Integrand size = 33, antiderivative size = 131 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=-\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b}+\frac {2 B n \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b}+\frac {2 B^2 n^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b} \]
-(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2*ln(1-b*(d*x+c)/d/(b*x+a))/b+2*B*n*(A+ B*ln(e*(b*x+a)^n/((d*x+c)^n)))*polylog(2,b*(d*x+c)/d/(b*x+a))/b+2*B^2*n^2* polylog(3,b*(d*x+c)/d/(b*x+a))/b
Leaf count is larger than twice the leaf count of optimal. \(269\) vs. \(2(131)=262\).
Time = 0.14 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.05 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\frac {-A B n \log ^2\left (\frac {-b c+a d}{d (a+b x)}\right )+A^2 \log (a+b x)-2 A B n \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {b (c+d x)}{b c-a d}\right )-2 A B \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )-B^2 \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log ^2\left (e (a+b x)^n (c+d x)^{-n}\right )+2 A B n \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )+2 B^2 n \log \left (e (a+b x)^n (c+d x)^{-n}\right ) \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )+2 B^2 n^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b} \]
(-(A*B*n*Log[(-(b*c) + a*d)/(d*(a + b*x))]^2) + A^2*Log[a + b*x] - 2*A*B*n *Log[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(b*(c + d*x))/(b*c - a*d)] - 2*A*B* Log[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(e*(a + b*x)^n)/(c + d*x)^n] - B^2*L og[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(e*(a + b*x)^n)/(c + d*x)^n]^2 + 2*A* B*n*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)] + 2*B^2*n*Log[(e*(a + b*x)^n) /(c + d*x)^n]*PolyLog[2, (b*(c + d*x))/(d*(a + b*x))] + 2*B^2*n^2*PolyLog[ 3, (b*(c + d*x))/(d*(a + b*x))])/b
Time = 0.57 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2973, 2949, 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.
| \(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{a+b x} \, dx\) |
| \(\Big \downarrow \) 2973 |
| \(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{a+b x}dx\) |
| \(\Big \downarrow \) 2949 |
| \(\displaystyle \int \frac {(c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{(a+b x) \left (b-\frac {d (a+b x)}{c+d x}\right )}d\frac {a+b x}{c+d x}\) |
| \(\Big \downarrow \) 2779 |
| \(\displaystyle \frac {2 B n \int \frac {(c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{a+b x}d\frac {a+b x}{c+d x}}{b}-\frac {\log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}\) |
| \(\Big \downarrow \) 2821 |
| \(\displaystyle \frac {2 B n \left (\operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )-B n \int \frac {(c+d x) \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{a+b x}d\frac {a+b x}{c+d x}\right )}{b}-\frac {\log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {2 B n \left (\operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+B n \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )\right )}{b}-\frac {\log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}\) |
-(((A + B*Log[e*((a + b*x)/(c + d*x))^n])^2*Log[1 - (b*(c + d*x))/(d*(a + b*x))])/b) + (2*B*n*((A + B*Log[e*((a + b*x)/(c + d*x))^n])*PolyLog[2, (b* (c + d*x))/(d*(a + b*x))] + B*n*PolyLog[3, (b*(c + d*x))/(d*(a + b*x))]))/ b
3.2.59.3.1 Defintions of rubi rules used
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_.), x_Symbol] :> Simp[(b*c - a*d)^(m + 1)*(g/b)^m Subst[Int[x^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && Ne Q[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[b*f - a*g, 0] && (GtQ[p, 0] || Lt Q[m, -1])
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]
\[\int \frac {{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{2}}{b x +a}d x\]
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{b x + a} \,d x } \]
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*x + a), x)
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int \frac {\left (A + B \log {\left (e \left (a + b x\right )^{n} \left (c + d x\right )^{- n} \right )}\right )^{2}}{a + b x}\, dx \]
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{b x + a} \,d x } \]
B^2*log(b*x + a)*log((d*x + c)^n)^2/b + A^2*log(b*x + a)/b - integrate(-(B ^2*b*c*log(e)^2 + 2*A*B*b*c*log(e) + (B^2*b*d*x + B^2*b*c)*log((b*x + a)^n )^2 + (B^2*b*d*log(e)^2 + 2*A*B*b*d*log(e))*x + 2*(B^2*b*c*log(e) + A*B*b* c + (B^2*b*d*log(e) + A*B*b*d)*x)*log((b*x + a)^n) - 2*(B^2*b*c*log(e) + A *B*b*c + (B^2*b*d*log(e) + A*B*b*d)*x + (B^2*b*d*n*x + B^2*a*d*n)*log(b*x + a) + (B^2*b*d*x + B^2*b*c)*log((b*x + a)^n))*log((d*x + c)^n))/(b^2*d*x^ 2 + a*b*c + (b^2*c + a*b*d)*x), x)
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{b x + a} \,d x } \]
Timed out. \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int \frac {{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}^2}{a+b\,x} \,d x \]