3.3.7 \(\int \frac {A+B \log (\frac {e (c+d x)^2}{(a+b x)^2})}{(a g+b g x)^3} \, dx\) [207]

3.3.7.1 Optimal result
3.3.7.2 Mathematica [A] (verified)
3.3.7.3 Rubi [A] (verified)
3.3.7.4 Maple [A] (verified)
3.3.7.5 Fricas [A] (verification not implemented)
3.3.7.6 Sympy [B] (verification not implemented)
3.3.7.7 Maxima [B] (verification not implemented)
3.3.7.8 Giac [A] (verification not implemented)
3.3.7.9 Mupad [B] (verification not implemented)

3.3.7.1 Optimal result

Integrand size = 32, antiderivative size = 139 \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{(a g+b g x)^3} \, dx=\frac {B}{2 b g^3 (a+b x)^2}-\frac {B d}{b (b c-a d) g^3 (a+b x)}-\frac {B d^2 \log (a+b x)}{b (b c-a d)^2 g^3}+\frac {B d^2 \log (c+d x)}{b (b c-a d)^2 g^3}-\frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{2 b g^3 (a+b x)^2} \]

output
1/2*B/b/g^3/(b*x+a)^2-B*d/b/(-a*d+b*c)/g^3/(b*x+a)-B*d^2*ln(b*x+a)/b/(-a*d 
+b*c)^2/g^3+B*d^2*ln(d*x+c)/b/(-a*d+b*c)^2/g^3+1/2*(-A-B*ln(e*(d*x+c)^2/(b 
*x+a)^2))/b/g^3/(b*x+a)^2
 
3.3.7.2 Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.92 \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{(a g+b g x)^3} \, dx=-\frac {2 B d^2 (a+b x)^2 \log (a+b x)-2 B d^2 (a+b x)^2 \log (c+d x)+(b c-a d) \left (A b c-b B c-a A d+3 a B d+2 b B d x+B (b c-a d) \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}{2 b (b c-a d)^2 g^3 (a+b x)^2} \]

input
Integrate[(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])/(a*g + b*g*x)^3,x]
 
output
-1/2*(2*B*d^2*(a + b*x)^2*Log[a + b*x] - 2*B*d^2*(a + b*x)^2*Log[c + d*x] 
+ (b*c - a*d)*(A*b*c - b*B*c - a*A*d + 3*a*B*d + 2*b*B*d*x + B*(b*c - a*d) 
*Log[(e*(c + d*x)^2)/(a + b*x)^2]))/(b*(b*c - a*d)^2*g^3*(a + b*x)^2)
 
3.3.7.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2948, 27, 54, 2009}

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.

\(\displaystyle \int \frac {B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A}{(a g+b g x)^3} \, dx\)

\(\Big \downarrow \) 2948

\(\displaystyle -\frac {B (b c-a d) \int \frac {1}{g^2 (a+b x)^3 (c+d x)}dx}{b g}-\frac {B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A}{2 b g^3 (a+b x)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {B (b c-a d) \int \frac {1}{(a+b x)^3 (c+d x)}dx}{b g^3}-\frac {B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A}{2 b g^3 (a+b x)^2}\)

\(\Big \downarrow \) 54

\(\displaystyle -\frac {B (b c-a d) \int \left (-\frac {d^3}{(b c-a d)^3 (c+d x)}+\frac {b d^2}{(b c-a d)^3 (a+b x)}-\frac {b d}{(b c-a d)^2 (a+b x)^2}+\frac {b}{(b c-a d) (a+b x)^3}\right )dx}{b g^3}-\frac {B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A}{2 b g^3 (a+b x)^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A}{2 b g^3 (a+b x)^2}-\frac {B (b c-a d) \left (\frac {d^2 \log (a+b x)}{(b c-a d)^3}-\frac {d^2 \log (c+d x)}{(b c-a d)^3}+\frac {d}{(a+b x) (b c-a d)^2}-\frac {1}{2 (a+b x)^2 (b c-a d)}\right )}{b g^3}\)

input
Int[(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])/(a*g + b*g*x)^3,x]
 
output
-((B*(b*c - a*d)*(-1/2*1/((b*c - a*d)*(a + b*x)^2) + d/((b*c - a*d)^2*(a + 
 b*x)) + (d^2*Log[a + b*x])/(b*c - a*d)^3 - (d^2*Log[c + d*x])/(b*c - a*d) 
^3))/(b*g^3)) - (A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])/(2*b*g^3*(a + b*x 
)^2)
 

3.3.7.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 54
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2948
Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ 
)]*(B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*( 
(A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])/(g*(m + 1))), x] - Simp[B*n*((b*c 
- a*d)/(g*(m + 1)))   Int[(f + g*x)^(m + 1)/((a + b*x)*(c + d*x)), x], x] / 
; FreeQ[{a, b, c, d, e, f, g, A, B, m, n}, x] && EqQ[n + mn, 0] && NeQ[b*c 
- a*d, 0] && NeQ[m, -1] &&  !(EqQ[m, -2] && IntegerQ[n])
 
3.3.7.4 Maple [A] (verified)

Time = 1.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.17

method result size
derivativedivides \(-\frac {\frac {A}{2 g^{3} \left (b x +a \right )^{2}}+\frac {B \left (\frac {\ln \left (\frac {e \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2}}{b^{2}}\right )}{2 \left (b x +a \right )^{2}}-\left (a d -c b \right ) \left (\frac {\frac {a d}{2 \left (b x +a \right )^{2}}-\frac {b c}{2 \left (b x +a \right )^{2}}+\frac {d}{b x +a}}{\left (a d -c b \right )^{2}}+\frac {d^{2} \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{\left (a d -c b \right )^{3}}\right )\right )}{g^{3}}}{b}\) \(162\)
default \(-\frac {\frac {A}{2 g^{3} \left (b x +a \right )^{2}}+\frac {B \left (\frac {\ln \left (\frac {e \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2}}{b^{2}}\right )}{2 \left (b x +a \right )^{2}}-\left (a d -c b \right ) \left (\frac {\frac {a d}{2 \left (b x +a \right )^{2}}-\frac {b c}{2 \left (b x +a \right )^{2}}+\frac {d}{b x +a}}{\left (a d -c b \right )^{2}}+\frac {d^{2} \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{\left (a d -c b \right )^{3}}\right )\right )}{g^{3}}}{b}\) \(162\)
parts \(-\frac {A}{2 g^{3} \left (b x +a \right )^{2} b}-\frac {B \left (\frac {\ln \left (\frac {e \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2}}{b^{2}}\right )}{2 \left (b x +a \right )^{2}}-\left (a d -c b \right ) \left (\frac {\frac {a d}{2 \left (b x +a \right )^{2}}-\frac {b c}{2 \left (b x +a \right )^{2}}+\frac {d}{b x +a}}{\left (a d -c b \right )^{2}}+\frac {d^{2} \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{\left (a d -c b \right )^{3}}\right )\right )}{g^{3} b}\) \(164\)
norman \(\frac {\frac {B d x}{g \left (a d -c b \right )}+\frac {B a \,d^{2} x \ln \left (\frac {e \left (d x +c \right )^{2}}{\left (b x +a \right )^{2}}\right )}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) g}-\frac {A a b d -A \,b^{2} c -3 B a b d +B \,b^{2} c}{2 g \,b^{2} \left (a d -c b \right )}+\frac {B c \left (2 a d -c b \right ) \ln \left (\frac {e \left (d x +c \right )^{2}}{\left (b x +a \right )^{2}}\right )}{2 g \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {B \,d^{2} b \,x^{2} \ln \left (\frac {e \left (d x +c \right )^{2}}{\left (b x +a \right )^{2}}\right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) g}}{g^{2} \left (b x +a \right )^{2}}\) \(232\)
parallelrisch \(-\frac {-2 B x a \,b^{4} d^{3}+2 B x \,b^{5} c \,d^{2}+A \,a^{2} b^{3} d^{3}+A \,b^{5} c^{2} d -3 B \,a^{2} b^{3} d^{3}-B \,b^{5} c^{2} d -2 A a \,b^{4} c \,d^{2}+4 B a \,b^{4} c \,d^{2}-2 B \ln \left (\frac {e \left (d x +c \right )^{2}}{\left (b x +a \right )^{2}}\right ) a \,b^{4} c \,d^{2}-B \,x^{2} \ln \left (\frac {e \left (d x +c \right )^{2}}{\left (b x +a \right )^{2}}\right ) b^{5} d^{3}+B \ln \left (\frac {e \left (d x +c \right )^{2}}{\left (b x +a \right )^{2}}\right ) b^{5} c^{2} d -2 B x \ln \left (\frac {e \left (d x +c \right )^{2}}{\left (b x +a \right )^{2}}\right ) a \,b^{4} d^{3}}{2 g^{3} \left (b x +a \right )^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{4} d}\) \(240\)
risch \(-\frac {B \ln \left (\frac {e \left (d x +c \right )^{2}}{\left (b x +a \right )^{2}}\right )}{2 b \,g^{3} \left (b x +a \right )^{2}}-\frac {2 B \ln \left (b x +a \right ) b^{2} d^{2} x^{2}-2 B \ln \left (-d x -c \right ) b^{2} d^{2} x^{2}+4 B \ln \left (b x +a \right ) a b \,d^{2} x -4 B \ln \left (-d x -c \right ) a b \,d^{2} x +2 B \,a^{2} \ln \left (b x +a \right ) d^{2}-2 B \ln \left (-d x -c \right ) a^{2} d^{2}-2 B a b \,d^{2} x +2 B \,b^{2} c d x +A \,a^{2} d^{2}-2 A a b c d +A \,b^{2} c^{2}-3 B \,a^{2} d^{2}+4 B a b c d -B \,b^{2} c^{2}}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) g^{3} \left (b x +a \right )^{2} b}\) \(245\)

input
int((A+B*ln(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g)^3,x,method=_RETURNVERBOSE)
 
output
-1/b*(1/2/g^3*A/(b*x+a)^2+1/g^3*B*(1/2/(b*x+a)^2*ln(e*(a*d/(b*x+a)-b*c/(b* 
x+a)-d)^2/b^2)-(a*d-b*c)*(1/(a*d-b*c)^2*(1/2*a*d/(b*x+a)^2-1/2*b*c/(b*x+a) 
^2+d/(b*x+a))+d^2/(a*d-b*c)^3*ln(a*d/(b*x+a)-b*c/(b*x+a)-d))))
 
3.3.7.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.73 \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{(a g+b g x)^3} \, dx=-\frac {{\left (A - B\right )} b^{2} c^{2} - 2 \, {\left (A - 2 \, B\right )} a b c d + {\left (A - 3 \, B\right )} a^{2} d^{2} + 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} x - {\left (B b^{2} d^{2} x^{2} + 2 \, B a b d^{2} x - B b^{2} c^{2} + 2 \, B a b c d\right )} \log \left (\frac {d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{2 \, {\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} g^{3} x + {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} g^{3}\right )}} \]

input
integrate((A+B*log(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g)^3,x, algorithm="fri 
cas")
 
output
-1/2*((A - B)*b^2*c^2 - 2*(A - 2*B)*a*b*c*d + (A - 3*B)*a^2*d^2 + 2*(B*b^2 
*c*d - B*a*b*d^2)*x - (B*b^2*d^2*x^2 + 2*B*a*b*d^2*x - B*b^2*c^2 + 2*B*a*b 
*c*d)*log((d^2*e*x^2 + 2*c*d*e*x + c^2*e)/(b^2*x^2 + 2*a*b*x + a^2)))/((b^ 
5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*g^3*x^2 + 2*(a*b^4*c^2 - 2*a^2*b^3*c*d 
+ a^3*b^2*d^2)*g^3*x + (a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^4*b*d^2)*g^3)
 
3.3.7.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (122) = 244\).

Time = 1.14 (sec) , antiderivative size = 418, normalized size of antiderivative = 3.01 \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{(a g+b g x)^3} \, dx=- \frac {B \log {\left (\frac {e \left (c + d x\right )^{2}}{\left (a + b x\right )^{2}} \right )}}{2 a^{2} b g^{3} + 4 a b^{2} g^{3} x + 2 b^{3} g^{3} x^{2}} + \frac {B d^{2} \log {\left (x + \frac {- \frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} + \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} - \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} + \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{b g^{3} \left (a d - b c\right )^{2}} - \frac {B d^{2} \log {\left (x + \frac {\frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} - \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} + \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} - \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{b g^{3} \left (a d - b c\right )^{2}} + \frac {- A a d + A b c + 3 B a d - B b c + 2 B b d x}{2 a^{3} b d g^{3} - 2 a^{2} b^{2} c g^{3} + x^{2} \cdot \left (2 a b^{3} d g^{3} - 2 b^{4} c g^{3}\right ) + x \left (4 a^{2} b^{2} d g^{3} - 4 a b^{3} c g^{3}\right )} \]

input
integrate((A+B*ln(e*(d*x+c)**2/(b*x+a)**2))/(b*g*x+a*g)**3,x)
 
output
-B*log(e*(c + d*x)**2/(a + b*x)**2)/(2*a**2*b*g**3 + 4*a*b**2*g**3*x + 2*b 
**3*g**3*x**2) + B*d**2*log(x + (-B*a**3*d**5/(a*d - b*c)**2 + 3*B*a**2*b* 
c*d**4/(a*d - b*c)**2 - 3*B*a*b**2*c**2*d**3/(a*d - b*c)**2 + B*a*d**3 + B 
*b**3*c**3*d**2/(a*d - b*c)**2 + B*b*c*d**2)/(2*B*b*d**3))/(b*g**3*(a*d - 
b*c)**2) - B*d**2*log(x + (B*a**3*d**5/(a*d - b*c)**2 - 3*B*a**2*b*c*d**4/ 
(a*d - b*c)**2 + 3*B*a*b**2*c**2*d**3/(a*d - b*c)**2 + B*a*d**3 - B*b**3*c 
**3*d**2/(a*d - b*c)**2 + B*b*c*d**2)/(2*B*b*d**3))/(b*g**3*(a*d - b*c)**2 
) + (-A*a*d + A*b*c + 3*B*a*d - B*b*c + 2*B*b*d*x)/(2*a**3*b*d*g**3 - 2*a* 
*2*b**2*c*g**3 + x**2*(2*a*b**3*d*g**3 - 2*b**4*c*g**3) + x*(4*a**2*b**2*d 
*g**3 - 4*a*b**3*c*g**3))
 
3.3.7.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (135) = 270\).

Time = 0.21 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.20 \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{(a g+b g x)^3} \, dx=-\frac {1}{2} \, B {\left (\frac {2 \, b d x - b c + 3 \, a d}{{\left (b^{4} c - a b^{3} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} g^{3} x + {\left (a^{2} b^{2} c - a^{3} b d\right )} g^{3}} + \frac {\log \left (\frac {d^{2} e x^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {2 \, c d e x}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}} + \frac {2 \, d^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}} - \frac {2 \, d^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}}\right )} - \frac {A}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} \]

input
integrate((A+B*log(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g)^3,x, algorithm="max 
ima")
 
output
-1/2*B*((2*b*d*x - b*c + 3*a*d)/((b^4*c - a*b^3*d)*g^3*x^2 + 2*(a*b^3*c - 
a^2*b^2*d)*g^3*x + (a^2*b^2*c - a^3*b*d)*g^3) + log(d^2*e*x^2/(b^2*x^2 + 2 
*a*b*x + a^2) + 2*c*d*e*x/(b^2*x^2 + 2*a*b*x + a^2) + c^2*e/(b^2*x^2 + 2*a 
*b*x + a^2))/(b^3*g^3*x^2 + 2*a*b^2*g^3*x + a^2*b*g^3) + 2*d^2*log(b*x + a 
)/((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*g^3) - 2*d^2*log(d*x + c)/((b^3*c^2 
 - 2*a*b^2*c*d + a^2*b*d^2)*g^3)) - 1/2*A/(b^3*g^3*x^2 + 2*a*b^2*g^3*x + a 
^2*b*g^3)
 
3.3.7.8 Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.93 \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{(a g+b g x)^3} \, dx=-\frac {B d^{2} \log \left (b x + a\right )}{b^{3} c^{2} g^{3} - 2 \, a b^{2} c d g^{3} + a^{2} b d^{2} g^{3}} + \frac {B d^{2} \log \left (d x + c\right )}{b^{3} c^{2} g^{3} - 2 \, a b^{2} c d g^{3} + a^{2} b d^{2} g^{3}} - \frac {B \log \left (\frac {d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} - \frac {2 \, B b d x + A b c - B b c - A a d + 3 \, B a d}{2 \, {\left (b^{4} c g^{3} x^{2} - a b^{3} d g^{3} x^{2} + 2 \, a b^{3} c g^{3} x - 2 \, a^{2} b^{2} d g^{3} x + a^{2} b^{2} c g^{3} - a^{3} b d g^{3}\right )}} \]

input
integrate((A+B*log(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g)^3,x, algorithm="gia 
c")
 
output
-B*d^2*log(b*x + a)/(b^3*c^2*g^3 - 2*a*b^2*c*d*g^3 + a^2*b*d^2*g^3) + B*d^ 
2*log(d*x + c)/(b^3*c^2*g^3 - 2*a*b^2*c*d*g^3 + a^2*b*d^2*g^3) - 1/2*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))/(b^3*g^3*x^2 + 
 2*a*b^2*g^3*x + a^2*b*g^3) - 1/2*(2*B*b*d*x + A*b*c - B*b*c - A*a*d + 3*B 
*a*d)/(b^4*c*g^3*x^2 - a*b^3*d*g^3*x^2 + 2*a*b^3*c*g^3*x - 2*a^2*b^2*d*g^3 
*x + a^2*b^2*c*g^3 - a^3*b*d*g^3)
 
3.3.7.9 Mupad [B] (verification not implemented)

Time = 2.54 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.48 \[ \int \frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{(a g+b g x)^3} \, dx=\frac {2\,B\,d^2\,\mathrm {atanh}\left (\frac {b^3\,c^2\,g^3-a^2\,b\,d^2\,g^3}{b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {B\,\ln \left (\frac {e\,{\left (c+d\,x\right )}^2}{{\left (a+b\,x\right )}^2}\right )}{2\,b^2\,g^3\,\left (2\,a\,x+b\,x^2+\frac {a^2}{b}\right )}-\frac {\frac {A\,a\,d-A\,b\,c-3\,B\,a\,d+B\,b\,c}{2\,\left (a\,d-b\,c\right )}-\frac {B\,b\,d\,x}{a\,d-b\,c}}{a^2\,b\,g^3+2\,a\,b^2\,g^3\,x+b^3\,g^3\,x^2} \]

input
int((A + B*log((e*(c + d*x)^2)/(a + b*x)^2))/(a*g + b*g*x)^3,x)
 
output
(2*B*d^2*atanh((b^3*c^2*g^3 - a^2*b*d^2*g^3)/(b*g^3*(a*d - b*c)^2) - (2*b* 
d*x)/(a*d - b*c)))/(b*g^3*(a*d - b*c)^2) - (B*log((e*(c + d*x)^2)/(a + b*x 
)^2))/(2*b^2*g^3*(2*a*x + b*x^2 + a^2/b)) - ((A*a*d - A*b*c - 3*B*a*d + B* 
b*c)/(2*(a*d - b*c)) - (B*b*d*x)/(a*d - b*c))/(a^2*b*g^3 + b^3*g^3*x^2 + 2 
*a*b^2*g^3*x)