\(\int \frac {(c i+d i x) (A+B \log (e (\frac {a+b x}{c+d x})^n))^2}{(a g+b g x)^3} \, dx\) [165]

Optimal result

Integrand size = 43, antiderivative size = 151 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^3} \, dx=-\frac {B^2 i n^2 (c+d x)^2}{4 (b c-a d) g^3 (a+b x)^2}-\frac {B i n (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 (b c-a d) g^3 (a+b x)^2}-\frac {i (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{2 (b c-a d) g^3 (a+b x)^2} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.94 (sec) , antiderivative size = 801, normalized size of antiderivative = 5.30 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^3} \, dx=-\frac {i \left (2 (b c-a d)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2-4 d (-b c+a d) (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2+4 B d n (a+b x) \left (2 (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+2 d (a+b x) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-2 d (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)+2 B n (b c-a d+d (a+b x) \log (a+b x)-d (a+b x) \log (c+d x))-B d n (a+b x) \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 )+B d n (a+b x) \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 )\right )+B n \left (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 d (-b c+a d) (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-4 d^2 (a+b 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 d^2 (a+b 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 d n (a+b x) (b c-a d+d (a+b x) \log (a+b x)-d (a+b x) \log (c+d x))+B n \left ((b c-a d)^2+2 d (-b c+a d) (a+b x)-2 d^2 (a+b x)^2 \log (a+b x)+2 d^2 (a+b x)^2 \log (c+d x)\right )+2 B d^2 n (a+b 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 d^2 n (a+b 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 )\right )\right )}{4 b^2 (b c-a d) g^3 (a+b x)^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2961, 2742, 2741}

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.

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

\(\Big \downarrow \) 2961

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

\(\Big \downarrow \) 2742

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

\(\Big \downarrow \) 2741

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2741

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> 
Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( 
m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
 

rule 2742

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbo 
l] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])^p/(d*(m + 1))), x] - Simp[b*n* 
(p/(m + 1))   Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b 
, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]
 

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. \(529\) vs. \(2(145)=290\).

Time = 4.81 (sec) , antiderivative size = 530, normalized size of antiderivative = 3.51

method	result	size
parallelrisch	\(-\frac {2 A B \,a^{2} b^{2} d^{3} i \,n^{2}-2 A B \,b^{4} c^{2} d i \,n^{2}+2 B^{2} x a \,b^{3} d^{3} i \,n^{3}-2 B^{2} x \,b^{4} c \,d^{2} i \,n^{3}-2 B^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} b^{4} c^{2} d i n -2 B^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} c^{2} d i \,n^{2}-8 A B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} c \,d^{2} i n -4 A B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} d^{3} i n -4 B^{2} x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} b^{4} c \,d^{2} i n -4 B^{2} x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} c \,d^{2} i \,n^{2}+4 A B x a \,b^{3} d^{3} i \,n^{2}-4 A B x \,b^{4} c \,d^{2} i \,n^{2}-4 A B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} c^{2} d i n +B^{2} a^{2} b^{2} d^{3} i \,n^{3}-B^{2} b^{4} c^{2} d i \,n^{3}+2 A^{2} a^{2} b^{2} d^{3} i n -2 A^{2} b^{4} c^{2} d i n +4 A^{2} x a \,b^{3} d^{3} i n -4 A^{2} x \,b^{4} c \,d^{2} i n -2 B^{2} x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} b^{4} d^{3} i n -2 B^{2} x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} d^{3} i \,n^{2}}{4 g^{3} \left (b x +a \right )^{2} b^{4} d n \left (d a -b c \right )}\)	\(530\)

Input:

int((d*i*x+c*i)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))^2/(b*g*x+a*g)^3,x,method=_ 
RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 600 vs. \(2 (145) = 290\).

Time = 0.10 (sec) , antiderivative size = 600, normalized size of antiderivative = 3.97 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^3} \, dx=-\frac {{\left (B^{2} b^{2} c^{2} - B^{2} a^{2} d^{2}\right )} i n^{2} + 2 \, {\left (A B b^{2} c^{2} - A B a^{2} d^{2}\right )} i n + 2 \, {\left (2 \, {\left (B^{2} b^{2} c d - B^{2} a b d^{2}\right )} i x + {\left (B^{2} b^{2} c^{2} - B^{2} a^{2} d^{2}\right )} i\right )} \log \left (e\right )^{2} + 2 \, {\left (B^{2} b^{2} d^{2} i n^{2} x^{2} + 2 \, B^{2} b^{2} c d i n^{2} x + B^{2} b^{2} c^{2} i n^{2}\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2} + 2 \, {\left (A^{2} b^{2} c^{2} - A^{2} a^{2} d^{2}\right )} i + 2 \, {\left ({\left (B^{2} b^{2} c d - B^{2} a b d^{2}\right )} i n^{2} + 2 \, {\left (A B b^{2} c d - A B a b d^{2}\right )} i n + 2 \, {\left (A^{2} b^{2} c d - A^{2} a b d^{2}\right )} i\right )} x + 2 \, {\left ({\left (B^{2} b^{2} c^{2} - B^{2} a^{2} d^{2}\right )} i n + 2 \, {\left (A B b^{2} c^{2} - A B a^{2} d^{2}\right )} i + 2 \, {\left ({\left (B^{2} b^{2} c d - B^{2} a b d^{2}\right )} i n + 2 \, {\left (A B b^{2} c d - A B a b d^{2}\right )} i\right )} x + 2 \, {\left (B^{2} b^{2} d^{2} i n x^{2} + 2 \, B^{2} b^{2} c d i n x + B^{2} b^{2} c^{2} i n\right )} \log \left (\frac {b x + a}{d x + c}\right )\right )} \log \left (e\right ) + 2 \, {\left (B^{2} b^{2} c^{2} i n^{2} + 2 \, A B b^{2} c^{2} i n + {\left (B^{2} b^{2} d^{2} i n^{2} + 2 \, A B b^{2} d^{2} i n\right )} x^{2} + 2 \, {\left (B^{2} b^{2} c d i n^{2} + 2 \, A B b^{2} c d i n\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left ({\left (b^{5} c - a b^{4} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c - a^{2} b^{3} d\right )} g^{3} x + {\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} g^{3}\right )}} \] Input:

integrate((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2/(b*g*x+a*g)^3,x, 
algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^3} \, dx=\frac {i \left (\int \frac {A^{2} c}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\, dx + \int \frac {A^{2} d x}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\, dx + \int \frac {B^{2} c \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}^{2}}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\, dx + \int \frac {2 A B c \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\, dx + \int \frac {B^{2} d x \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}^{2}}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\, dx + \int \frac {2 A B d x \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\, dx\right )}{g^{3}} \] Input:

integrate((d*i*x+c*i)*(A+B*ln(e*((b*x+a)/(d*x+c))**n))**2/(b*g*x+a*g)**3,x 
)
 

Output:

i*(Integral(A**2*c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3), x) + I 
ntegral(A**2*d*x/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3), x) + Int 
egral(B**2*c*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)**2/(a**3 + 3*a**2*b*x 
 + 3*a*b**2*x**2 + b**3*x**3), x) + Integral(2*A*B*c*log(e*(a/(c + d*x) + 
b*x/(c + d*x))**n)/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3), x) + I 
ntegral(B**2*d*x*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)**2/(a**3 + 3*a**2 
*b*x + 3*a*b**2*x**2 + b**3*x**3), x) + Integral(2*A*B*d*x*log(e*(a/(c + d 
*x) + b*x/(c + d*x))**n)/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3), 
x))/g**3
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2017 vs. \(2 (145) = 290\).

Time = 0.14 (sec) , antiderivative size = 2017, normalized size of antiderivative = 13.36 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^3} \, dx=\text {Too large to display} \] Input:

integrate((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2/(b*g*x+a*g)^3,x, 
algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.29 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^3} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, {\left (d x + c\right )}^{2} B^{2} i n^{2} \log \left (\frac {b x + a}{d x + c}\right )^{2}}{{\left (b x + a\right )}^{2} g^{3}} + \frac {2 \, {\left (B^{2} i n^{2} + 2 \, B^{2} i n \log \left (e\right ) + 2 \, A B i n\right )} {\left (d x + c\right )}^{2} \log \left (\frac {b x + a}{d x + c}\right )}{{\left (b x + a\right )}^{2} g^{3}} + \frac {{\left (B^{2} i n^{2} + 2 \, B^{2} i n \log \left (e\right ) + 2 \, B^{2} i \log \left (e\right )^{2} + 2 \, A B i n + 4 \, A B i \log \left (e\right ) + 2 \, A^{2} i\right )} {\left (d x + c\right )}^{2}}{{\left (b x + a\right )}^{2} g^{3}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \] Input:

integrate((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2/(b*g*x+a*g)^3,x, 
algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 27.52 (sec) , antiderivative size = 561, normalized size of antiderivative = 3.72 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^3} \, dx=-{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2\,\left (\frac {\frac {B^2\,c\,i}{2\,b}+\frac {B^2\,d\,i\,x}{b}+\frac {B^2\,a\,d\,i}{2\,b^2}}{a^2\,g^3+2\,a\,b\,g^3\,x+b^2\,g^3\,x^2}-\frac {B^2\,d^2\,i}{2\,b^2\,g^3\,\left (a\,d-b\,c\right )}\right )-\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {A\,B\,a\,d\,i+A\,B\,b\,c\,i-B^2\,a\,d\,i\,n+B^2\,b\,c\,i\,n+2\,A\,B\,b\,d\,i\,x}{a^2\,b^2\,g^3+2\,a\,b^3\,g^3\,x+b^4\,g^3\,x^2}+\frac {B^2\,d^2\,i\,\left (\frac {a\,b^2\,g^3\,n\,\left (a\,d-b\,c\right )}{2\,d}+\frac {b^3\,g^3\,n\,x\,\left (a\,d-b\,c\right )}{d}+\frac {b^2\,g^3\,n\,\left (a\,d-b\,c\right )\,\left (2\,a\,d-b\,c\right )}{2\,d^2}\right )}{b^2\,g^3\,\left (a\,d-b\,c\right )\,\left (a^2\,b^2\,g^3+2\,a\,b^3\,g^3\,x+b^4\,g^3\,x^2\right )}\right )-\frac {x\,\left (2\,b\,d\,i\,A^2+2\,b\,d\,i\,A\,B\,n+b\,d\,i\,B^2\,n^2\right )+A^2\,a\,d\,i+A^2\,b\,c\,i+\frac {B^2\,a\,d\,i\,n^2}{2}+\frac {B^2\,b\,c\,i\,n^2}{2}+A\,B\,a\,d\,i\,n+A\,B\,b\,c\,i\,n}{2\,a^2\,b^2\,g^3+4\,a\,b^3\,g^3\,x+2\,b^4\,g^3\,x^2}-\frac {B\,d^2\,i\,n\,\mathrm {atan}\left (\frac {B\,d^2\,i\,n\,\left (2\,A+B\,n\right )\,\left (\frac {c\,b^3\,g^3+a\,d\,b^2\,g^3}{b^2\,g^3}+2\,b\,d\,x\right )\,1{}\mathrm {i}}{\left (a\,d-b\,c\right )\,\left (i\,B^2\,d^2\,n^2+2\,A\,i\,B\,d^2\,n\right )}\right )\,\left (2\,A+B\,n\right )\,1{}\mathrm {i}}{b^2\,g^3\,\left (a\,d-b\,c\right )} \] Input:

int(((c*i + d*i*x)*(A + B*log(e*((a + b*x)/(c + d*x))^n))^2)/(a*g + b*g*x) 
^3,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 750, normalized size of antiderivative = 4.97 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^3} \, dx =\text {Too large to display} \] Input:

int((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2/(b*g*x+a*g)^3,x)
 

Output:

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