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

3.1.2.1 Optimal result
3.1.2.2 Mathematica [A] (verified)
3.1.2.3 Rubi [A] (verified)
3.1.2.4 Maple [B] (verified)
3.1.2.5 Fricas [B] (verification not implemented)
3.1.2.6 Sympy [F(-1)]
3.1.2.7 Maxima [B] (verification not implemented)
3.1.2.8 Giac [B] (verification not implemented)
3.1.2.9 Mupad [B] (verification not implemented)

3.1.2.1 Optimal result

Integrand size = 33, antiderivative size = 156 \[ \int (a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=-\frac {B (b c-a d)^3 g^3 n x}{4 d^3}+\frac {B (b c-a d)^2 g^3 n (a+b x)^2}{8 b d^2}-\frac {B (b c-a d) g^3 n (a+b x)^3}{12 b d}+\frac {g^3 (a+b x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{4 b}+\frac {B (b c-a d)^4 g^3 n \log (c+d x)}{4 b d^4} \]

output
-1/4*B*(-a*d+b*c)^3*g^3*n*x/d^3+1/8*B*(-a*d+b*c)^2*g^3*n*(b*x+a)^2/b/d^2-1 
/12*B*(-a*d+b*c)*g^3*n*(b*x+a)^3/b/d+1/4*g^3*(b*x+a)^4*(A+B*ln(e*((b*x+a)/ 
(d*x+c))^n))/b+1/4*B*(-a*d+b*c)^4*g^3*n*ln(d*x+c)/b/d^4
 
3.1.2.2 Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.79 \[ \int (a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {g^3 \left ((a+b x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-\frac {B (b c-a d) n \left (6 b d (b c-a d)^2 x+3 d^2 (-b c+a d) (a+b x)^2+2 d^3 (a+b x)^3-6 (b c-a d)^3 \log (c+d x)\right )}{6 d^4}\right )}{4 b} \]

input
Integrate[(a*g + b*g*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n]),x]
 
output
(g^3*((a + b*x)^4*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) - (B*(b*c - a*d)* 
n*(6*b*d*(b*c - a*d)^2*x + 3*d^2*(-(b*c) + a*d)*(a + b*x)^2 + 2*d^3*(a + b 
*x)^3 - 6*(b*c - a*d)^3*Log[c + d*x]))/(6*d^4)))/(4*b)
 
3.1.2.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2947, 27, 49, 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.

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

\(\Big \downarrow \) 2947

\(\displaystyle \frac {g^3 (a+b x)^4 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{4 b}-\frac {B n (b c-a d) \int \frac {g^4 (a+b x)^3}{c+d x}dx}{4 b g}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {g^3 (a+b x)^4 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{4 b}-\frac {B g^3 n (b c-a d) \int \frac {(a+b x)^3}{c+d x}dx}{4 b}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {g^3 (a+b x)^4 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{4 b}-\frac {B g^3 n (b c-a d) \int \left (\frac {(a d-b c)^3}{d^3 (c+d x)}+\frac {b (b c-a d)^2}{d^3}+\frac {b (a+b x)^2}{d}-\frac {b (b c-a d) (a+b x)}{d^2}\right )dx}{4 b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {g^3 (a+b x)^4 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{4 b}-\frac {B g^3 n (b c-a d) \left (-\frac {(b c-a d)^3 \log (c+d x)}{d^4}+\frac {b x (b c-a d)^2}{d^3}-\frac {(a+b x)^2 (b c-a d)}{2 d^2}+\frac {(a+b x)^3}{3 d}\right )}{4 b}\)

input
Int[(a*g + b*g*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n]),x]
 
output
(g^3*(a + b*x)^4*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(4*b) - (B*(b*c - 
 a*d)*g^3*n*((b*(b*c - a*d)^2*x)/d^3 - ((b*c - a*d)*(a + b*x)^2)/(2*d^2) + 
 (a + b*x)^3/(3*d) - ((b*c - a*d)^3*Log[c + d*x])/d^4))/(4*b)
 

3.1.2.3.1 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 49
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

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

rule 2947
Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*((A + 
 B*Log[e*((a + b*x)/(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] /; Free 
Q[{a, b, c, d, e, f, g, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] 
&& NeQ[m, -2]
 
3.1.2.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(754\) vs. \(2(146)=292\).

Time = 7.22 (sec) , antiderivative size = 755, normalized size of antiderivative = 4.84

method result size
parallelrisch \(\frac {-36 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} b^{2} c^{2} d^{2} g^{3} n -6 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} c^{4} g^{3} n +6 B \ln \left (b x +a \right ) a^{4} d^{4} g^{3} n^{2}+6 B \ln \left (b x +a \right ) b^{4} c^{4} g^{3} n^{2}+9 B \,a^{3} b c \,d^{3} g^{3} n^{2}+24 B \,a^{2} b^{2} c^{2} d^{2} g^{3} n^{2}-21 B a \,b^{3} c^{3} d \,g^{3} n^{2}-60 A \,a^{3} b c \,d^{3} g^{3} n +6 B \,x^{4} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} d^{4} g^{3} n +2 B \,x^{3} a \,b^{3} d^{4} g^{3} n^{2}-2 B \,x^{3} b^{4} c \,d^{3} g^{3} n^{2}+24 A \,x^{3} a \,b^{3} d^{4} g^{3} n +9 B \,x^{2} a^{2} b^{2} d^{4} g^{3} n^{2}+3 B \,x^{2} b^{4} c^{2} d^{2} g^{3} n^{2}+36 A \,x^{2} a^{2} b^{2} d^{4} g^{3} n +18 B x \,a^{3} b \,d^{4} g^{3} n^{2}-6 B x \,b^{4} c^{3} d \,g^{3} n^{2}+24 A x \,a^{3} b \,d^{4} g^{3} n +24 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a \,b^{3} c^{3} d \,g^{3} n -24 B \ln \left (b x +a \right ) a^{3} b c \,d^{3} g^{3} n^{2}+36 B \ln \left (b x +a \right ) a^{2} b^{2} c^{2} d^{2} g^{3} n^{2}-24 B \ln \left (b x +a \right ) a \,b^{3} c^{3} d \,g^{3} n^{2}+24 B \,x^{3} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a \,b^{3} d^{4} g^{3} n +36 B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} b^{2} d^{4} g^{3} n -12 B \,x^{2} a \,b^{3} c \,d^{3} g^{3} n^{2}+24 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{3} b \,d^{4} g^{3} n -36 B x \,a^{2} b^{2} c \,d^{3} g^{3} n^{2}+24 B x a \,b^{3} c^{2} d^{2} g^{3} n^{2}+24 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{3} b c \,d^{3} g^{3} n -18 B \,a^{4} d^{4} g^{3} n^{2}+6 B \,b^{4} c^{4} g^{3} n^{2}-24 A \,a^{4} d^{4} g^{3} n +6 A \,x^{4} b^{4} d^{4} g^{3} n}{24 d^{4} n b}\) \(755\)

input
int((b*g*x+a*g)^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x,method=_RETURNVERBOSE)
 
output
1/24*(-36*B*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b^2*c^2*d^2*g^3*n-6*B*ln(e*((b*x 
+a)/(d*x+c))^n)*b^4*c^4*g^3*n+6*B*ln(b*x+a)*a^4*d^4*g^3*n^2+6*B*ln(b*x+a)* 
b^4*c^4*g^3*n^2+9*B*a^3*b*c*d^3*g^3*n^2+24*B*a^2*b^2*c^2*d^2*g^3*n^2-21*B* 
a*b^3*c^3*d*g^3*n^2-60*A*a^3*b*c*d^3*g^3*n+6*B*x^4*ln(e*((b*x+a)/(d*x+c))^ 
n)*b^4*d^4*g^3*n+2*B*x^3*a*b^3*d^4*g^3*n^2-2*B*x^3*b^4*c*d^3*g^3*n^2+24*A* 
x^3*a*b^3*d^4*g^3*n+9*B*x^2*a^2*b^2*d^4*g^3*n^2+3*B*x^2*b^4*c^2*d^2*g^3*n^ 
2+36*A*x^2*a^2*b^2*d^4*g^3*n+18*B*x*a^3*b*d^4*g^3*n^2-6*B*x*b^4*c^3*d*g^3* 
n^2+24*A*x*a^3*b*d^4*g^3*n+24*B*ln(e*((b*x+a)/(d*x+c))^n)*a*b^3*c^3*d*g^3* 
n-24*B*ln(b*x+a)*a^3*b*c*d^3*g^3*n^2+36*B*ln(b*x+a)*a^2*b^2*c^2*d^2*g^3*n^ 
2-24*B*ln(b*x+a)*a*b^3*c^3*d*g^3*n^2+24*B*x^3*ln(e*((b*x+a)/(d*x+c))^n)*a* 
b^3*d^4*g^3*n+36*B*x^2*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b^2*d^4*g^3*n-12*B*x^ 
2*a*b^3*c*d^3*g^3*n^2+24*B*x*ln(e*((b*x+a)/(d*x+c))^n)*a^3*b*d^4*g^3*n-36* 
B*x*a^2*b^2*c*d^3*g^3*n^2+24*B*x*a*b^3*c^2*d^2*g^3*n^2+24*B*ln(e*((b*x+a)/ 
(d*x+c))^n)*a^3*b*c*d^3*g^3*n-18*B*a^4*d^4*g^3*n^2+6*B*b^4*c^4*g^3*n^2-24* 
A*a^4*d^4*g^3*n+6*A*x^4*b^4*d^4*g^3*n)/d^4/n/b
 
3.1.2.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (146) = 292\).

Time = 0.32 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.73 \[ \int (a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {6 \, A b^{4} d^{4} g^{3} x^{4} + 6 \, B a^{4} d^{4} g^{3} n \log \left (b x + a\right ) + 6 \, {\left (B b^{4} c^{4} - 4 \, B a b^{3} c^{3} d + 6 \, B a^{2} b^{2} c^{2} d^{2} - 4 \, B a^{3} b c d^{3}\right )} g^{3} n \log \left (d x + c\right ) + 2 \, {\left (12 \, A a b^{3} d^{4} g^{3} - {\left (B b^{4} c d^{3} - B a b^{3} d^{4}\right )} g^{3} n\right )} x^{3} + 3 \, {\left (12 \, A a^{2} b^{2} d^{4} g^{3} + {\left (B b^{4} c^{2} d^{2} - 4 \, B a b^{3} c d^{3} + 3 \, B a^{2} b^{2} d^{4}\right )} g^{3} n\right )} x^{2} + 6 \, {\left (4 \, A a^{3} b d^{4} g^{3} - {\left (B b^{4} c^{3} d - 4 \, B a b^{3} c^{2} d^{2} + 6 \, B a^{2} b^{2} c d^{3} - 3 \, B a^{3} b d^{4}\right )} g^{3} n\right )} x + 6 \, {\left (B b^{4} d^{4} g^{3} x^{4} + 4 \, B a b^{3} d^{4} g^{3} x^{3} + 6 \, B a^{2} b^{2} d^{4} g^{3} x^{2} + 4 \, B a^{3} b d^{4} g^{3} x\right )} \log \left (e\right ) + 6 \, {\left (B b^{4} d^{4} g^{3} n x^{4} + 4 \, B a b^{3} d^{4} g^{3} n x^{3} + 6 \, B a^{2} b^{2} d^{4} g^{3} n x^{2} + 4 \, B a^{3} b d^{4} g^{3} n x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{24 \, b d^{4}} \]

input
integrate((b*g*x+a*g)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="fri 
cas")
 
output
1/24*(6*A*b^4*d^4*g^3*x^4 + 6*B*a^4*d^4*g^3*n*log(b*x + a) + 6*(B*b^4*c^4 
- 4*B*a*b^3*c^3*d + 6*B*a^2*b^2*c^2*d^2 - 4*B*a^3*b*c*d^3)*g^3*n*log(d*x + 
 c) + 2*(12*A*a*b^3*d^4*g^3 - (B*b^4*c*d^3 - B*a*b^3*d^4)*g^3*n)*x^3 + 3*( 
12*A*a^2*b^2*d^4*g^3 + (B*b^4*c^2*d^2 - 4*B*a*b^3*c*d^3 + 3*B*a^2*b^2*d^4) 
*g^3*n)*x^2 + 6*(4*A*a^3*b*d^4*g^3 - (B*b^4*c^3*d - 4*B*a*b^3*c^2*d^2 + 6* 
B*a^2*b^2*c*d^3 - 3*B*a^3*b*d^4)*g^3*n)*x + 6*(B*b^4*d^4*g^3*x^4 + 4*B*a*b 
^3*d^4*g^3*x^3 + 6*B*a^2*b^2*d^4*g^3*x^2 + 4*B*a^3*b*d^4*g^3*x)*log(e) + 6 
*(B*b^4*d^4*g^3*n*x^4 + 4*B*a*b^3*d^4*g^3*n*x^3 + 6*B*a^2*b^2*d^4*g^3*n*x^ 
2 + 4*B*a^3*b*d^4*g^3*n*x)*log((b*x + a)/(d*x + c)))/(b*d^4)
 
3.1.2.6 Sympy [F(-1)]

Timed out. \[ \int (a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\text {Timed out} \]

input
integrate((b*g*x+a*g)**3*(A+B*ln(e*((b*x+a)/(d*x+c))**n)),x)
 
output
Timed out
 
3.1.2.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 479 vs. \(2 (146) = 292\).

Time = 0.21 (sec) , antiderivative size = 479, normalized size of antiderivative = 3.07 \[ \int (a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {1}{4} \, B b^{3} g^{3} x^{4} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {1}{4} \, A b^{3} g^{3} x^{4} + B a b^{2} g^{3} x^{3} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A a b^{2} g^{3} x^{3} + \frac {3}{2} \, B a^{2} b g^{3} x^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {3}{2} \, A a^{2} b g^{3} x^{2} - \frac {1}{24} \, B b^{3} g^{3} n {\left (\frac {6 \, a^{4} \log \left (b x + a\right )}{b^{4}} - \frac {6 \, c^{4} \log \left (d x + c\right )}{d^{4}} + \frac {2 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{3} - 3 \, {\left (b^{3} c^{2} d - a^{2} b d^{3}\right )} x^{2} + 6 \, {\left (b^{3} c^{3} - a^{3} d^{3}\right )} x}{b^{3} d^{3}}\right )} + \frac {1}{2} \, B a b^{2} g^{3} n {\left (\frac {2 \, a^{3} \log \left (b x + a\right )}{b^{3}} - \frac {2 \, c^{3} \log \left (d x + c\right )}{d^{3}} - \frac {{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} - \frac {3}{2} \, B a^{2} b g^{3} n {\left (\frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} + B a^{3} g^{3} n {\left (\frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} + B a^{3} g^{3} x \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A a^{3} g^{3} x \]

input
integrate((b*g*x+a*g)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="max 
ima")
 
output
1/4*B*b^3*g^3*x^4*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 1/4*A*b^3*g^3*x 
^4 + B*a*b^2*g^3*x^3*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + A*a*b^2*g^3* 
x^3 + 3/2*B*a^2*b*g^3*x^2*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 3/2*A*a 
^2*b*g^3*x^2 - 1/24*B*b^3*g^3*n*(6*a^4*log(b*x + a)/b^4 - 6*c^4*log(d*x + 
c)/d^4 + (2*(b^3*c*d^2 - a*b^2*d^3)*x^3 - 3*(b^3*c^2*d - a^2*b*d^3)*x^2 + 
6*(b^3*c^3 - a^3*d^3)*x)/(b^3*d^3)) + 1/2*B*a*b^2*g^3*n*(2*a^3*log(b*x + a 
)/b^3 - 2*c^3*log(d*x + c)/d^3 - ((b^2*c*d - a*b*d^2)*x^2 - 2*(b^2*c^2 - a 
^2*d^2)*x)/(b^2*d^2)) - 3/2*B*a^2*b*g^3*n*(a^2*log(b*x + a)/b^2 - c^2*log( 
d*x + c)/d^2 + (b*c - a*d)*x/(b*d)) + B*a^3*g^3*n*(a*log(b*x + a)/b - c*lo 
g(d*x + c)/d) + B*a^3*g^3*x*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + A*a^3 
*g^3*x
 
3.1.2.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3034 vs. \(2 (146) = 292\).

Time = 0.81 (sec) , antiderivative size = 3034, normalized size of antiderivative = 19.45 \[ \int (a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\text {Too large to display} \]

input
integrate((b*g*x+a*g)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="gia 
c")
 
output
-1/24*(6*(B*b^8*c^5*g^3*n - 5*B*a*b^7*c^4*d*g^3*n - 4*(b*x + a)*B*b^7*c^5* 
d*g^3*n/(d*x + c) + 10*B*a^2*b^6*c^3*d^2*g^3*n + 20*(b*x + a)*B*a*b^6*c^4* 
d^2*g^3*n/(d*x + c) + 6*(b*x + a)^2*B*b^6*c^5*d^2*g^3*n/(d*x + c)^2 - 10*B 
*a^3*b^5*c^2*d^3*g^3*n - 40*(b*x + a)*B*a^2*b^5*c^3*d^3*g^3*n/(d*x + c) - 
30*(b*x + a)^2*B*a*b^5*c^4*d^3*g^3*n/(d*x + c)^2 - 4*(b*x + a)^3*B*b^5*c^5 
*d^3*g^3*n/(d*x + c)^3 + 5*B*a^4*b^4*c*d^4*g^3*n + 40*(b*x + a)*B*a^3*b^4* 
c^2*d^4*g^3*n/(d*x + c) + 60*(b*x + a)^2*B*a^2*b^4*c^3*d^4*g^3*n/(d*x + c) 
^2 + 20*(b*x + a)^3*B*a*b^4*c^4*d^4*g^3*n/(d*x + c)^3 - B*a^5*b^3*d^5*g^3* 
n - 20*(b*x + a)*B*a^4*b^3*c*d^5*g^3*n/(d*x + c) - 60*(b*x + a)^2*B*a^3*b^ 
3*c^2*d^5*g^3*n/(d*x + c)^2 - 40*(b*x + a)^3*B*a^2*b^3*c^3*d^5*g^3*n/(d*x 
+ c)^3 + 4*(b*x + a)*B*a^5*b^2*d^6*g^3*n/(d*x + c) + 30*(b*x + a)^2*B*a^4* 
b^2*c*d^6*g^3*n/(d*x + c)^2 + 40*(b*x + a)^3*B*a^3*b^2*c^2*d^6*g^3*n/(d*x 
+ c)^3 - 6*(b*x + a)^2*B*a^5*b*d^7*g^3*n/(d*x + c)^2 - 20*(b*x + a)^3*B*a^ 
4*b*c*d^7*g^3*n/(d*x + c)^3 + 4*(b*x + a)^3*B*a^5*d^8*g^3*n/(d*x + c)^3)*l 
og((b*x + a)/(d*x + c))/(b^4*d^4 - 4*(b*x + a)*b^3*d^5/(d*x + c) + 6*(b*x 
+ a)^2*b^2*d^6/(d*x + c)^2 - 4*(b*x + a)^3*b*d^7/(d*x + c)^3 + (b*x + a)^4 
*d^8/(d*x + c)^4) + (11*B*b^8*c^5*g^3*n - 55*B*a*b^7*c^4*d*g^3*n - 38*(b*x 
 + a)*B*b^7*c^5*d*g^3*n/(d*x + c) + 110*B*a^2*b^6*c^3*d^2*g^3*n + 190*(b*x 
 + a)*B*a*b^6*c^4*d^2*g^3*n/(d*x + c) + 45*(b*x + a)^2*B*b^6*c^5*d^2*g^3*n 
/(d*x + c)^2 - 110*B*a^3*b^5*c^2*d^3*g^3*n - 380*(b*x + a)*B*a^2*b^5*c^...
 
3.1.2.9 Mupad [B] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 588, normalized size of antiderivative = 3.77 \[ \int (a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=x^3\,\left (\frac {b^2\,g^3\,\left (16\,A\,a\,d+4\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{12\,d}-\frac {A\,b^2\,g^3\,\left (4\,a\,d+4\,b\,c\right )}{12\,d}\right )-x^2\,\left (\frac {\left (\frac {b^2\,g^3\,\left (16\,A\,a\,d+4\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{4\,d}-\frac {A\,b^2\,g^3\,\left (4\,a\,d+4\,b\,c\right )}{4\,d}\right )\,\left (4\,a\,d+4\,b\,c\right )}{8\,b\,d}-\frac {a\,b\,g^3\,\left (6\,A\,a\,d+4\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{2\,d}+\frac {A\,a\,b^2\,c\,g^3}{2\,d}\right )+\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (B\,a^3\,g^3\,x+\frac {3\,B\,a^2\,b\,g^3\,x^2}{2}+B\,a\,b^2\,g^3\,x^3+\frac {B\,b^3\,g^3\,x^4}{4}\right )+x\,\left (\frac {\left (4\,a\,d+4\,b\,c\right )\,\left (\frac {\left (\frac {b^2\,g^3\,\left (16\,A\,a\,d+4\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{4\,d}-\frac {A\,b^2\,g^3\,\left (4\,a\,d+4\,b\,c\right )}{4\,d}\right )\,\left (4\,a\,d+4\,b\,c\right )}{4\,b\,d}-\frac {a\,b\,g^3\,\left (6\,A\,a\,d+4\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{d}+\frac {A\,a\,b^2\,c\,g^3}{d}\right )}{4\,b\,d}+\frac {a^2\,g^3\,\left (8\,A\,a\,d+12\,A\,b\,c+3\,B\,a\,d\,n-3\,B\,b\,c\,n\right )}{2\,d}-\frac {a\,c\,\left (\frac {b^2\,g^3\,\left (16\,A\,a\,d+4\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{4\,d}-\frac {A\,b^2\,g^3\,\left (4\,a\,d+4\,b\,c\right )}{4\,d}\right )}{b\,d}\right )+\frac {\ln \left (c+d\,x\right )\,\left (-4\,B\,n\,a^3\,c\,d^3\,g^3+6\,B\,n\,a^2\,b\,c^2\,d^2\,g^3-4\,B\,n\,a\,b^2\,c^3\,d\,g^3+B\,n\,b^3\,c^4\,g^3\right )}{4\,d^4}+\frac {A\,b^3\,g^3\,x^4}{4}+\frac {B\,a^4\,g^3\,n\,\ln \left (a+b\,x\right )}{4\,b} \]

input
int((a*g + b*g*x)^3*(A + B*log(e*((a + b*x)/(c + d*x))^n)),x)
 
output
x^3*((b^2*g^3*(16*A*a*d + 4*A*b*c + B*a*d*n - B*b*c*n))/(12*d) - (A*b^2*g^ 
3*(4*a*d + 4*b*c))/(12*d)) - x^2*((((b^2*g^3*(16*A*a*d + 4*A*b*c + B*a*d*n 
 - B*b*c*n))/(4*d) - (A*b^2*g^3*(4*a*d + 4*b*c))/(4*d))*(4*a*d + 4*b*c))/( 
8*b*d) - (a*b*g^3*(6*A*a*d + 4*A*b*c + B*a*d*n - B*b*c*n))/(2*d) + (A*a*b^ 
2*c*g^3)/(2*d)) + log(e*((a + b*x)/(c + d*x))^n)*((B*b^3*g^3*x^4)/4 + B*a^ 
3*g^3*x + (3*B*a^2*b*g^3*x^2)/2 + B*a*b^2*g^3*x^3) + x*(((4*a*d + 4*b*c)*( 
(((b^2*g^3*(16*A*a*d + 4*A*b*c + B*a*d*n - B*b*c*n))/(4*d) - (A*b^2*g^3*(4 
*a*d + 4*b*c))/(4*d))*(4*a*d + 4*b*c))/(4*b*d) - (a*b*g^3*(6*A*a*d + 4*A*b 
*c + B*a*d*n - B*b*c*n))/d + (A*a*b^2*c*g^3)/d))/(4*b*d) + (a^2*g^3*(8*A*a 
*d + 12*A*b*c + 3*B*a*d*n - 3*B*b*c*n))/(2*d) - (a*c*((b^2*g^3*(16*A*a*d + 
 4*A*b*c + B*a*d*n - B*b*c*n))/(4*d) - (A*b^2*g^3*(4*a*d + 4*b*c))/(4*d))) 
/(b*d)) + (log(c + d*x)*(B*b^3*c^4*g^3*n - 4*B*a^3*c*d^3*g^3*n - 4*B*a*b^2 
*c^3*d*g^3*n + 6*B*a^2*b*c^2*d^2*g^3*n))/(4*d^4) + (A*b^3*g^3*x^4)/4 + (B* 
a^4*g^3*n*log(a + b*x))/(4*b)