*PolyLog[2, (d*( a + b*x))/(-(b*c) + a*d)]))/2 + n*(Log[e*((a + b*x)/(c + d*x))^n] - n*Log[ (a + b*x)/(c + d*x)])*(Log[c/d + x]^3 + 3*Log[c/d + x]^2*(-Log[a/b + x] + Log[(d*(a + b*x))/(-(b*c) + a*d)]) + 3*(-Log[a/b + x] + Log[c/d + x] + Log [(a + b*x)/(c + d*x)])^2*Log[c + d*x] + 3*Log[a/b + x]^2*Log[(b*(c + d*x)) /(b*c - a*d)] + 6*Log[a/b + x]*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)] + 3*(Log[a/b + x] - Log[c/d + x] - Log[(a + b*x)/(c + d*x)])*(Log[c/d + x]^2 - 2*(Log[a/b + x]*Log[(b*(c + d*x))/(b*c - a*d)] + PolyLog[2, (d*(a + b*x ))/(-(b*c) + a*d)])) + 6*Log[c/d + x]*PolyLog[2, (b*(c + d*x))/(b*c - a*d) ] - 6*PolyLog[3, (d*(a + b*x))/(-(b*c) + a*d)] - 6*PolyLog[3, (b*(c + d*x) )/(b*c - a*d)]) - n^2*(Log[(a + b*x)/(c + d*x)]^3*Log[(b*c - a*d)/(b*c + b *d*x)] + 3*Log[(a + b*x)/(c + d*x)]^2*PolyLog[2, (d*(a + b*x))/(b*(c + d*x ))] - 6*Log[(a + b*x)/(c + d*x)]*PolyLog[3, (d*(a + b*x))/(b*(c + d*x))] + 6*PolyLog[4, (d*(a + b*x))/(b*(c + d*x))]))/(3*(b*c - a*d)*g)
Time = 1.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, 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.
\(\displaystyle \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c+d x) (a g+b g x)} \, dx\) |
\(\Big \downarrow \) 2988 |
\(\displaystyle \frac {2 n \int \frac {\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)}dx}{g}-\frac {\operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}\) |
\(\Big \downarrow \) 2990 |
\(\displaystyle \frac {2 n \left (\frac {\operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right ) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b c-a d}-n \int \frac {\operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)}dx\right )}{g}-\frac {\operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle \frac {2 n \left (\frac {\operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right ) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b c-a d}-\frac {n \operatorname {PolyLog}\left (4,1-\frac {b c-a d}{b (c+d x)}\right )}{b c-a d}\right )}{g}-\frac {\operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}\) |
Input:
Int[(Log[e*((a + b*x)/(c + d*x))^n]^2*Log[(b*c - a*d)/(b*(c + d*x))])/((c + d*x)*(a*g + b*g*x)),x]
Output:
-((Log[e*((a + b*x)/(c + d*x))^n]^2*PolyLog[2, 1 - (b*c - a*d)/(b*(c + d*x ))])/((b*c - a*d)*g)) + (2*n*((Log[e*((a + b*x)/(c + d*x))^n]*PolyLog[3, 1 - (b*c - a*d)/(b*(c + d*x))])/(b*c - a*d) - (n*PolyLog[4, 1 - (b*c - a*d) /(b*(c + d*x))])/(b*c - a*d)))/g
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]
\[\int \frac {\ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} \ln \left (\frac {-d a +b c}{b \left (d x +c \right )}\right )}{\left (d x +c \right ) \left (b g x +a g \right )}d x\]
Input:
int(ln(e*((b*x+a)/(d*x+c))^n)^2*ln((-a*d+b*c)/b/(d*x+c))/(d*x+c)/(b*g*x+a* g),x)
Output:
int(ln(e*((b*x+a)/(d*x+c))^n)^2*ln((-a*d+b*c)/b/(d*x+c))/(d*x+c)/(b*g*x+a* g),x)
\[ \int \frac {\log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx=\int { \frac {\log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}} \,d x } \] Input:
integrate(log(e*((b*x+a)/(d*x+c))^n)^2*log((-a*d+b*c)/b/(d*x+c))/(d*x+c)/( b*g*x+a*g),x, algorithm="fricas")
Output:
integral(log(e*((b*x + a)/(d*x + c))^n)^2*log((b*c - a*d)/(b*d*x + b*c))/( b*d*g*x^2 + a*c*g + (b*c + a*d)*g*x), x)
\[ \int \frac {\log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx=\frac {\int \frac {\log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}^{2} \log {\left (- \frac {a d}{b c + b d x} + \frac {b c}{b c + b d x} \right )}}{a c + a d x + b c x + b d x^{2}}\, dx}{g} \] Input:
integrate(ln(e*((b*x+a)/(d*x+c))**n)**2*ln((-a*d+b*c)/b/(d*x+c))/(d*x+c)/( b*g*x+a*g),x)
Output:
Integral(log(e*(a/(c + d*x) + b*x/(c + d*x))**n)**2*log(-a*d/(b*c + b*d*x) + b*c/(b*c + b*d*x))/(a*c + a*d*x + b*c*x + b*d*x**2), x)/g
\[ \int \frac {\log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx=\int { \frac {\log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}} \,d x } \] Input:
integrate(log(e*((b*x+a)/(d*x+c))^n)^2*log((-a*d+b*c)/b/(d*x+c))/(d*x+c)/( b*g*x+a*g),x, algorithm="maxima")
Output:
integrate(log(e*((b*x + a)/(d*x + c))^n)^2*log((b*c - a*d)/((d*x + c)*b))/ ((b*g*x + a*g)*(d*x + c)), x)
\[ \int \frac {\log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx=\int { \frac {\log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}} \,d x } \] Input:
integrate(log(e*((b*x+a)/(d*x+c))^n)^2*log((-a*d+b*c)/b/(d*x+c))/(d*x+c)/( b*g*x+a*g),x, algorithm="giac")
Output:
integrate(log(e*((b*x + a)/(d*x + c))^n)^2*log((b*c - a*d)/((d*x + c)*b))/ ((b*g*x + a*g)*(d*x + c)), x)
Timed out. \[ \int \frac {\log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx=\int \frac {\ln \left (-\frac {a\,d-b\,c}{b\,\left (c+d\,x\right )}\right )\,{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2}{\left (a\,g+b\,g\,x\right )\,\left (c+d\,x\right )} \,d x \] Input:
int((log(-(a*d - b*c)/(b*(c + d*x)))*log(e*((a + b*x)/(c + d*x))^n)^2)/((a *g + b*g*x)*(c + d*x)),x)
Output:
int((log(-(a*d - b*c)/(b*(c + d*x)))*log(e*((a + b*x)/(c + d*x))^n)^2)/((a *g + b*g*x)*(c + d*x)), x)
\[ \int \frac {\log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx=\frac {\int \frac {\mathrm {log}\left (\frac {-a d +b c}{b d x +b c}\right ) \mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right )^{2}}{b d \,x^{2}+a d x +b c x +a c}d x}{g} \] Input:
int(log(e*((b*x+a)/(d*x+c))^n)^2*log((-a*d+b*c)/b/(d*x+c))/(d*x+c)/(b*g*x+ a*g),x)
Output:
int((log(( - a*d + b*c)/(b*c + b*d*x))*log(((a + b*x)**n*e)/(c + d*x)**n)* *2)/(a*c + a*d*x + b*c*x + b*d*x**2),x)/g