\(\int \frac {(a g+b g x) (A+B \log (e (\frac {a+b x}{c+d x})^n))^2}{(c i+d i x)^3} \, dx\) [204]

Optimal result

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

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

Mathematica [C] (verified)

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

Time = 1.15 (sec) , antiderivative size = 803, normalized size of antiderivative = 5.32 \[ \int \frac {(a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^3} \, dx=\frac {g \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 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 )^2+4 b B n (c+d 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 b (c+d x) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-2 b (c+d 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+b (c+d x) \log (a+b x)-b (c+d x) \log (c+d x))-b B n (c+d 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 B n (c+d 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 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 )\right )\right )}{4 d^2 (b c-a d) i^3 (c+d x)^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 123, 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.

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

\(\Big \downarrow \) 2961

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

\(\Big \downarrow \) 2742

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

\(\Big \downarrow \) 2741

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

Input:

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

Output:

(g*(((a + b*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(2*(c + d*x)^2) 
 - B*n*(-1/4*(B*n*(a + b*x)^2)/(c + d*x)^2 + ((a + b*x)^2*(A + B*Log[e*((a 
 + b*x)/(c + d*x))^n]))/(2*(c + d*x)^2))))/((b*c - a*d)*i^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 = 4.75 (sec) , antiderivative size = 530, normalized size of antiderivative = 3.51

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

Input:

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

Output:

-1/4*(8*A*B*x*ln(e*((b*x+a)/(d*x+c))^n)*a*b^2*d^4*g*n+4*A*B*x^2*ln(e*((b*x 
+a)/(d*x+c))^n)*b^3*d^4*g*n+4*B^2*x*ln(e*((b*x+a)/(d*x+c))^n)^2*a*b^2*d^4* 
g*n-4*B^2*x*ln(e*((b*x+a)/(d*x+c))^n)*a*b^2*d^4*g*n^2-4*A*B*x*a*b^2*d^4*g* 
n^2+4*A*B*x*b^3*c*d^3*g*n^2+4*A*B*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b*d^4*g*n+ 
2*B^2*x^2*ln(e*((b*x+a)/(d*x+c))^n)^2*b^3*d^4*g*n-2*B^2*x^2*ln(e*((b*x+a)/ 
(d*x+c))^n)*b^3*d^4*g*n^2+2*B^2*x*a*b^2*d^4*g*n^3-2*B^2*x*b^3*c*d^3*g*n^3+ 
2*B^2*ln(e*((b*x+a)/(d*x+c))^n)^2*a^2*b*d^4*g*n-2*B^2*ln(e*((b*x+a)/(d*x+c 
))^n)*a^2*b*d^4*g*n^2+4*A^2*x*a*b^2*d^4*g*n-4*A^2*x*b^3*c*d^3*g*n-2*A*B*a^ 
2*b*d^4*g*n^2+2*A*B*b^3*c^2*d^2*g*n^2+B^2*a^2*b*d^4*g*n^3-B^2*b^3*c^2*d^2* 
g*n^3+2*A^2*a^2*b*d^4*g*n-2*A^2*b^3*c^2*d^2*g*n)/i^3/(d*x+c)^2/b/d^4/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 \[ \int \frac {(a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^3} \, dx=-\frac {{\left (B^{2} b^{2} c^{2} - B^{2} a^{2} d^{2}\right )} g n^{2} - 2 \, {\left (A B b^{2} c^{2} - A B a^{2} d^{2}\right )} g n + 2 \, {\left (2 \, {\left (B^{2} b^{2} c d - B^{2} a b d^{2}\right )} g x + {\left (B^{2} b^{2} c^{2} - B^{2} a^{2} d^{2}\right )} g\right )} \log \left (e\right )^{2} - 2 \, {\left (B^{2} b^{2} d^{2} g n^{2} x^{2} + 2 \, B^{2} a b d^{2} g n^{2} x + B^{2} a^{2} d^{2} g n^{2}\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2} + 2 \, {\left (A^{2} b^{2} c^{2} - A^{2} a^{2} d^{2}\right )} g + 2 \, {\left ({\left (B^{2} b^{2} c d - B^{2} a b d^{2}\right )} g n^{2} - 2 \, {\left (A B b^{2} c d - A B a b d^{2}\right )} g n + 2 \, {\left (A^{2} b^{2} c d - A^{2} a b d^{2}\right )} g\right )} x - 2 \, {\left ({\left (B^{2} b^{2} c^{2} - B^{2} a^{2} d^{2}\right )} g n - 2 \, {\left (A B b^{2} c^{2} - A B a^{2} d^{2}\right )} g + 2 \, {\left ({\left (B^{2} b^{2} c d - B^{2} a b d^{2}\right )} g n - 2 \, {\left (A B b^{2} c d - A B a b d^{2}\right )} g\right )} x + 2 \, {\left (B^{2} b^{2} d^{2} g n x^{2} + 2 \, B^{2} a b d^{2} g n x + B^{2} a^{2} d^{2} g n\right )} \log \left (\frac {b x + a}{d x + c}\right )\right )} \log \left (e\right ) + 2 \, {\left (B^{2} a^{2} d^{2} g n^{2} - 2 \, A B a^{2} d^{2} g n + {\left (B^{2} b^{2} d^{2} g n^{2} - 2 \, A B b^{2} d^{2} g n\right )} x^{2} + 2 \, {\left (B^{2} a b d^{2} g n^{2} - 2 \, A B a b d^{2} g n\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left ({\left (b c d^{4} - a d^{5}\right )} i^{3} x^{2} + 2 \, {\left (b c^{2} d^{3} - a c d^{4}\right )} i^{3} x + {\left (b c^{3} d^{2} - a c^{2} d^{3}\right )} i^{3}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1995 vs. \(2 (145) = 290\).

Time = 0.13 (sec) , antiderivative size = 1995, normalized size of antiderivative = 13.21 \[ \int \frac {(a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.29 \[ \int \frac {(a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^3} \, dx=\frac {1}{4} \, {\left (\frac {2 \, {\left (b x + a\right )}^{2} B^{2} g n^{2} \log \left (\frac {b x + a}{d x + c}\right )^{2}}{{\left (d x + c\right )}^{2} i^{3}} - \frac {2 \, {\left (B^{2} g n^{2} - 2 \, B^{2} g n \log \left (e\right ) - 2 \, A B g n\right )} {\left (b x + a\right )}^{2} \log \left (\frac {b x + a}{d x + c}\right )}{{\left (d x + c\right )}^{2} i^{3}} + \frac {{\left (B^{2} g n^{2} - 2 \, B^{2} g n \log \left (e\right ) + 2 \, B^{2} g \log \left (e\right )^{2} - 2 \, A B g n + 4 \, A B g \log \left (e\right ) + 2 \, A^{2} g\right )} {\left (b x + a\right )}^{2}}{{\left (d x + c\right )}^{2} i^{3}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 29.01 (sec) , antiderivative size = 565, normalized size of antiderivative = 3.74 \[ \int \frac {(a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^3} \, dx=-{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2\,\left (\frac {\frac {B^2\,a\,g}{2\,d}+\frac {B^2\,b\,g\,x}{d}+\frac {B^2\,b\,c\,g}{2\,d^2}}{c^2\,i^3+2\,c\,d\,i^3\,x+d^2\,i^3\,x^2}+\frac {B^2\,b^2\,g}{2\,d^2\,i^3\,\left (a\,d-b\,c\right )}\right )-\frac {x\,\left (2\,b\,d\,g\,A^2-2\,b\,d\,g\,A\,B\,n+b\,d\,g\,B^2\,n^2\right )+A^2\,a\,d\,g+A^2\,b\,c\,g+\frac {B^2\,a\,d\,g\,n^2}{2}+\frac {B^2\,b\,c\,g\,n^2}{2}-A\,B\,a\,d\,g\,n-A\,B\,b\,c\,g\,n}{2\,c^2\,d^2\,i^3+4\,c\,d^3\,i^3\,x+2\,d^4\,i^3\,x^2}-\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {A\,B\,a\,d\,g+A\,B\,b\,c\,g-B^2\,a\,d\,g\,n+B^2\,b\,c\,g\,n+2\,A\,B\,b\,d\,g\,x}{c^2\,d^2\,i^3+2\,c\,d^3\,i^3\,x+d^4\,i^3\,x^2}-\frac {B^2\,b^2\,g\,\left (\frac {c\,d^2\,i^3\,n\,\left (a\,d-b\,c\right )}{2\,b}+\frac {d^3\,i^3\,n\,x\,\left (a\,d-b\,c\right )}{b}-\frac {d^2\,i^3\,n\,\left (a\,d-b\,c\right )\,\left (a\,d-2\,b\,c\right )}{2\,b^2}\right )}{d^2\,i^3\,\left (a\,d-b\,c\right )\,\left (c^2\,d^2\,i^3+2\,c\,d^3\,i^3\,x+d^4\,i^3\,x^2\right )}\right )-\frac {B\,b^2\,g\,n\,\mathrm {atan}\left (\frac {B\,b^2\,g\,n\,\left (2\,A-B\,n\right )\,\left (\frac {a\,d^3\,i^3+b\,c\,d^2\,i^3}{d^2\,i^3}+2\,b\,d\,x\right )\,1{}\mathrm {i}}{\left (a\,d-b\,c\right )\,\left (B^2\,b^2\,g\,n^2-2\,A\,B\,b^2\,g\,n\right )}\right )\,\left (2\,A-B\,n\right )\,1{}\mathrm {i}}{d^2\,i^3\,\left (a\,d-b\,c\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 761, normalized size of antiderivative = 5.04 \[ \int \frac {(a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

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