Integrand size = 24, antiderivative size = 135 \[ \int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\frac {(a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{b}+\frac {2 B (b c-a d) n \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b d}+\frac {2 B^2 (b c-a d) n^2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d} \]
(b*x+a)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))^2/b+2*B*(-a*d+b*c)*n*(A+B*ln(e*((b *x+a)/(d*x+c))^n))*ln((-a*d+b*c)/b/(d*x+c))/b/d+2*B^2*(-a*d+b*c)*n^2*polyl og(2,d*(b*x+a)/b/(d*x+c))/b/d
Time = 0.11 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.67 \[ \int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=x \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2+\frac {B n \left (2 a d \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-2 b c \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)-a B d n \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )+b B c n \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{b d} \]
x*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2 + (B*n*(2*a*d*Log[a + b*x]*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) - 2*b*c*(A + B*Log[e*((a + b*x)/(c + d* x))^n])*Log[c + d*x] - a*B*d*n*(Log[a + b*x]*(Log[a + b*x] - 2*Log[(b*(c + d*x))/(b*c - a*d)]) - 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)]) + b*B*c *n*((2*Log[(d*(a + b*x))/(-(b*c) + a*d)] - Log[c + d*x])*Log[c + d*x] + 2* PolyLog[2, (b*(c + d*x))/(b*c - a*d)])))/(b*d)
Time = 0.60 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2935, 2943, 2858, 27, 25, 2778, 2005, 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.
\(\displaystyle \int \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2 \, dx\) |
\(\Big \downarrow \) 2935 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}-\frac {2 B n (b c-a d) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c+d x}dx}{b}\) |
\(\Big \downarrow \) 2943 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}-\frac {2 B n (b c-a d) \left (\frac {B n (b c-a d) \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)}dx}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 2858 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}-\frac {2 B n (b c-a d) \left (\frac {B n (b c-a d) \int \frac {d \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) \left (\left (a-\frac {b c}{d}\right ) d+b (c+d x)\right )}d(c+d x)}{d^2}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}-\frac {2 B n (b c-a d) \left (\frac {B n (b c-a d) \int -\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (b c-a d-b (c+d x))}d(c+d x)}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}-\frac {2 B n (b c-a d) \left (-\frac {B n (b c-a d) \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (b c-a d-b (c+d x))}d(c+d x)}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 2778 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}-\frac {2 B n (b c-a d) \left (\frac {B n (b c-a d) \int \frac {(c+d x) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b c-a d-b (c+d x)}d\frac {1}{c+d x}}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}-\frac {2 B n (b c-a d) \left (\frac {B n (b c-a d) \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{\frac {b c-a d}{c+d x}-b}d\frac {1}{c+d x}}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b}-\frac {2 B n (b c-a d) \left (-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d}-\frac {B n \operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right )}{d}\right )}{b}\) |
((a + b*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/b - (2*B*(b*c - a*d)* n*(-(((A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[(b*c - a*d)/(b*(c + d*x)) ])/d) - (B*n*PolyLog[2, 1 - (b*c - a*d)/(b*(c + d*x))])/d))/b
3.1.70.3.1 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[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
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]
Int[((a_.) + Log[(c_.)*(x_)^(n_)]*(b_.))/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[1/n Subst[Int[(a + b*Log[c*x])/(x*(d + e*x^(r/n))), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IntegerQ[r/n]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_ .)*(x_))^(q_.)*((h_.) + (i_.)*(x_))^(r_.), x_Symbol] :> Simp[1/e Subst[In t[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[r, 0]) && IntegerQ[2*r]
Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( B_.))^(p_.), x_Symbol] :> Simp[(a + b*x)*((A + B*Log[e*((a + b*x)/(c + d*x) )^n])^p/b), x] - Simp[B*n*p*((b*c - a*d)/b) Int[(A + B*Log[e*((a + b*x)/( c + d*x))^n])^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0]
Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( B_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(-Log[(b*c - a*d)/(b*(c + d*x ))])*((A + B*Log[e*((a + b*x)/(c + d*x))^n])/g), x] + Simp[B*n*((b*c - a*d) /g) Int[Log[(b*c - a*d)/(b*(c + d*x))]/((a + b*x)*(c + d*x)), x], x] /; F reeQ[{a, b, c, d, e, f, g, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[d*f - c *g, 0]
\[\int {\left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}^{2}d x\]
\[ \int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\int { {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2} \,d x } \]
integral(B^2*log(e*((b*x + a)/(d*x + c))^n)^2 + 2*A*B*log(e*((b*x + a)/(d* x + c))^n) + A^2, x)
\[ \int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\int \left (A + B \log {\left (e \left (\frac {a + b x}{c + d x}\right )^{n} \right )}\right )^{2}\, dx \]
\[ \int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\int { {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2} \,d x } \]
2*A*B*n*(a*log(b*x + a)/b - c*log(d*x + c)/d) + 2*A*B*x*log(e*((b*x + a)/( d*x + c))^n) + A^2*x + B^2*((2*b*c*n^2*log(b*x + a)*log(d*x + c) - b*c*n^2 *log(d*x + c)^2 + b*d*x*log((b*x + a)^n)^2 + b*d*x*log((d*x + c)^n)^2 + 2* (a*d*n*log(b*x + a) - b*c*n*log(d*x + c) + b*d*x*log(e))*log((b*x + a)^n) - 2*(a*d*n*log(b*x + a) - b*c*n*log(d*x + c) + b*d*x*log((b*x + a)^n) + b* d*x*log(e))*log((d*x + c)^n))/(b*d) - integrate(-(b^2*d*x^2*log(e)^2 + a*b *c*log(e)^2 - ((2*n*log(e) - log(e)^2)*b^2*c - (2*n*log(e) + log(e)^2)*a*b *d)*x - 2*(b^2*c*n^2*x + 2*a*b*c*n^2 - a^2*d*n^2)*log(b*x + a))/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x), x))
\[ \int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\int { {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2} \,d x } \]
Timed out. \[ \int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \, dx=\int {\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^2 \,d x \]