Integrand size = 32, antiderivative size = 83 \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx=-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{b g}-\frac {2 B \operatorname {PolyLog}\left (2,1+\frac {b c-a d}{d (a+b x)}\right )}{b g} \] Output:
-ln(-(-a*d+b*c)/d/(b*x+a))*(A+B*ln(e*(d*x+c)^2/(b*x+a)^2))/b/g-2*B*polylog (2,1+(-a*d+b*c)/d/(b*x+a))/b/g
Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05 \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx=\frac {\log (a+b x) \left (A+B \log (a+b x)-2 B \log \left (\frac {b (c+d x)}{b c-a d}\right )+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )-2 B \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )}{b g} \] Input:
Integrate[(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])/(a*g + b*g*x),x]
Output:
(Log[a + b*x]*(A + B*Log[a + b*x] - 2*B*Log[(b*(c + d*x))/(b*c - a*d)] + B *Log[(e*(c + d*x)^2)/(a + b*x)^2]) - 2*B*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)])/(b*g)
Time = 0.53 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2942, 2858, 27, 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.
\(\displaystyle \int \frac {B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A}{a g+b g x} \, dx\) |
\(\Big \downarrow \) 2942 |
\(\displaystyle -\frac {2 B (b c-a d) \int \frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right )}{(a+b x) (c+d x)}dx}{b g}-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g}\) |
\(\Big \downarrow \) 2858 |
\(\displaystyle -\frac {2 B (b c-a d) \int \frac {b \log \left (-\frac {b c-a d}{d (a+b x)}\right )}{(a+b x) \left (b \left (c-\frac {a d}{b}\right )+d (a+b x)\right )}d(a+b x)}{b^2 g}-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 B (b c-a d) \int \frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right )}{(a+b x) (b c-a d+d (a+b x))}d(a+b x)}{b g}-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g}\) |
\(\Big \downarrow \) 2778 |
\(\displaystyle \frac {2 B (b c-a d) \int \frac {(a+b x) \log \left (-\frac {b c-a d}{d (a+b x)}\right )}{b c-a d+d (a+b x)}d\frac {1}{a+b x}}{b g}-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g}\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \frac {2 B (b c-a d) \int \frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right )}{d+\frac {b c-a d}{a+b x}}d\frac {1}{a+b x}}{b g}-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle -\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{b g}-\frac {2 B \operatorname {PolyLog}\left (2,\frac {b c-a d}{d (a+b x)}+1\right )}{b g}\) |
Input:
Int[(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])/(a*g + b*g*x),x]
Output:
-((Log[-((b*c - a*d)/(d*(a + b*x)))]*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^ 2]))/(b*g)) - (2*B*PolyLog[2, 1 + (b*c - a*d)/(d*(a + b*x))])/(b*g)
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_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(-Log[-(b*c - a*d)/(d*(a + b*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)/(d*(a + b*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[b*f - a*g, 0]
Time = 1.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.95
method | result | size |
derivativedivides | \(-\frac {\frac {A \ln \left (\frac {1}{b x +a}\right )}{g}+\frac {B \left (\ln \left (\frac {1}{b x +a}\right ) \ln \left (\frac {e \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2}}{b^{2}}\right )-\left (2 d a -2 b c \right ) \left (\frac {\operatorname {dilog}\left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right )}{d a -b c}+\frac {\ln \left (\frac {1}{b x +a}\right ) \ln \left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right )}{d a -b c}\right )\right )}{g}}{b}\) | \(162\) |
default | \(-\frac {\frac {A \ln \left (\frac {1}{b x +a}\right )}{g}+\frac {B \left (\ln \left (\frac {1}{b x +a}\right ) \ln \left (\frac {e \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2}}{b^{2}}\right )-\left (2 d a -2 b c \right ) \left (\frac {\operatorname {dilog}\left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right )}{d a -b c}+\frac {\ln \left (\frac {1}{b x +a}\right ) \ln \left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right )}{d a -b c}\right )\right )}{g}}{b}\) | \(162\) |
parts | \(\frac {A \ln \left (b x +a \right )}{g b}-\frac {B \left (\ln \left (\frac {1}{b x +a}\right ) \ln \left (\frac {e \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2}}{b^{2}}\right )-\left (2 d a -2 b c \right ) \left (\frac {\operatorname {dilog}\left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right )}{d a -b c}+\frac {\ln \left (\frac {1}{b x +a}\right ) \ln \left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right )}{d a -b c}\right )\right )}{g b}\) | \(162\) |
risch | \(\frac {A \ln \left (b x +a \right )}{g b}-\frac {B \ln \left (\frac {1}{b x +a}\right ) \ln \left (\frac {e \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2}}{b^{2}}\right )}{b g}+\frac {2 B \operatorname {dilog}\left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right ) d a}{b g \left (d a -b c \right )}-\frac {2 B \operatorname {dilog}\left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right ) c}{g \left (d a -b c \right )}+\frac {2 B \ln \left (\frac {1}{b x +a}\right ) \ln \left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right ) d a}{b g \left (d a -b c \right )}-\frac {2 B \ln \left (\frac {1}{b x +a}\right ) \ln \left (-\frac {\frac {d a -b c}{b x +a}-d}{d}\right ) c}{g \left (d a -b c \right )}\) | \(262\) |
Input:
int((A+B*ln(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g),x,method=_RETURNVERBOSE)
Output:
-1/b*(1/g*A*ln(1/(b*x+a))+1/g*B*(ln(1/(b*x+a))*ln(e*(a*d/(b*x+a)-b*c/(b*x+ a)-d)^2/b^2)-(2*a*d-2*b*c)*(dilog(-(1/(b*x+a)*(a*d-b*c)-d)/d)/(a*d-b*c)+ln (1/(b*x+a))*ln(-(1/(b*x+a)*(a*d-b*c)-d)/d)/(a*d-b*c))))
\[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx=\int { \frac {B \log \left (\frac {{\left (d x + c\right )}^{2} e}{{\left (b x + a\right )}^{2}}\right ) + A}{b g x + a g} \,d x } \] Input:
integrate((A+B*log(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g),x, algorithm="frica s")
Output:
integral((B*log((d^2*e*x^2 + 2*c*d*e*x + c^2*e)/(b^2*x^2 + 2*a*b*x + a^2)) + A)/(b*g*x + a*g), x)
\[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx=\frac {\int \frac {A}{a + b x}\, dx + \int \frac {B \log {\left (\frac {c^{2} e}{a^{2} + 2 a b x + b^{2} x^{2}} + \frac {2 c d e x}{a^{2} + 2 a b x + b^{2} x^{2}} + \frac {d^{2} e x^{2}}{a^{2} + 2 a b x + b^{2} x^{2}} \right )}}{a + b x}\, dx}{g} \] Input:
integrate((A+B*ln(e*(d*x+c)**2/(b*x+a)**2))/(b*g*x+a*g),x)
Output:
(Integral(A/(a + b*x), x) + Integral(B*log(c**2*e/(a**2 + 2*a*b*x + b**2*x **2) + 2*c*d*e*x/(a**2 + 2*a*b*x + b**2*x**2) + d**2*e*x**2/(a**2 + 2*a*b* x + b**2*x**2))/(a + b*x), x))/g
\[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx=\int { \frac {B \log \left (\frac {{\left (d x + c\right )}^{2} e}{{\left (b x + a\right )}^{2}}\right ) + A}{b g x + a g} \,d x } \] Input:
integrate((A+B*log(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g),x, algorithm="maxim a")
Output:
B*(2*log(b*x + a)*log(d*x + c)/(b*g) - integrate(-(b*d*x*log(e) + b*c*log( e) - 2*(2*b*d*x + b*c + a*d)*log(b*x + a))/(b^2*d*g*x^2 + a*b*c*g + (b^2*c *g + a*b*d*g)*x), x)) + A*log(b*g*x + a*g)/(b*g)
\[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx=\int { \frac {B \log \left (\frac {{\left (d x + c\right )}^{2} e}{{\left (b x + a\right )}^{2}}\right ) + A}{b g x + a g} \,d x } \] Input:
integrate((A+B*log(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g),x, algorithm="giac" )
Output:
integrate((B*log((d*x + c)^2*e/(b*x + a)^2) + A)/(b*g*x + a*g), x)
Timed out. \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx=\int \frac {A+B\,\ln \left (\frac {e\,{\left (c+d\,x\right )}^2}{{\left (a+b\,x\right )}^2}\right )}{a\,g+b\,g\,x} \,d x \] Input:
int((A + B*log((e*(c + d*x)^2)/(a + b*x)^2))/(a*g + b*g*x),x)
Output:
int((A + B*log((e*(c + d*x)^2)/(a + b*x)^2))/(a*g + b*g*x), x)
\[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{a g+b g x} \, dx=\frac {\left (\int \frac {\mathrm {log}\left (\frac {d^{2} e \,x^{2}+2 c d e x +c^{2} e}{b^{2} x^{2}+2 a b x +a^{2}}\right )}{b x +a}d x \right ) b^{2}+\mathrm {log}\left (b x +a \right ) a}{b g} \] Input:
int((A+B*log(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g),x)
Output:
(int(log((c**2*e + 2*c*d*e*x + d**2*e*x**2)/(a**2 + 2*a*b*x + b**2*x**2))/ (a + b*x),x)*b**2 + log(a + b*x)*a)/(b*g)