3.2.57 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(a g+b g x)^3 (c i+d i x)^3} \, dx\) [157]

3.2.57.1 Optimal result

Integrand size = 43, antiderivative size = 483 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3 (c i+d i x)^3} \, dx=-\frac {B d^4 n (a+b x)^2}{4 (b c-a d)^5 g^3 i^3 (c+d x)^2}+\frac {4 b B d^3 n (a+b x)}{(b c-a d)^5 g^3 i^3 (c+d x)}+\frac {4 b^3 B d n (c+d x)}{(b c-a d)^5 g^3 i^3 (a+b x)}-\frac {b^4 B n (c+d x)^2}{4 (b c-a d)^5 g^3 i^3 (a+b x)^2}+\frac {d^4 (a+b x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 (b c-a d)^5 g^3 i^3 (c+d x)^2}-\frac {4 b d^3 (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d)^5 g^3 i^3 (c+d x)}+\frac {4 b^3 d (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d)^5 g^3 i^3 (a+b x)}-\frac {b^4 (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 (b c-a d)^5 g^3 i^3 (a+b x)^2}+\frac {6 b^2 d^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {a+b x}{c+d x}\right )}{(b c-a d)^5 g^3 i^3}-\frac {3 b^2 B d^2 n \log ^2\left (\frac {a+b x}{c+d x}\right )}{(b c-a d)^5 g^3 i^3} \]

output

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

3.2.57.2 Mathematica [C] (verified)

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

Time = 0.68 (sec) , antiderivative size = 561, normalized size of antiderivative = 1.16 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3 (c i+d i x)^3} \, dx=-\frac {\frac {b^2 B (b c-a d)^2 n}{(a+b x)^2}-\frac {12 b^3 B c d n}{a+b x}+\frac {12 a b^2 B d^2 n}{a+b x}-\frac {2 b^2 B d (b c-a d) n}{a+b x}+\frac {B d^2 (b c-a d)^2 n}{(c+d x)^2}+\frac {12 b^2 B c d^2 n}{c+d x}-\frac {12 a b B d^3 n}{c+d x}+\frac {2 b B d^2 (b c-a d) n}{c+d x}+\frac {2 b^2 (b c-a d)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a+b x)^2}-\frac {12 b^2 d (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{a+b x}-\frac {2 d^2 (b c-a d)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c+d x)^2}-\frac {12 b d^2 (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{c+d x}-24 b^2 d^2 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+24 b^2 d^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)+12 b^2 B d^2 n \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )-12 b^2 B d^2 n \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )}{4 (b c-a d)^5 g^3 i^3} \]

input

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

output

-1/4*((b^2*B*(b*c - a*d)^2*n)/(a + b*x)^2 - (12*b^3*B*c*d*n)/(a + b*x) + ( 
12*a*b^2*B*d^2*n)/(a + b*x) - (2*b^2*B*d*(b*c - a*d)*n)/(a + b*x) + (B*d^2 
*(b*c - a*d)^2*n)/(c + d*x)^2 + (12*b^2*B*c*d^2*n)/(c + d*x) - (12*a*b*B*d 
^3*n)/(c + d*x) + (2*b*B*d^2*(b*c - a*d)*n)/(c + d*x) + (2*b^2*(b*c - a*d) 
^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(a + b*x)^2 - (12*b^2*d*(b*c - 
a*d)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(a + b*x) - (2*d^2*(b*c - a*d 
)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(c + d*x)^2 - (12*b*d^2*(b*c - 
 a*d)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(c + d*x) - 24*b^2*d^2*Log[a 
 + b*x]*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + 24*b^2*d^2*(A + B*Log[e*( 
(a + b*x)/(c + d*x))^n])*Log[c + d*x] + 12*b^2*B*d^2*n*(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)]) - 12*b^2*B*d^2*n*((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 
*c - a*d)^5*g^3*i^3)
 

3.2.57.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 335, normalized size of antiderivative = 0.69, 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.

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

\(\Big \downarrow \) 2961

\(\displaystyle \frac {\int \frac {(c+d x)^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a+b x)^3}d\frac {a+b x}{c+d x}}{g^3 i^3 (b c-a d)^5}\)

\(\Big \downarrow \) 2772

\(\displaystyle \frac {-B n \int \left (\frac {6 b^2 (c+d x) \log \left (\frac {a+b x}{c+d x}\right ) d^2}{a+b x}+\frac {1}{2} \left (-\frac {(c+d x)^3 b^4}{(a+b x)^3}+\frac {8 d (c+d x)^2 b^3}{(a+b x)^2}-8 d^3 b+\frac {d^4 (a+b x)}{c+d x}\right )\right )d\frac {a+b x}{c+d x}-\frac {b^4 (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 (a+b x)^2}+\frac {4 b^3 d (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{a+b x}+6 b^2 d^2 \log \left (\frac {a+b x}{c+d x}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+\frac {d^4 (a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 (c+d x)^2}-\frac {4 b d^3 (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{c+d x}}{g^3 i^3 (b c-a d)^5}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {b^4 (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 (a+b x)^2}+\frac {4 b^3 d (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{a+b x}+6 b^2 d^2 \log \left (\frac {a+b x}{c+d x}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+\frac {d^4 (a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 (c+d x)^2}-\frac {4 b d^3 (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{c+d x}-B n \left (\frac {b^4 (c+d x)^2}{4 (a+b x)^2}-\frac {4 b^3 d (c+d x)}{a+b x}+3 b^2 d^2 \log ^2\left (\frac {a+b x}{c+d x}\right )+\frac {d^4 (a+b x)^2}{4 (c+d x)^2}-\frac {4 b d^3 (a+b x)}{c+d x}\right )}{g^3 i^3 (b c-a d)^5}\)

input

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

output

((d^4*(a + b*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*(c + d*x)^2) 
- (4*b*d^3*(a + b*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(c + d*x) + ( 
4*b^3*d*(c + d*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(a + b*x) - (b^4 
*(c + d*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*(a + b*x)^2) + 6*b 
^2*d^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[(a + b*x)/(c + d*x)] - B 
*n*((d^4*(a + b*x)^2)/(4*(c + d*x)^2) - (4*b*d^3*(a + b*x))/(c + d*x) - (4 
*b^3*d*(c + d*x))/(a + b*x) + (b^4*(c + d*x)^2)/(4*(a + b*x)^2) + 3*b^2*d^ 
2*Log[(a + b*x)/(c + d*x)]^2))/((b*c - a*d)^5*g^3*i^3)
 

3.2.57.3.1 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]
 

3.2.57.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1449\) vs. \(2(475)=950\).

Time = 54.00 (sec) , antiderivative size = 1450, normalized size of antiderivative = 3.00

method	result	size
parallelrisch	\(\text {Expression too large to display}\)	\(1450\)

input

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

output

-1/4*(12*B*x^4*ln(e*((b*x+a)/(d*x+c))^n)^2*b^8*d^8+24*A*x^4*ln(e*((b*x+a)/ 
(d*x+c))^n)*b^8*d^8-24*B*x^3*ln(e*((b*x+a)/(d*x+c))^n)*a*b^7*d^8*n+24*B*x^ 
3*ln(e*((b*x+a)/(d*x+c))^n)*b^8*c*d^7*n+48*B*x^2*ln(e*((b*x+a)/(d*x+c))^n) 
^2*a*b^7*c*d^7-36*B*x^2*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b^6*d^8*n+36*B*x^2*l 
n(e*((b*x+a)/(d*x+c))^n)*b^8*c^2*d^6*n-24*B*x^2*a*b^7*c*d^7*n^2+96*A*x^2*l 
n(e*((b*x+a)/(d*x+c))^n)*a*b^7*c*d^7+24*B*x*ln(e*((b*x+a)/(d*x+c))^n)^2*a^ 
2*b^6*c*d^7+24*B*x*ln(e*((b*x+a)/(d*x+c))^n)^2*a*b^7*c^2*d^6-8*B*x*ln(e*(( 
b*x+a)/(d*x+c))^n)*a^3*b^5*d^8*n+8*B*x*ln(e*((b*x+a)/(d*x+c))^n)*b^8*c^3*d 
^5*n-12*B*x*a^2*b^6*c*d^7*n^2-12*B*x*a*b^7*c^2*d^6*n^2+48*A*x*ln(e*((b*x+a 
)/(d*x+c))^n)*a^2*b^6*c*d^7+48*A*x*ln(e*((b*x+a)/(d*x+c))^n)*a*b^7*c^2*d^6 
-48*A*x*a^2*b^6*c*d^7*n+48*A*x*a*b^7*c^2*d^6*n-16*B*ln(e*((b*x+a)/(d*x+c)) 
^n)*a^3*b^5*c*d^7*n+16*B*ln(e*((b*x+a)/(d*x+c))^n)*a*b^7*c^3*d^5*n+24*A*x^ 
2*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b^6*d^8+24*A*x^2*ln(e*((b*x+a)/(d*x+c))^n) 
*b^8*c^2*d^6-36*A*x^2*a^2*b^6*d^8*n+36*A*x^2*b^8*c^2*d^6*n+12*B*x*a^3*b^5* 
d^8*n^2+12*B*x*b^8*c^3*d^5*n^2-8*A*x*a^3*b^5*d^8*n+8*A*x*b^8*c^3*d^5*n+12* 
B*ln(e*((b*x+a)/(d*x+c))^n)^2*a^2*b^6*c^2*d^6+2*B*ln(e*((b*x+a)/(d*x+c))^n 
)*a^4*b^4*d^8*n-2*B*ln(e*((b*x+a)/(d*x+c))^n)*b^8*c^4*d^4*n+24*A*ln(e*((b* 
x+a)/(d*x+c))^n)*a^2*b^6*c^2*d^6+16*B*a^3*b^5*c*d^7*n^2-30*B*a^2*b^6*c^2*d 
^6*n^2+16*B*a*b^7*c^3*d^5*n^2-16*A*a^3*b^5*c*d^7*n+16*A*a*b^7*c^3*d^5*n-48 
*B*x*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b^6*c*d^7*n+48*B*x*ln(e*((b*x+a)/(d*...
 

3.2.57.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1416 vs. \(2 (475) = 950\).

Time = 0.43 (sec) , antiderivative size = 1416, normalized size of antiderivative = 2.93 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3 (c i+d i x)^3} \, dx=\text {Too large to display} \]

input

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

output

-1/4*(2*A*b^4*c^4 - 16*A*a*b^3*c^3*d + 16*A*a^3*b*c*d^3 - 2*A*a^4*d^4 - 24 
*(A*b^4*c*d^3 - A*a*b^3*d^4)*x^3 - 12*(3*A*b^4*c^2*d^2 - 3*A*a^2*b^2*d^4 + 
 (B*b^4*c^2*d^2 - 2*B*a*b^3*c*d^3 + B*a^2*b^2*d^4)*n)*x^2 - 12*(B*b^4*d^4* 
n*x^4 + B*a^2*b^2*c^2*d^2*n + 2*(B*b^4*c*d^3 + B*a*b^3*d^4)*n*x^3 + (B*b^4 
*c^2*d^2 + 4*B*a*b^3*c*d^3 + B*a^2*b^2*d^4)*n*x^2 + 2*(B*a*b^3*c^2*d^2 + B 
*a^2*b^2*c*d^3)*n*x)*log((b*x + a)/(d*x + c))^2 + (B*b^4*c^4 - 16*B*a*b^3* 
c^3*d + 30*B*a^2*b^2*c^2*d^2 - 16*B*a^3*b*c*d^3 + B*a^4*d^4)*n - 4*(2*A*b^ 
4*c^3*d + 12*A*a*b^3*c^2*d^2 - 12*A*a^2*b^2*c*d^3 - 2*A*a^3*b*d^4 + 3*(B*b 
^4*c^3*d - B*a*b^3*c^2*d^2 - B*a^2*b^2*c*d^3 + B*a^3*b*d^4)*n)*x + 2*(B*b^ 
4*c^4 - 8*B*a*b^3*c^3*d + 8*B*a^3*b*c*d^3 - B*a^4*d^4 - 12*(B*b^4*c*d^3 - 
B*a*b^3*d^4)*x^3 - 18*(B*b^4*c^2*d^2 - B*a^2*b^2*d^4)*x^2 - 4*(B*b^4*c^3*d 
 + 6*B*a*b^3*c^2*d^2 - 6*B*a^2*b^2*c*d^3 - B*a^3*b*d^4)*x - 12*(B*b^4*d^4* 
x^4 + B*a^2*b^2*c^2*d^2 + 2*(B*b^4*c*d^3 + B*a*b^3*d^4)*x^3 + (B*b^4*c^2*d 
^2 + 4*B*a*b^3*c*d^3 + B*a^2*b^2*d^4)*x^2 + 2*(B*a*b^3*c^2*d^2 + B*a^2*b^2 
*c*d^3)*x)*log((b*x + a)/(d*x + c)))*log(e) - 2*(12*A*b^4*d^4*x^4 + 12*A*a 
^2*b^2*c^2*d^2 + 12*(2*A*b^4*c*d^3 + 2*A*a*b^3*d^4 + (B*b^4*c*d^3 - B*a*b^ 
3*d^4)*n)*x^3 + 6*(2*A*b^4*c^2*d^2 + 8*A*a*b^3*c*d^3 + 2*A*a^2*b^2*d^4 + 3 
*(B*b^4*c^2*d^2 - B*a^2*b^2*d^4)*n)*x^2 - (B*b^4*c^4 - 8*B*a*b^3*c^3*d + 8 
*B*a^3*b*c*d^3 - B*a^4*d^4)*n + 4*(6*A*a*b^3*c^2*d^2 + 6*A*a^2*b^2*c*d^3 + 
 (B*b^4*c^3*d + 6*B*a*b^3*c^2*d^2 - 6*B*a^2*b^2*c*d^3 - B*a^3*b*d^4)*n)...
 

3.2.57.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.2.57.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2383 vs. \(2 (475) = 950\).

Time = 0.33 (sec) , antiderivative size = 2383, normalized size of antiderivative = 4.93 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3 (c i+d i x)^3} \, dx=\text {Too large to display} \]

input

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

output

1/2*B*((12*b^3*d^3*x^3 - b^3*c^3 + 7*a*b^2*c^2*d + 7*a^2*b*c*d^2 - a^3*d^3 
 + 18*(b^3*c*d^2 + a*b^2*d^3)*x^2 + 4*(b^3*c^2*d + 7*a*b^2*c*d^2 + a^2*b*d 
^3)*x)/((b^6*c^4*d^2 - 4*a*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d 
^5 + a^4*b^2*d^6)*g^3*i^3*x^4 + 2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2*b^4 
*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*g^3*i^3*x^3 + 
(b^6*c^6 - 9*a^2*b^4*c^4*d^2 + 16*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^ 
6*d^6)*g^3*i^3*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3*b^3*c^4*d^2 + 
2*a^4*b^2*c^3*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5)*g^3*i^3*x + (a^2*b^4*c^6 
- 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2*d^4)*g^3 
*i^3) + 12*b^2*d^2*log(b*x + a)/((b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3 
*d^2 - 10*a^3*b^2*c^2*d^3 + 5*a^4*b*c*d^4 - a^5*d^5)*g^3*i^3) - 12*b^2*d^2 
*log(d*x + c)/((b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2* 
c^2*d^3 + 5*a^4*b*c*d^4 - a^5*d^5)*g^3*i^3))*log(e*(b*x/(d*x + c) + a/(d*x 
 + c))^n) - 1/4*(b^4*c^4 - 16*a*b^3*c^3*d + 30*a^2*b^2*c^2*d^2 - 16*a^3*b* 
c*d^3 + a^4*d^4 - 12*(b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*x^2 + 12* 
(b^4*d^4*x^4 + a^2*b^2*c^2*d^2 + 2*(b^4*c*d^3 + a*b^3*d^4)*x^3 + (b^4*c^2* 
d^2 + 4*a*b^3*c*d^3 + a^2*b^2*d^4)*x^2 + 2*(a*b^3*c^2*d^2 + a^2*b^2*c*d^3) 
*x)*log(b*x + a)^2 - 24*(b^4*d^4*x^4 + a^2*b^2*c^2*d^2 + 2*(b^4*c*d^3 + a* 
b^3*d^4)*x^3 + (b^4*c^2*d^2 + 4*a*b^3*c*d^3 + a^2*b^2*d^4)*x^2 + 2*(a*b^3* 
c^2*d^2 + a^2*b^2*c*d^3)*x)*log(b*x + a)*log(d*x + c) + 12*(b^4*d^4*x^4...
 

3.2.57.8 Giac [F]

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

input

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

output

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

3.2.57.9 Mupad [B] (verification not implemented)

Time = 4.60 (sec) , antiderivative size = 1341, normalized size of antiderivative = 2.78 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a g+b g x)^3 (c i+d i x)^3} \, dx=\frac {\frac {2\,x\,\left (2\,A\,a^2\,b\,d^3+2\,A\,b^3\,c^2\,d+14\,A\,a\,b^2\,c\,d^2-3\,B\,a^2\,b\,d^3\,n+3\,B\,b^3\,c^2\,d\,n\right )}{a\,d-b\,c}-\frac {2\,A\,a^3\,d^3+2\,A\,b^3\,c^3-B\,a^3\,d^3\,n+B\,b^3\,c^3\,n-14\,A\,a\,b^2\,c^2\,d-14\,A\,a^2\,b\,c\,d^2-15\,B\,a\,b^2\,c^2\,d\,n+15\,B\,a^2\,b\,c\,d^2\,n}{2\,\left (a\,d-b\,c\right )}+\frac {6\,x^2\,\left (3\,A\,a\,b^2\,d^3+3\,A\,b^3\,c\,d^2-B\,a\,b^2\,d^3\,n+B\,b^3\,c\,d^2\,n\right )}{a\,d-b\,c}+\frac {12\,A\,b^3\,d^3\,x^3}{a\,d-b\,c}}{x^4\,\left (2\,a^3\,b^2\,d^5\,g^3\,i^3-6\,a^2\,b^3\,c\,d^4\,g^3\,i^3+6\,a\,b^4\,c^2\,d^3\,g^3\,i^3-2\,b^5\,c^3\,d^2\,g^3\,i^3\right )-x\,\left (-4\,a^5\,c\,d^4\,g^3\,i^3+8\,a^4\,b\,c^2\,d^3\,g^3\,i^3-8\,a^2\,b^3\,c^4\,d\,g^3\,i^3+4\,a\,b^4\,c^5\,g^3\,i^3\right )+x^3\,\left (4\,a^4\,b\,d^5\,g^3\,i^3-8\,a^3\,b^2\,c\,d^4\,g^3\,i^3+8\,a\,b^4\,c^3\,d^2\,g^3\,i^3-4\,b^5\,c^4\,d\,g^3\,i^3\right )+x^2\,\left (2\,a^5\,d^5\,g^3\,i^3+2\,a^4\,b\,c\,d^4\,g^3\,i^3-16\,a^3\,b^2\,c^2\,d^3\,g^3\,i^3+16\,a^2\,b^3\,c^3\,d^2\,g^3\,i^3-2\,a\,b^4\,c^4\,d\,g^3\,i^3-2\,b^5\,c^5\,g^3\,i^3\right )-2\,a^2\,b^3\,c^5\,g^3\,i^3+2\,a^5\,c^2\,d^3\,g^3\,i^3+6\,a^3\,b^2\,c^4\,d\,g^3\,i^3-6\,a^4\,b\,c^3\,d^2\,g^3\,i^3}+\frac {\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (x\,\left (\frac {3\,B\,b\,d\,{\left (a\,d+b\,c\right )}^2}{{\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}^2}-\frac {B\,b\,d}{a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}+\frac {6\,B\,a\,b^2\,c\,d^2}{{\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}^2}\right )-\frac {B\,\left (a\,d+b\,c\right )}{2\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {6\,B\,b^3\,d^3\,x^3}{{\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}^2}+\frac {9\,B\,b^2\,d^2\,x^2\,\left (a\,d+b\,c\right )}{{\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}^2}+\frac {3\,B\,a\,b\,c\,d\,\left (a\,d+b\,c\right )}{{\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}^2}\right )}{x\,\left (2\,d\,a^2\,c\,g^3\,i^3+2\,b\,a\,c^2\,g^3\,i^3\right )+x^3\,\left (2\,c\,b^2\,d\,g^3\,i^3+2\,a\,b\,d^2\,g^3\,i^3\right )+x^2\,\left (a^2\,d^2\,g^3\,i^3+4\,a\,b\,c\,d\,g^3\,i^3+b^2\,c^2\,g^3\,i^3\right )+a^2\,c^2\,g^3\,i^3+b^2\,d^2\,g^3\,i^3\,x^4}-\frac {3\,B\,b^2\,d^2\,{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2}{g^3\,i^3\,n\,{\left (a\,d-b\,c\right )}^5}+\frac {A\,b^2\,d^2\,\mathrm {atan}\left (\frac {\left (a^5\,d^5\,g^3\,i^3-3\,a^4\,b\,c\,d^4\,g^3\,i^3+2\,a^3\,b^2\,c^2\,d^3\,g^3\,i^3+2\,a^2\,b^3\,c^3\,d^2\,g^3\,i^3-3\,a\,b^4\,c^4\,d\,g^3\,i^3+b^5\,c^5\,g^3\,i^3\right )\,1{}\mathrm {i}}{g^3\,i^3\,{\left (a\,d-b\,c\right )}^5}+\frac {b\,d\,x\,\left (a^4\,d^4\,g^3\,i^3-4\,a^3\,b\,c\,d^3\,g^3\,i^3+6\,a^2\,b^2\,c^2\,d^2\,g^3\,i^3-4\,a\,b^3\,c^3\,d\,g^3\,i^3+b^4\,c^4\,g^3\,i^3\right )\,2{}\mathrm {i}}{g^3\,i^3\,{\left (a\,d-b\,c\right )}^5}\right )\,12{}\mathrm {i}}{g^3\,i^3\,{\left (a\,d-b\,c\right )}^5} \]

input

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

output

((2*x*(2*A*a^2*b*d^3 + 2*A*b^3*c^2*d + 14*A*a*b^2*c*d^2 - 3*B*a^2*b*d^3*n 
+ 3*B*b^3*c^2*d*n))/(a*d - b*c) - (2*A*a^3*d^3 + 2*A*b^3*c^3 - B*a^3*d^3*n 
 + B*b^3*c^3*n - 14*A*a*b^2*c^2*d - 14*A*a^2*b*c*d^2 - 15*B*a*b^2*c^2*d*n 
+ 15*B*a^2*b*c*d^2*n)/(2*(a*d - b*c)) + (6*x^2*(3*A*a*b^2*d^3 + 3*A*b^3*c* 
d^2 - B*a*b^2*d^3*n + B*b^3*c*d^2*n))/(a*d - b*c) + (12*A*b^3*d^3*x^3)/(a* 
d - b*c))/(x^4*(2*a^3*b^2*d^5*g^3*i^3 - 2*b^5*c^3*d^2*g^3*i^3 + 6*a*b^4*c^ 
2*d^3*g^3*i^3 - 6*a^2*b^3*c*d^4*g^3*i^3) - x*(4*a*b^4*c^5*g^3*i^3 - 4*a^5* 
c*d^4*g^3*i^3 - 8*a^2*b^3*c^4*d*g^3*i^3 + 8*a^4*b*c^2*d^3*g^3*i^3) + x^3*( 
4*a^4*b*d^5*g^3*i^3 - 4*b^5*c^4*d*g^3*i^3 + 8*a*b^4*c^3*d^2*g^3*i^3 - 8*a^ 
3*b^2*c*d^4*g^3*i^3) + x^2*(2*a^5*d^5*g^3*i^3 - 2*b^5*c^5*g^3*i^3 - 2*a*b^ 
4*c^4*d*g^3*i^3 + 2*a^4*b*c*d^4*g^3*i^3 + 16*a^2*b^3*c^3*d^2*g^3*i^3 - 16* 
a^3*b^2*c^2*d^3*g^3*i^3) - 2*a^2*b^3*c^5*g^3*i^3 + 2*a^5*c^2*d^3*g^3*i^3 + 
 6*a^3*b^2*c^4*d*g^3*i^3 - 6*a^4*b*c^3*d^2*g^3*i^3) + (log(e*((a + b*x)/(c 
 + d*x))^n)*(x*((3*B*b*d*(a*d + b*c)^2)/(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)^2 
- (B*b*d)/(a^2*d^2 + b^2*c^2 - 2*a*b*c*d) + (6*B*a*b^2*c*d^2)/(a^2*d^2 + b 
^2*c^2 - 2*a*b*c*d)^2) - (B*(a*d + b*c))/(2*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d 
)) + (6*B*b^3*d^3*x^3)/(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)^2 + (9*B*b^2*d^2*x^ 
2*(a*d + b*c))/(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)^2 + (3*B*a*b*c*d*(a*d + b*c 
))/(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)^2))/(x*(2*a*b*c^2*g^3*i^3 + 2*a^2*c*d*g 
^3*i^3) + x^3*(2*a*b*d^2*g^3*i^3 + 2*b^2*c*d*g^3*i^3) + x^2*(a^2*d^2*g^...