Integrand size = 43, antiderivative size = 203 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(f+g x) (a h+b h x)} \, dx=-\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \log \left (1-\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )}{(b f-a g) h}+\frac {2 B n \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \operatorname {PolyLog}\left (2,\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )}{(b f-a g) h}+\frac {2 B^2 n^2 \operatorname {PolyLog}\left (3,\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )}{(b f-a g) h} \]
-(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2*ln(1-(-a*g+b*f)*(d*x+c)/(-c*g+d*f)/(b *x+a))/(-a*g+b*f)/h+2*B*n*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))*polylog(2,(-a* g+b*f)*(d*x+c)/(-c*g+d*f)/(b*x+a))/(-a*g+b*f)/h+2*B^2*n^2*polylog(3,(-a*g+ b*f)*(d*x+c)/(-c*g+d*f)/(b*x+a))/(-a*g+b*f)/h
Leaf count is larger than twice the leaf count of optimal. \(1415\) vs. \(2(203)=406\).
Time = 0.78 (sec) , antiderivative size = 1415, normalized size of antiderivative = 6.97 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(f+g x) (a h+b h x)} \, dx =\text {Too large to display} \]
(3*Log[a + b*x]*(A + B*(-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[(e*(a + b *x)^n)/(c + d*x)^n]))^2 - 3*(A + B*(-(n*Log[a + b*x]) + n*Log[c + d*x] + L og[(e*(a + b*x)^n)/(c + d*x)^n]))^2*Log[f + g*x] + 3*B*n*(A + B*(-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[(e*(a + b*x)^n)/(c + d*x)^n]))*(Log[a + b* x]^2 - 2*(Log[a + b*x]*Log[(b*(f + g*x))/(b*f - a*g)] + PolyLog[2, (g*(a + b*x))/(-(b*f) + a*g)])) - 6*A*B*n*(Log[c + d*x]*(Log[(d*(a + b*x))/(-(b*c ) + a*d)] - Log[(d*(f + g*x))/(d*f - c*g)]) + PolyLog[2, (b*(c + d*x))/(b* c - a*d)] - PolyLog[2, (g*(c + d*x))/(-(d*f) + c*g)]) + 6*B^2*n*(n*Log[a + b*x] - n*Log[c + d*x] - Log[(e*(a + b*x)^n)/(c + d*x)^n])*(Log[c + d*x]*( Log[(d*(a + b*x))/(-(b*c) + a*d)] - Log[(d*(f + g*x))/(d*f - c*g)]) + Poly Log[2, (b*(c + d*x))/(b*c - a*d)] - PolyLog[2, (g*(c + d*x))/(-(d*f) + c*g )]) + B^2*n^2*(Log[a + b*x]^2*(Log[a + b*x] - 3*Log[(b*(f + g*x))/(b*f - a *g)]) - 6*Log[a + b*x]*PolyLog[2, (g*(a + b*x))/(-(b*f) + a*g)] + 6*PolyLo g[3, (g*(a + b*x))/(-(b*f) + a*g)]) + 3*B^2*n^2*(Log[(d*(a + b*x))/(-(b*c) + a*d)]*Log[c + d*x]^2 - Log[c + d*x]^2*Log[(d*(f + g*x))/(d*f - c*g)] + 2*Log[c + d*x]*PolyLog[2, (b*(c + d*x))/(b*c - a*d)] - 2*Log[c + d*x]*Poly Log[2, (g*(c + d*x))/(-(d*f) + c*g)] - 2*PolyLog[3, (b*(c + d*x))/(b*c - a *d)] + 2*PolyLog[3, (g*(c + d*x))/(-(d*f) + c*g)]) - 6*B^2*n^2*((Log[a + b *x]^2*(Log[c + d*x] - Log[(b*(c + d*x))/(b*c - a*d)]))/2 - Log[a + b*x]*Lo g[c + d*x]*Log[(b*(f + g*x))/(b*f - a*g)] - (Log[(g*(c + d*x))/(-(d*f) ...
Time = 0.87 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {2973, 2967, 27, 2779, 2821, 7143}
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.
| \(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{(f+g x) (a h+b h x)} \, dx\) |
| \(\Big \downarrow \) 2973 |
| \(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{(f+g x) (a h+b h x)}dx\) |
| \(\Big \downarrow \) 2967 |
| \(\displaystyle (b c-a d) \int \frac {(c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b c-a d) h (a+b x) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}d\frac {a+b x}{c+d x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a+b x) \left (b f-a g-\frac {(d f-c g) (a+b x)}{c+d x}\right )}d\frac {a+b x}{c+d x}}{h}\) |
| \(\Big \downarrow \) 2779 |
| \(\displaystyle \frac {\frac {2 B n \int \frac {(c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (1-\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )}{a+b x}d\frac {a+b x}{c+d x}}{b f-a g}-\frac {\log \left (1-\frac {(c+d x) (b f-a g)}{(a+b x) (d f-c g)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b f-a g}}{h}\) |
| \(\Big \downarrow \) 2821 |
| \(\displaystyle \frac {\frac {2 B n \left (\operatorname {PolyLog}\left (2,\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )-B n \int \frac {(c+d x) \operatorname {PolyLog}\left (2,\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )}{a+b x}d\frac {a+b x}{c+d x}\right )}{b f-a g}-\frac {\log \left (1-\frac {(c+d x) (b f-a g)}{(a+b x) (d f-c g)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b f-a g}}{h}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {\frac {2 B n \left (\operatorname {PolyLog}\left (2,\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+B n \operatorname {PolyLog}\left (3,\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )\right )}{b f-a g}-\frac {\log \left (1-\frac {(c+d x) (b f-a g)}{(a+b x) (d f-c g)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b f-a g}}{h}\) |
(-(((A + B*Log[e*((a + b*x)/(c + d*x))^n])^2*Log[1 - ((b*f - a*g)*(c + d*x ))/((d*f - c*g)*(a + b*x))])/(b*f - a*g)) + (2*B*n*((A + B*Log[e*((a + b*x )/(c + d*x))^n])*PolyLog[2, ((b*f - a*g)*(c + d*x))/((d*f - c*g)*(a + b*x) )] + B*n*PolyLog[3, ((b*f - a*g)*(c + d*x))/((d*f - c*g)*(a + b*x))]))/(b* f - a*g))/h
3.3.51.3.1 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[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b _.))^(p_.))/(x_), x_Symbol] :> Simp[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c *x^n])^p/m), x] + Simp[b*n*(p/m) Int[PolyLog[2, (-d)*f*x^m]*((a + b*Log[c *x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]
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) Subst[Int[(b*f - a*g - (d*f - c*g)*x)^m*(b*h - a*i - (d*h - c*i)*x)^q*((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]
Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] :> Subst[Int[w*(A + B*Log[e*(u/v)^n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; Fr eeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] && !Intege rQ[n]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
\[\int \frac {{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{2}}{\left (g x +f \right ) \left (b h x +a h \right )}d x\]
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(f+g x) (a h+b h x)} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{{\left (b h x + a h\right )} {\left (g x + f\right )}} \,d x } \]
integral((B^2*log((b*x + a)^n*e/(d*x + c)^n)^2 + 2*A*B*log((b*x + a)^n*e/( d*x + c)^n) + A^2)/(b*g*h*x^2 + a*f*h + (b*f + a*g)*h*x), x)
Timed out. \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(f+g x) (a h+b h x)} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(f+g x) (a h+b h x)} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{{\left (b h x + a h\right )} {\left (g x + f\right )}} \,d x } \]
A^2*(log(b*x + a)/((b*f - a*g)*h) - log(g*x + f)/((b*f - a*g)*h)) + integr ate((B^2*log((b*x + a)^n)^2 + B^2*log((d*x + c)^n)^2 + B^2*log(e)^2 + 2*A* B*log(e) + 2*(B^2*log(e) + A*B)*log((b*x + a)^n) - 2*(B^2*log((b*x + a)^n) + B^2*log(e) + A*B)*log((d*x + c)^n))/(b*g*h*x^2 + a*f*h + (b*f*h + a*g*h )*x), x)
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(f+g x) (a h+b h x)} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{{\left (b h x + a h\right )} {\left (g x + f\right )}} \,d x } \]
Timed out. \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(f+g x) (a h+b h x)} \, dx=\int \frac {{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}^2}{\left (f+g\,x\right )\,\left (a\,h+b\,h\,x\right )} \,d x \]