\(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(a g+b g x) (c i+d i x)^3} \, dx\) [155]

Optimal result

Integrand size = 43, antiderivative size = 254 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x) (c i+d i x)^3} \, dx=-\frac {B n \left (4 b-\frac {d (a+b x)}{c+d x}\right )^2}{4 (b c-a d)^3 g i^3}+\frac {d^2 (a+b x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 (b c-a d)^3 g i^3 (c+d x)^2}-\frac {2 b d (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d)^3 g i^3 (c+d x)}+\frac {b^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {a+b x}{c+d x}\right )}{(b c-a d)^3 g i^3}-\frac {b^2 B n \log ^2\left (\frac {a+b x}{c+d x}\right )}{2 (b c-a d)^3 g i^3} \] Output:

-1/4*B*n*(4*b-d*(b*x+a)/(d*x+c))^2/(-a*d+b*c)^3/g/i^3+1/2*d^2*(b*x+a)^2*(A 
+B*ln(e*((b*x+a)/(d*x+c))^n))/(-a*d+b*c)^3/g/i^3/(d*x+c)^2-2*b*d*(b*x+a)*( 
A+B*ln(e*((b*x+a)/(d*x+c))^n))/(-a*d+b*c)^3/g/i^3/(d*x+c)+b^2*(A+B*ln(e*(( 
b*x+a)/(d*x+c))^n))*ln((b*x+a)/(d*x+c))/(-a*d+b*c)^3/g/i^3-1/2*b^2*B*n*ln( 
(b*x+a)/(d*x+c))^2/(-a*d+b*c)^3/g/i^3
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.38 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.71 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x) (c i+d i x)^3} \, dx=\frac {2 (b c-a d)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+4 b (b c-a d) (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+4 b^2 (c+d x)^2 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-4 b^2 (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)-4 b B n (c+d x) (b c-a d+b (c+d x) \log (a+b x)-b (c+d x) \log (c+d x))-B n \left ((b c-a d)^2+2 b (b c-a d) (c+d x)+2 b^2 (c+d x)^2 \log (a+b x)-2 b^2 (c+d x)^2 \log (c+d x)\right )-2 b^2 B n (c+d x)^2 \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )+2 b^2 B n (c+d x)^2 \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )}{4 (b c-a d)^3 g i^3 (c+d x)^2} \] Input:

Integrate[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/((a*g + b*g*x)*(c*i + d*i 
*x)^3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2961, 2772, 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.

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

\(\Big \downarrow \) 2961

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

\(\Big \downarrow \) 2772

\(\displaystyle \frac {-B n \int \left (\frac {b^2 (c+d x) \log \left (\frac {a+b x}{c+d x}\right )}{a+b x}-\frac {1}{2} d \left (4 b-\frac {d (a+b x)}{c+d x}\right )\right )d\frac {a+b x}{c+d x}+b^2 \log \left (\frac {a+b x}{c+d x}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+\frac {d^2 (a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 (c+d x)^2}-\frac {2 b d (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{c+d x}}{g i^3 (b c-a d)^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 \log \left (\frac {a+b x}{c+d x}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+\frac {d^2 (a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 (c+d x)^2}-\frac {2 b d (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{c+d x}-B n \left (\frac {1}{2} b^2 \log ^2\left (\frac {a+b x}{c+d x}\right )+\frac {1}{4} \left (4 b-\frac {d (a+b x)}{c+d x}\right )^2\right )}{g i^3 (b c-a d)^3}\)

Input:

Int[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/((a*g + b*g*x)*(c*i + d*i*x)^3) 
,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2772

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_ 
.))^(q_.), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^r)^q, x]}, Simp[(a + 
 b*Log[c*x^n])   u, x] - Simp[b*n   Int[SimplifyIntegrand[u/x, x], x], x]] 
/; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q 
, 1] && EqQ[m, -1])
 

rule 2961

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol 
] :> Simp[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q   Subst[Int[x^m*((A + B*L 
og[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, i, A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(632\) vs. \(2(248)=496\).

Time = 13.38 (sec) , antiderivative size = 633, normalized size of antiderivative = 2.49

method	result	size
parallelrisch	\(-\frac {2 B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} a^{2} b^{2} c^{4} d^{2}-8 B \,x^{2} a^{3} b \,c^{3} d^{3} n^{2}+7 B \,x^{2} a^{2} b^{2} c^{4} d^{2} n^{2}+4 A \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} b^{2} c^{4} d^{2}+8 A \,x^{2} a^{3} b \,c^{3} d^{3} n -6 A \,x^{2} a^{2} b^{2} c^{4} d^{2} n -8 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} b^{2} c^{5} d n -4 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{3} b \,c^{4} d^{2} n +2 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} a^{2} b^{2} c^{6}+4 A \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} b^{2} c^{6}+4 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} a^{2} b^{2} c^{5} d -10 B x \,a^{3} b \,c^{4} d^{2} n^{2}+8 B x \,a^{2} b^{2} c^{5} d \,n^{2}+8 A x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} b^{2} c^{5} d +12 A x \,a^{3} b \,c^{4} d^{2} n -8 A x \,a^{2} b^{2} c^{5} d n -8 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{3} b \,c^{5} d n -6 B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} b^{2} c^{4} d^{2} n +B \,x^{2} a^{4} c^{2} d^{4} n^{2}-2 A \,x^{2} a^{4} c^{2} d^{4} n +2 B x \,a^{4} c^{3} d^{3} n^{2}-4 A x \,a^{4} c^{3} d^{3} n +2 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{4} c^{4} d^{2} n}{4 i^{3} g \left (d x +c \right )^{2} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) a^{2} c^{4} n}\)	\(633\)

Input:

int((A+B*ln(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)/(d*i*x+c*i)^3,x,method=_RE 
TURNVERBOSE)
 

Output:

-1/4*(2*B*x^2*ln(e*((b*x+a)/(d*x+c))^n)^2*a^2*b^2*c^4*d^2-8*B*x^2*a^3*b*c^ 
3*d^3*n^2+7*B*x^2*a^2*b^2*c^4*d^2*n^2+4*A*x^2*ln(e*((b*x+a)/(d*x+c))^n)*a^ 
2*b^2*c^4*d^2+8*A*x^2*a^3*b*c^3*d^3*n-6*A*x^2*a^2*b^2*c^4*d^2*n-8*B*x*ln(e 
*((b*x+a)/(d*x+c))^n)*a^2*b^2*c^5*d*n-4*B*x*ln(e*((b*x+a)/(d*x+c))^n)*a^3* 
b*c^4*d^2*n+2*B*ln(e*((b*x+a)/(d*x+c))^n)^2*a^2*b^2*c^6+4*A*ln(e*((b*x+a)/ 
(d*x+c))^n)*a^2*b^2*c^6+4*B*x*ln(e*((b*x+a)/(d*x+c))^n)^2*a^2*b^2*c^5*d-10 
*B*x*a^3*b*c^4*d^2*n^2+8*B*x*a^2*b^2*c^5*d*n^2+8*A*x*ln(e*((b*x+a)/(d*x+c) 
)^n)*a^2*b^2*c^5*d+12*A*x*a^3*b*c^4*d^2*n-8*A*x*a^2*b^2*c^5*d*n-8*B*ln(e*( 
(b*x+a)/(d*x+c))^n)*a^3*b*c^5*d*n-6*B*x^2*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b^ 
2*c^4*d^2*n+B*x^2*a^4*c^2*d^4*n^2-2*A*x^2*a^4*c^2*d^4*n+2*B*x*a^4*c^3*d^3* 
n^2-4*A*x*a^4*c^3*d^3*n+2*B*ln(e*((b*x+a)/(d*x+c))^n)*a^4*c^4*d^2*n)/i^3/g 
/(d*x+c)^2/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/a^2/c^4/n
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 487, normalized size of antiderivative = 1.92 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x) (c i+d i x)^3} \, dx=\frac {6 \, A b^{2} c^{2} - 8 \, A a b c d + 2 \, A a^{2} d^{2} + 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B b^{2} c d n x + B b^{2} c^{2} n\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2} - {\left (7 \, B b^{2} c^{2} - 8 \, B a b c d + B a^{2} d^{2}\right )} n + 2 \, {\left (2 \, A b^{2} c d - 2 \, A a b d^{2} - 3 \, {\left (B b^{2} c d - B a b d^{2}\right )} n\right )} x + 2 \, {\left (3 \, B b^{2} c^{2} - 4 \, B a b c d + B a^{2} d^{2} + 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} x + 2 \, {\left (B b^{2} d^{2} x^{2} + 2 \, B b^{2} c d x + B b^{2} c^{2}\right )} \log \left (\frac {b x + a}{d x + c}\right )\right )} \log \left (e\right ) + 2 \, {\left (2 \, A b^{2} c^{2} - {\left (3 \, B b^{2} d^{2} n - 2 \, A b^{2} d^{2}\right )} x^{2} - {\left (4 \, B a b c d - B a^{2} d^{2}\right )} n + 2 \, {\left (2 \, A b^{2} c d - {\left (2 \, B b^{2} c d + B a b d^{2}\right )} n\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left ({\left (b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - a^{3} d^{5}\right )} g i^{3} x^{2} + 2 \, {\left (b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} + 3 \, a^{2} b c^{2} d^{3} - a^{3} c d^{4}\right )} g i^{3} x + {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} g i^{3}\right )}} \] Input:

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)/(d*i*x+c*i)^3,x, al 
gorithm="fricas")
 

Output:

1/4*(6*A*b^2*c^2 - 8*A*a*b*c*d + 2*A*a^2*d^2 + 2*(B*b^2*d^2*n*x^2 + 2*B*b^ 
2*c*d*n*x + B*b^2*c^2*n)*log((b*x + a)/(d*x + c))^2 - (7*B*b^2*c^2 - 8*B*a 
*b*c*d + B*a^2*d^2)*n + 2*(2*A*b^2*c*d - 2*A*a*b*d^2 - 3*(B*b^2*c*d - B*a* 
b*d^2)*n)*x + 2*(3*B*b^2*c^2 - 4*B*a*b*c*d + B*a^2*d^2 + 2*(B*b^2*c*d - B* 
a*b*d^2)*x + 2*(B*b^2*d^2*x^2 + 2*B*b^2*c*d*x + B*b^2*c^2)*log((b*x + a)/( 
d*x + c)))*log(e) + 2*(2*A*b^2*c^2 - (3*B*b^2*d^2*n - 2*A*b^2*d^2)*x^2 - ( 
4*B*a*b*c*d - B*a^2*d^2)*n + 2*(2*A*b^2*c*d - (2*B*b^2*c*d + B*a*b*d^2)*n) 
*x)*log((b*x + a)/(d*x + c)))/((b^3*c^3*d^2 - 3*a*b^2*c^2*d^3 + 3*a^2*b*c* 
d^4 - a^3*d^5)*g*i^3*x^2 + 2*(b^3*c^4*d - 3*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^ 
3 - a^3*c*d^4)*g*i^3*x + (b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 - a^3* 
c^2*d^3)*g*i^3)
 

Sympy [F(-1)]

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

integrate((A+B*ln(e*((b*x+a)/(d*x+c))**n))/(b*g*x+a*g)/(d*i*x+c*i)**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 888 vs. \(2 (248) = 496\).

Time = 0.09 (sec) , antiderivative size = 888, normalized size of antiderivative = 3.50 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x) (c i+d i x)^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)/(d*i*x+c*i)^3,x, al 
gorithm="maxima")
 

Output:

1/2*B*((2*b*d*x + 3*b*c - a*d)/((b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*g*i^ 
3*x^2 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*g*i^3*x + (b^2*c^4 - 2*a 
*b*c^3*d + a^2*c^2*d^2)*g*i^3) + 2*b^2*log(b*x + a)/((b^3*c^3 - 3*a*b^2*c^ 
2*d + 3*a^2*b*c*d^2 - a^3*d^3)*g*i^3) - 2*b^2*log(d*x + c)/((b^3*c^3 - 3*a 
*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*g*i^3))*log(e*(b*x/(d*x + c) + a/(d* 
x + c))^n) - 1/4*(7*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(b^2*d^2*x^2 + 2*b^2 
*c*d*x + b^2*c^2)*log(b*x + a)^2 + 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2) 
*log(d*x + c)^2 + 6*(b^2*c*d - a*b*d^2)*x + 6*(b^2*d^2*x^2 + 2*b^2*c*d*x + 
 b^2*c^2)*log(b*x + a) - 2*(3*b^2*d^2*x^2 + 6*b^2*c*d*x + 3*b^2*c^2 + 2*(b 
^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(b*x + a))*log(d*x + c))*B*n/(b^3*c 
^5*g*i^3 - 3*a*b^2*c^4*d*g*i^3 + 3*a^2*b*c^3*d^2*g*i^3 - a^3*c^2*d^3*g*i^3 
 + (b^3*c^3*d^2*g*i^3 - 3*a*b^2*c^2*d^3*g*i^3 + 3*a^2*b*c*d^4*g*i^3 - a^3* 
d^5*g*i^3)*x^2 + 2*(b^3*c^4*d*g*i^3 - 3*a*b^2*c^3*d^2*g*i^3 + 3*a^2*b*c^2* 
d^3*g*i^3 - a^3*c*d^4*g*i^3)*x) + 1/2*A*((2*b*d*x + 3*b*c - a*d)/((b^2*c^2 
*d^2 - 2*a*b*c*d^3 + a^2*d^4)*g*i^3*x^2 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a 
^2*c*d^3)*g*i^3*x + (b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*g*i^3) + 2*b^2*l 
og(b*x + a)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*g*i^3) - 
2*b^2*log(d*x + c)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*g* 
i^3))
 

Giac [A] (verification not implemented)

Time = 1.30 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.65 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x) (c i+d i x)^3} \, dx=\frac {1}{4} \, {\left (\frac {2 \, B b^{2} n \log \left (\frac {b x + a}{d x + c}\right )^{2}}{b^{2} c^{2} g i^{3} - 2 \, a b c d g i^{3} + a^{2} d^{2} g i^{3}} - 2 \, {\left (\frac {4 \, {\left (b x + a\right )} B b d n}{{\left (b^{2} c^{2} g i^{3} - 2 \, a b c d g i^{3} + a^{2} d^{2} g i^{3}\right )} {\left (d x + c\right )}} - \frac {{\left (b x + a\right )}^{2} B d^{2} n}{{\left (b^{2} c^{2} g i^{3} - 2 \, a b c d g i^{3} + a^{2} d^{2} g i^{3}\right )} {\left (d x + c\right )}^{2}}\right )} \log \left (\frac {b x + a}{d x + c}\right ) + \frac {4 \, {\left (B b^{2} \log \left (e\right ) + A b^{2}\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} c^{2} g i^{3} - 2 \, a b c d g i^{3} + a^{2} d^{2} g i^{3}} - \frac {{\left (B d^{2} n - 2 \, B d^{2} \log \left (e\right ) - 2 \, A d^{2}\right )} {\left (b x + a\right )}^{2}}{{\left (b^{2} c^{2} g i^{3} - 2 \, a b c d g i^{3} + a^{2} d^{2} g i^{3}\right )} {\left (d x + c\right )}^{2}} + \frac {8 \, {\left (B b d n - B b d \log \left (e\right ) - A b d\right )} {\left (b x + a\right )}}{{\left (b^{2} c^{2} g i^{3} - 2 \, a b c d g i^{3} + a^{2} d^{2} g i^{3}\right )} {\left (d x + c\right )}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \] Input:

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)/(d*i*x+c*i)^3,x, al 
gorithm="giac")
 

Output:

1/4*(2*B*b^2*n*log((b*x + a)/(d*x + c))^2/(b^2*c^2*g*i^3 - 2*a*b*c*d*g*i^3 
 + a^2*d^2*g*i^3) - 2*(4*(b*x + a)*B*b*d*n/((b^2*c^2*g*i^3 - 2*a*b*c*d*g*i 
^3 + a^2*d^2*g*i^3)*(d*x + c)) - (b*x + a)^2*B*d^2*n/((b^2*c^2*g*i^3 - 2*a 
*b*c*d*g*i^3 + a^2*d^2*g*i^3)*(d*x + c)^2))*log((b*x + a)/(d*x + c)) + 4*( 
B*b^2*log(e) + A*b^2)*log((b*x + a)/(d*x + c))/(b^2*c^2*g*i^3 - 2*a*b*c*d* 
g*i^3 + a^2*d^2*g*i^3) - (B*d^2*n - 2*B*d^2*log(e) - 2*A*d^2)*(b*x + a)^2/ 
((b^2*c^2*g*i^3 - 2*a*b*c*d*g*i^3 + a^2*d^2*g*i^3)*(d*x + c)^2) + 8*(B*b*d 
*n - B*b*d*log(e) - A*b*d)*(b*x + a)/((b^2*c^2*g*i^3 - 2*a*b*c*d*g*i^3 + a 
^2*d^2*g*i^3)*(d*x + c)))*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)
 

Mupad [B] (verification not implemented)

Time = 27.88 (sec) , antiderivative size = 573, normalized size of antiderivative = 2.26 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x) (c i+d i x)^3} \, dx=\frac {B\,b^2\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {c\,g\,i^3\,n\,\left (a\,d-b\,c\right )}{2\,b}-\frac {g\,i^3\,n\,\left (a\,d-b\,c\right )\,\left (a\,d-2\,b\,c\right )}{2\,b^2}+\frac {d\,g\,i^3\,n\,x\,\left (a\,d-b\,c\right )}{b}\right )}{g\,i^3\,n\,\left (a\,d-b\,c\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )\,\left (g\,c^2\,i^3+2\,g\,c\,d\,i^3\,x+g\,d^2\,i^3\,x^2\right )}-\frac {B\,b^2\,{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2}{2\,g\,i^3\,n\,\left (a\,d-b\,c\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}-\frac {\frac {2\,A\,a\,d-6\,A\,b\,c-B\,a\,d\,n+7\,B\,b\,c\,n}{2\,\left (a\,d-b\,c\right )}-\frac {b\,x\,\left (2\,A\,d-3\,B\,d\,n\right )}{a\,d-b\,c}}{x^2\,\left (2\,a\,d^3\,g\,i^3-2\,b\,c\,d^2\,g\,i^3\right )+x\,\left (4\,a\,c\,d^2\,g\,i^3-4\,b\,c^2\,d\,g\,i^3\right )-2\,b\,c^3\,g\,i^3+2\,a\,c^2\,d\,g\,i^3}+\frac {b^2\,\mathrm {atan}\left (\frac {b^2\,\left (A-\frac {3\,B\,n}{2}\right )\,\left (2\,g\,a^3\,d^3\,i^3-2\,g\,a^2\,b\,c\,d^2\,i^3-2\,g\,a\,b^2\,c^2\,d\,i^3+2\,g\,b^3\,c^3\,i^3\right )\,1{}\mathrm {i}}{g\,i^3\,\left (2\,A\,b^2-3\,B\,b^2\,n\right )\,{\left (a\,d-b\,c\right )}^3}+\frac {b^3\,d\,x\,\left (A-\frac {3\,B\,n}{2}\right )\,\left (g\,a^2\,d^2\,i^3-2\,g\,a\,b\,c\,d\,i^3+g\,b^2\,c^2\,i^3\right )\,4{}\mathrm {i}}{g\,i^3\,\left (2\,A\,b^2-3\,B\,b^2\,n\right )\,{\left (a\,d-b\,c\right )}^3}\right )\,\left (A-\frac {3\,B\,n}{2}\right )\,2{}\mathrm {i}}{g\,i^3\,{\left (a\,d-b\,c\right )}^3} \] Input:

int((A + B*log(e*((a + b*x)/(c + d*x))^n))/((a*g + b*g*x)*(c*i + d*i*x)^3) 
,x)
 

Output:

(b^2*atan((b^2*(A - (3*B*n)/2)*(2*a^3*d^3*g*i^3 + 2*b^3*c^3*g*i^3 - 2*a*b^ 
2*c^2*d*g*i^3 - 2*a^2*b*c*d^2*g*i^3)*1i)/(g*i^3*(2*A*b^2 - 3*B*b^2*n)*(a*d 
 - b*c)^3) + (b^3*d*x*(A - (3*B*n)/2)*(a^2*d^2*g*i^3 + b^2*c^2*g*i^3 - 2*a 
*b*c*d*g*i^3)*4i)/(g*i^3*(2*A*b^2 - 3*B*b^2*n)*(a*d - b*c)^3))*(A - (3*B*n 
)/2)*2i)/(g*i^3*(a*d - b*c)^3) - ((2*A*a*d - 6*A*b*c - B*a*d*n + 7*B*b*c*n 
)/(2*(a*d - b*c)) - (b*x*(2*A*d - 3*B*d*n))/(a*d - b*c))/(x^2*(2*a*d^3*g*i 
^3 - 2*b*c*d^2*g*i^3) + x*(4*a*c*d^2*g*i^3 - 4*b*c^2*d*g*i^3) - 2*b*c^3*g* 
i^3 + 2*a*c^2*d*g*i^3) - (B*b^2*log(e*((a + b*x)/(c + d*x))^n)^2)/(2*g*i^3 
*n*(a*d - b*c)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (B*b^2*log(e*((a + b*x)/ 
(c + d*x))^n)*((c*g*i^3*n*(a*d - b*c))/(2*b) - (g*i^3*n*(a*d - b*c)*(a*d - 
 2*b*c))/(2*b^2) + (d*g*i^3*n*x*(a*d - b*c))/b))/(g*i^3*n*(a*d - b*c)*(a^2 
*d^2 + b^2*c^2 - 2*a*b*c*d)*(c^2*g*i^3 + d^2*g*i^3*x^2 + 2*c*d*g*i^3*x))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 742, normalized size of antiderivative = 2.92 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x) (c i+d i x)^3} \, dx =\text {Too large to display} \] Input:

int((A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)/(d*i*x+c*i)^3,x)
 

Output:

(i*( - 4*log(a + b*x)*a*b**2*c**3*n - 8*log(a + b*x)*a*b**2*c**2*d*n*x - 4 
*log(a + b*x)*a*b**2*c*d**2*n*x**2 + 6*log(a + b*x)*b**3*c**3*n**2 + 12*lo 
g(a + b*x)*b**3*c**2*d*n**2*x + 6*log(a + b*x)*b**3*c*d**2*n**2*x**2 + 4*l 
og(c + d*x)*a*b**2*c**3*n + 8*log(c + d*x)*a*b**2*c**2*d*n*x + 4*log(c + d 
*x)*a*b**2*c*d**2*n*x**2 - 6*log(c + d*x)*b**3*c**3*n**2 - 12*log(c + d*x) 
*b**3*c**2*d*n**2*x - 6*log(c + d*x)*b**3*c*d**2*n**2*x**2 - 2*log(((a + b 
*x)**n*e)/(c + d*x)**n)**2*b**3*c**3 - 4*log(((a + b*x)**n*e)/(c + d*x)**n 
)**2*b**3*c**2*d*x - 2*log(((a + b*x)**n*e)/(c + d*x)**n)**2*b**3*c*d**2*x 
**2 - 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b*c*d**2*n + 8*log(((a + b 
*x)**n*e)/(c + d*x)**n)*a*b**2*c**2*d*n + 4*log(((a + b*x)**n*e)/(c + d*x) 
**n)*a*b**2*c*d**2*n*x - 6*log(((a + b*x)**n*e)/(c + d*x)**n)*b**3*c**3*n 
- 4*log(((a + b*x)**n*e)/(c + d*x)**n)*b**3*c**2*d*n*x - 2*a**3*c*d**2*n + 
 6*a**2*b*c**2*d*n + a**2*b*c*d**2*n**2 - 2*a**2*b*d**3*n*x**2 - 4*a*b**2* 
c**3*n - 5*a*b**2*c**2*d*n**2 + 2*a*b**2*c*d**2*n*x**2 + 3*a*b**2*d**3*n** 
2*x**2 + 4*b**3*c**3*n**2 - 3*b**3*c*d**2*n**2*x**2))/(4*c*g*n*(a**3*c**2* 
d**3 + 2*a**3*c*d**4*x + a**3*d**5*x**2 - 3*a**2*b*c**3*d**2 - 6*a**2*b*c* 
*2*d**3*x - 3*a**2*b*c*d**4*x**2 + 3*a*b**2*c**4*d + 6*a*b**2*c**3*d**2*x 
+ 3*a*b**2*c**2*d**3*x**2 - b**3*c**5 - 2*b**3*c**4*d*x - b**3*c**3*d**2*x 
**2))