Integrand size = 21, antiderivative size = 125 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\frac {2 B (b c-a d) \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b d}+\frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{b}+\frac {2 B^2 (b c-a d) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d} \]
2*B*(-a*d+b*c)*ln((-a*d+b*c)/b/(d*x+c))*(A+B*ln(e*(b*x+a)/(d*x+c)))/b/d+(b *x+a)*(A+B*ln(e*(b*x+a)/(d*x+c)))^2/b+2*B^2*(-a*d+b*c)*polylog(2,d*(b*x+a) /b/(d*x+c))/b/d
Time = 0.11 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.71 \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=x \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2+\frac {B \left (2 a d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )-2 b c \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)-a B d \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )+b B c \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{b d} \]
x*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2 + (B*(2*a*d*Log[a + b*x]*(A + B*L og[(e*(a + b*x))/(c + d*x)]) - 2*b*c*(A + B*Log[(e*(a + b*x))/(c + d*x)])* Log[c + d*x] - a*B*d*(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*((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 = 120, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2936, 2944, 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.
\(\displaystyle \int \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2 \, dx\) |
\(\Big \downarrow \) 2936 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{b}-\frac {2 B (b c-a d) \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{c+d x}dx}{b}\) |
\(\Big \downarrow \) 2944 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{b}-\frac {2 B (b c-a d) \left (\frac {B (b c-a d) \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)}dx}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 2858 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{b}-\frac {2 B (b c-a d) \left (\frac {B (b c-a d) \int \frac {d \log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) \left (\left (a-\frac {b c}{d}\right ) d+b (c+d x)\right )}d(c+d x)}{d^2}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{b}-\frac {2 B (b c-a d) \left (\frac {B (b c-a d) \int -\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (b c-a d-b (c+d x))}d(c+d x)}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{b}-\frac {2 B (b c-a d) \left (-\frac {B (b c-a d) \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{(c+d x) (b c-a d-b (c+d x))}d(c+d x)}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 2778 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{b}-\frac {2 B (b c-a d) \left (\frac {B (b c-a d) \int \frac {(c+d x) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b c-a d-b (c+d x)}d\frac {1}{c+d x}}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{b}-\frac {2 B (b c-a d) \left (\frac {B (b c-a d) \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{\frac {b c-a d}{c+d x}-b}d\frac {1}{c+d x}}{d}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d}\right )}{b}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {(a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{b}-\frac {2 B (b c-a d) \left (-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d}-\frac {B \operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right )}{d}\right )}{b}\) |
((a + b*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2)/b - (2*B*(b*c - a*d)*(- ((Log[(b*c - a*d)/(b*(c + d*x))]*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/d) - (B*PolyLog[2, 1 - (b*c - a*d)/(b*(c + d*x))])/d))/b
3.3.43.3.1 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_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ )]*(B_.))^(p_.), x_Symbol] :> Simp[(a + b*x)*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])^p/b), x] - Simp[B*n*p*((b*c - a*d)/b) Int[(A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && IGtQ[p, 0]
Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ )]*(B_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(-Log[(b*c - a*d)/(b*(c + d*x))])*((A + B*Log[e*((a + b*x)^n/(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 ] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && EqQ[d*f - c*g, 0]
\[\int \left (A +B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )\right )^{2}d x\]
\[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\int { {\left (B \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}^{2} \,d x } \]
Timed out. \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\text {Timed out} \]
\[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\int { {\left (B \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}^{2} \,d x } \]
2*(x*log((b*x + a)*e/(d*x + c)) + (a*e*log(b*x + a)/b - c*e*log(d*x + c)/d )/e)*A*B + A^2*x + B^2*((b*d*x*log(b*x + a)^2 + (b*d*x + b*c)*log(d*x + c) ^2 - 2*(b*d*x*log(e) + (b*d*x + a*d)*log(b*x + a))*log(d*x + c))/(b*d) + i ntegrate(((log(e)^2 + 2*log(e))*b^2*d*x^2 + a*b*c*log(e)^2 + (b^2*c*log(e) ^2 + (log(e)^2 + 2*log(e))*a*b*d)*x + 2*(b^2*d*x^2*log(e) + a*b*c*log(e) + a^2*d + (a*b*d*(log(e) + 2) + b^2*c*(log(e) - 1))*x)*log(b*x + a))/(b^2*d *x^2 + a*b*c + (b^2*c + a*b*d)*x), x))
\[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\int { {\left (B \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}^{2} \,d x } \]
Timed out. \[ \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx=\int {\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}^2 \,d x \]