\(\int (a g+b g x)^2 (c i+d i x)^2 (A+B \log (e (\frac {a+b x}{c+d x})^n)) \, dx\) [118]

Optimal result

Integrand size = 43, antiderivative size = 352 \[ \int (a g+b g x)^2 (c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=-\frac {B (b c-a d)^4 g^2 i^2 n x}{30 b^2 d^2}-\frac {B (b c-a d)^3 g^2 i^2 n (c+d x)^2}{60 b d^3}+\frac {B (b c-a d)^2 g^2 i^2 n (c+d x)^3}{10 d^3}-\frac {b B (b c-a d) g^2 i^2 n (c+d x)^4}{20 d^3}+\frac {(b c-a d)^2 g^2 i^2 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 d^3}-\frac {b (b c-a d) g^2 i^2 (c+d x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 d^3}+\frac {b^2 g^2 i^2 (c+d x)^5 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{5 d^3}-\frac {B (b c-a d)^5 g^2 i^2 n \log \left (\frac {a+b x}{c+d x}\right )}{30 b^3 d^3}-\frac {B (b c-a d)^5 g^2 i^2 n \log (c+d x)}{30 b^3 d^3} \] Output:

-1/30*B*(-a*d+b*c)^4*g^2*i^2*n*x/b^2/d^2-1/60*B*(-a*d+b*c)^3*g^2*i^2*n*(d* 
x+c)^2/b/d^3+1/10*B*(-a*d+b*c)^2*g^2*i^2*n*(d*x+c)^3/d^3-1/20*b*B*(-a*d+b* 
c)*g^2*i^2*n*(d*x+c)^4/d^3+1/3*(-a*d+b*c)^2*g^2*i^2*(d*x+c)^3*(A+B*ln(e*(( 
b*x+a)/(d*x+c))^n))/d^3-1/2*b*(-a*d+b*c)*g^2*i^2*(d*x+c)^4*(A+B*ln(e*((b*x 
+a)/(d*x+c))^n))/d^3+1/5*b^2*g^2*i^2*(d*x+c)^5*(A+B*ln(e*((b*x+a)/(d*x+c)) 
^n))/d^3-1/30*B*(-a*d+b*c)^5*g^2*i^2*n*ln((b*x+a)/(d*x+c))/b^3/d^3-1/30*B* 
(-a*d+b*c)^5*g^2*i^2*n*ln(d*x+c)/b^3/d^3
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.06 \[ \int (a g+b g x)^2 (c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {g^2 i^2 \left (20 d^3 (b c-a d)^2 (a+b x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+30 d^4 (b c-a d) (a+b x)^4 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+12 d^5 (a+b x)^5 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+10 B (b c-a d)^3 n \left (2 b d (b c-a d) x-d^2 (a+b x)^2-2 (b c-a d)^2 \log (c+d x)\right )-5 B (b c-a d)^2 n \left (6 b d (b c-a d)^2 x+3 d^2 (-b c+a d) (a+b x)^2+2 d^3 (a+b x)^3-6 (b c-a d)^3 \log (c+d x)\right )+B (b c-a d) n \left (12 b d (b c-a d)^3 x-6 d^2 (b c-a d)^2 (a+b x)^2+4 d^3 (b c-a d) (a+b x)^3-3 d^4 (a+b x)^4-12 (b c-a d)^4 \log (c+d x)\right )\right )}{60 b^3 d^3} \] Input:

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

Output:

(g^2*i^2*(20*d^3*(b*c - a*d)^2*(a + b*x)^3*(A + B*Log[e*((a + b*x)/(c + d* 
x))^n]) + 30*d^4*(b*c - a*d)*(a + b*x)^4*(A + B*Log[e*((a + b*x)/(c + d*x) 
)^n]) + 12*d^5*(a + b*x)^5*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + 10*B*( 
b*c - a*d)^3*n*(2*b*d*(b*c - a*d)*x - d^2*(a + b*x)^2 - 2*(b*c - a*d)^2*Lo 
g[c + d*x]) - 5*B*(b*c - a*d)^2*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]) + B* 
(b*c - a*d)*n*(12*b*d*(b*c - a*d)^3*x - 6*d^2*(b*c - a*d)^2*(a + b*x)^2 + 
4*d^3*(b*c - a*d)*(a + b*x)^3 - 3*d^4*(a + b*x)^4 - 12*(b*c - a*d)^4*Log[c 
 + d*x])))/(60*b^3*d^3)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {2961, 2782, 27, 1195, 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.

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

\(\Big \downarrow \) 2961

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

\(\Big \downarrow \) 2782

\(\displaystyle g^2 i^2 (b c-a d)^5 \left (-B n \int \frac {(c+d x) \left (b^2-\frac {5 d (a+b x) b}{c+d x}+\frac {10 d^2 (a+b x)^2}{(c+d x)^2}\right )}{30 d^3 (a+b x) \left (b-\frac {d (a+b x)}{c+d x}\right )^5}d\frac {a+b x}{c+d x}+\frac {b^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{5 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^5}-\frac {b \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^4}+\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{3 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle g^2 i^2 (b c-a d)^5 \left (-\frac {B n \int \frac {(c+d x) \left (b^2-\frac {5 d (a+b x) b}{c+d x}+\frac {10 d^2 (a+b x)^2}{(c+d x)^2}\right )}{(a+b x) \left (b-\frac {d (a+b x)}{c+d x}\right )^5}d\frac {a+b x}{c+d x}}{30 d^3}+\frac {b^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{5 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^5}-\frac {b \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^4}+\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{3 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^3}\right )\)

\(\Big \downarrow \) 1195

\(\displaystyle g^2 i^2 (b c-a d)^5 \left (-\frac {B n \int \left (\frac {d}{b^3 \left (b-\frac {d (a+b x)}{c+d x}\right )}+\frac {d}{b^2 \left (b-\frac {d (a+b x)}{c+d x}\right )^2}+\frac {d}{b \left (b-\frac {d (a+b x)}{c+d x}\right )^3}-\frac {9 d}{\left (b-\frac {d (a+b x)}{c+d x}\right )^4}+\frac {6 b d}{\left (b-\frac {d (a+b x)}{c+d x}\right )^5}+\frac {c+d x}{b^3 (a+b x)}\right )d\frac {a+b x}{c+d x}}{30 d^3}+\frac {b^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{5 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^5}-\frac {b \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^4}+\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{3 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^3}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle g^2 i^2 (b c-a d)^5 \left (\frac {b^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{5 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^5}-\frac {b \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^4}+\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{3 d^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^3}-\frac {B n \left (\frac {\log \left (\frac {a+b x}{c+d x}\right )}{b^3}-\frac {\log \left (b-\frac {d (a+b x)}{c+d x}\right )}{b^3}+\frac {1}{b^2 \left (b-\frac {d (a+b x)}{c+d x}\right )}+\frac {3 b}{2 \left (b-\frac {d (a+b x)}{c+d x}\right )^4}-\frac {3}{\left (b-\frac {d (a+b x)}{c+d x}\right )^3}+\frac {1}{2 b \left (b-\frac {d (a+b x)}{c+d x}\right )^2}\right )}{30 d^3}\right )\)

Input:

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

Output:

(b*c - a*d)^5*g^2*i^2*((b^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(5*d^3 
*(b - (d*(a + b*x))/(c + d*x))^5) - (b*(A + B*Log[e*((a + b*x)/(c + d*x))^ 
n]))/(2*d^3*(b - (d*(a + b*x))/(c + d*x))^4) + (A + B*Log[e*((a + b*x)/(c 
+ d*x))^n])/(3*d^3*(b - (d*(a + b*x))/(c + d*x))^3) - (B*n*((3*b)/(2*(b - 
(d*(a + b*x))/(c + d*x))^4) - 3/(b - (d*(a + b*x))/(c + d*x))^3 + 1/(2*b*( 
b - (d*(a + b*x))/(c + d*x))^2) + 1/(b^2*(b - (d*(a + b*x))/(c + d*x))) + 
Log[(a + b*x)/(c + d*x)]/b^3 - Log[b - (d*(a + b*x))/(c + d*x)]/b^3))/(30* 
d^3))
 

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 1195

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && IGtQ[p, 0]
 

rule 2009

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

rule 2782

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_))^(q 
_), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x)^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}, x] && ILtQ[m + q + 2, 0] && IGtQ[m, 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. \(1273\) vs. \(2(334)=668\).

Time = 11.69 (sec) , antiderivative size = 1274, normalized size of antiderivative = 3.62

method	result	size
parallelrisch	\(\text {Expression too large to display}\)	\(1274\)

Input:

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

Output:

1/60*(12*B*x^5*ln(e*((b*x+a)/(d*x+c))^n)*b^5*d^5*g^2*i^2*n+30*B*x^4*ln(e*( 
(b*x+a)/(d*x+c))^n)*a*b^4*d^5*g^2*i^2*n+30*B*x^4*ln(e*((b*x+a)/(d*x+c))^n) 
*b^5*c*d^4*g^2*i^2*n+20*B*x^3*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b^3*d^5*g^2*i^ 
2*n+20*B*x^3*ln(e*((b*x+a)/(d*x+c))^n)*b^5*c^2*d^3*g^2*i^2*n+80*A*x^3*a*b^ 
4*c*d^4*g^2*i^2*n+15*B*x^2*a^2*b^3*c*d^4*g^2*i^2*n^2-15*B*x^2*a*b^4*c^2*d^ 
3*g^2*i^2*n^2+60*A*x^2*a^2*b^3*c*d^4*g^2*i^2*n+B*x^2*a^3*b^2*d^5*g^2*i^2*n 
^2-B*x^2*b^5*c^3*d^2*g^2*i^2*n^2-2*B*x*a^4*b*d^5*g^2*i^2*n^2+2*B*x*b^5*c^4 
*d*g^2*i^2*n^2+80*B*x^3*ln(e*((b*x+a)/(d*x+c))^n)*a*b^4*c*d^4*g^2*i^2*n+10 
*B*x*a^3*b^2*c*d^4*g^2*i^2*n^2-10*B*x*a*b^4*c^3*d^2*g^2*i^2*n^2+60*A*x*a^2 
*b^3*c^2*d^3*g^2*i^2*n+2*B*a^5*d^5*g^2*i^2*n^2-2*B*b^5*c^5*g^2*i^2*n^2-25* 
B*a^3*b^2*c^2*d^3*g^2*i^2*n^2+25*B*a^2*b^3*c^3*d^2*g^2*i^2*n^2+9*B*a*b^4*c 
^4*d*g^2*i^2*n^2-120*A*a^3*b^2*c^2*d^3*g^2*i^2*n-120*A*a^2*b^3*c^3*d^2*g^2 
*i^2*n+60*B*x^2*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b^3*c*d^4*g^2*i^2*n+60*B*x^2 
*ln(e*((b*x+a)/(d*x+c))^n)*a*b^4*c^2*d^3*g^2*i^2*n+60*B*x*ln(e*((b*x+a)/(d 
*x+c))^n)*a^2*b^3*c^2*d^3*g^2*i^2*n+20*B*ln(e*((b*x+a)/(d*x+c))^n)*a^2*b^3 
*c^3*d^2*g^2*i^2*n-10*B*ln(e*((b*x+a)/(d*x+c))^n)*a*b^4*c^4*d*g^2*i^2*n+10 
*B*ln(b*x+a)*a*b^4*c^4*d*g^2*i^2*n^2-10*B*ln(b*x+a)*a^4*b*c*d^4*g^2*i^2*n^ 
2+20*B*ln(b*x+a)*a^3*b^2*c^2*d^3*g^2*i^2*n^2-20*B*ln(b*x+a)*a^2*b^3*c^3*d^ 
2*g^2*i^2*n^2+60*A*x^2*a*b^4*c^2*d^3*g^2*i^2*n+3*B*x^4*a*b^4*d^5*g^2*i^2*n 
^2-3*B*x^4*b^5*c*d^4*g^2*i^2*n^2+30*A*x^4*a*b^4*d^5*g^2*i^2*n+30*A*x^4*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 774 vs. \(2 (334) = 668\).

Time = 0.18 (sec) , antiderivative size = 774, normalized size of antiderivative = 2.20 \[ \int (a g+b g x)^2 (c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {12 \, A b^{5} d^{5} g^{2} i^{2} x^{5} + 2 \, {\left (10 \, B a^{3} b^{2} c^{2} d^{3} - 5 \, B a^{4} b c d^{4} + B a^{5} d^{5}\right )} g^{2} i^{2} n \log \left (b x + a\right ) - 2 \, {\left (B b^{5} c^{5} - 5 \, B a b^{4} c^{4} d + 10 \, B a^{2} b^{3} c^{3} d^{2}\right )} g^{2} i^{2} n \log \left (d x + c\right ) - 3 \, {\left ({\left (B b^{5} c d^{4} - B a b^{4} d^{5}\right )} g^{2} i^{2} n - 10 \, {\left (A b^{5} c d^{4} + A a b^{4} d^{5}\right )} g^{2} i^{2}\right )} x^{4} - 2 \, {\left (3 \, {\left (B b^{5} c^{2} d^{3} - B a^{2} b^{3} d^{5}\right )} g^{2} i^{2} n - 10 \, {\left (A b^{5} c^{2} d^{3} + 4 \, A a b^{4} c d^{4} + A a^{2} b^{3} d^{5}\right )} g^{2} i^{2}\right )} x^{3} - {\left ({\left (B b^{5} c^{3} d^{2} + 15 \, B a b^{4} c^{2} d^{3} - 15 \, B a^{2} b^{3} c d^{4} - B a^{3} b^{2} d^{5}\right )} g^{2} i^{2} n - 60 \, {\left (A a b^{4} c^{2} d^{3} + A a^{2} b^{3} c d^{4}\right )} g^{2} i^{2}\right )} x^{2} + 2 \, {\left (30 \, A a^{2} b^{3} c^{2} d^{3} g^{2} i^{2} + {\left (B b^{5} c^{4} d - 5 \, B a b^{4} c^{3} d^{2} + 5 \, B a^{3} b^{2} c d^{4} - B a^{4} b d^{5}\right )} g^{2} i^{2} n\right )} x + 2 \, {\left (6 \, B b^{5} d^{5} g^{2} i^{2} x^{5} + 30 \, B a^{2} b^{3} c^{2} d^{3} g^{2} i^{2} x + 15 \, {\left (B b^{5} c d^{4} + B a b^{4} d^{5}\right )} g^{2} i^{2} x^{4} + 10 \, {\left (B b^{5} c^{2} d^{3} + 4 \, B a b^{4} c d^{4} + B a^{2} b^{3} d^{5}\right )} g^{2} i^{2} x^{3} + 30 \, {\left (B a b^{4} c^{2} d^{3} + B a^{2} b^{3} c d^{4}\right )} g^{2} i^{2} x^{2}\right )} \log \left (e\right ) + 2 \, {\left (6 \, B b^{5} d^{5} g^{2} i^{2} n x^{5} + 30 \, B a^{2} b^{3} c^{2} d^{3} g^{2} i^{2} n x + 15 \, {\left (B b^{5} c d^{4} + B a b^{4} d^{5}\right )} g^{2} i^{2} n x^{4} + 10 \, {\left (B b^{5} c^{2} d^{3} + 4 \, B a b^{4} c d^{4} + B a^{2} b^{3} d^{5}\right )} g^{2} i^{2} n x^{3} + 30 \, {\left (B a b^{4} c^{2} d^{3} + B a^{2} b^{3} c d^{4}\right )} g^{2} i^{2} n x^{2}\right )} \log \left (\frac {b x + a}{d x + c}\right )}{60 \, b^{3} d^{3}} \] Input:

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

Output:

1/60*(12*A*b^5*d^5*g^2*i^2*x^5 + 2*(10*B*a^3*b^2*c^2*d^3 - 5*B*a^4*b*c*d^4 
 + B*a^5*d^5)*g^2*i^2*n*log(b*x + a) - 2*(B*b^5*c^5 - 5*B*a*b^4*c^4*d + 10 
*B*a^2*b^3*c^3*d^2)*g^2*i^2*n*log(d*x + c) - 3*((B*b^5*c*d^4 - B*a*b^4*d^5 
)*g^2*i^2*n - 10*(A*b^5*c*d^4 + A*a*b^4*d^5)*g^2*i^2)*x^4 - 2*(3*(B*b^5*c^ 
2*d^3 - B*a^2*b^3*d^5)*g^2*i^2*n - 10*(A*b^5*c^2*d^3 + 4*A*a*b^4*c*d^4 + A 
*a^2*b^3*d^5)*g^2*i^2)*x^3 - ((B*b^5*c^3*d^2 + 15*B*a*b^4*c^2*d^3 - 15*B*a 
^2*b^3*c*d^4 - B*a^3*b^2*d^5)*g^2*i^2*n - 60*(A*a*b^4*c^2*d^3 + A*a^2*b^3* 
c*d^4)*g^2*i^2)*x^2 + 2*(30*A*a^2*b^3*c^2*d^3*g^2*i^2 + (B*b^5*c^4*d - 5*B 
*a*b^4*c^3*d^2 + 5*B*a^3*b^2*c*d^4 - B*a^4*b*d^5)*g^2*i^2*n)*x + 2*(6*B*b^ 
5*d^5*g^2*i^2*x^5 + 30*B*a^2*b^3*c^2*d^3*g^2*i^2*x + 15*(B*b^5*c*d^4 + B*a 
*b^4*d^5)*g^2*i^2*x^4 + 10*(B*b^5*c^2*d^3 + 4*B*a*b^4*c*d^4 + B*a^2*b^3*d^ 
5)*g^2*i^2*x^3 + 30*(B*a*b^4*c^2*d^3 + B*a^2*b^3*c*d^4)*g^2*i^2*x^2)*log(e 
) + 2*(6*B*b^5*d^5*g^2*i^2*n*x^5 + 30*B*a^2*b^3*c^2*d^3*g^2*i^2*n*x + 15*( 
B*b^5*c*d^4 + B*a*b^4*d^5)*g^2*i^2*n*x^4 + 10*(B*b^5*c^2*d^3 + 4*B*a*b^4*c 
*d^4 + B*a^2*b^3*d^5)*g^2*i^2*n*x^3 + 30*(B*a*b^4*c^2*d^3 + B*a^2*b^3*c*d^ 
4)*g^2*i^2*n*x^2)*log((b*x + a)/(d*x + c)))/(b^3*d^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1336 vs. \(2 (334) = 668\).

Time = 0.07 (sec) , antiderivative size = 1336, normalized size of antiderivative = 3.80 \[ \int (a g+b g x)^2 (c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/5*B*b^2*d^2*g^2*i^2*x^5*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 1/5*A*b 
^2*d^2*g^2*i^2*x^5 + 1/2*B*b^2*c*d*g^2*i^2*x^4*log(e*(b*x/(d*x + c) + a/(d 
*x + c))^n) + 1/2*B*a*b*d^2*g^2*i^2*x^4*log(e*(b*x/(d*x + c) + a/(d*x + c) 
)^n) + 1/2*A*b^2*c*d*g^2*i^2*x^4 + 1/2*A*a*b*d^2*g^2*i^2*x^4 + 1/3*B*b^2*c 
^2*g^2*i^2*x^3*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 4/3*B*a*b*c*d*g^2* 
i^2*x^3*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 1/3*B*a^2*d^2*g^2*i^2*x^3 
*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 1/3*A*b^2*c^2*g^2*i^2*x^3 + 4/3* 
A*a*b*c*d*g^2*i^2*x^3 + 1/3*A*a^2*d^2*g^2*i^2*x^3 + B*a*b*c^2*g^2*i^2*x^2* 
log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + B*a^2*c*d*g^2*i^2*x^2*log(e*(b*x/ 
(d*x + c) + a/(d*x + c))^n) + A*a*b*c^2*g^2*i^2*x^2 + A*a^2*c*d*g^2*i^2*x^ 
2 + 1/60*B*b^2*d^2*g^2*i^2*n*(12*a^5*log(b*x + a)/b^5 - 12*c^5*log(d*x + c 
)/d^5 - (3*(b^4*c*d^3 - a*b^3*d^4)*x^4 - 4*(b^4*c^2*d^2 - a^2*b^2*d^4)*x^3 
 + 6*(b^4*c^3*d - a^3*b*d^4)*x^2 - 12*(b^4*c^4 - a^4*d^4)*x)/(b^4*d^4)) - 
1/12*B*b^2*c*d*g^2*i^2*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)) - 1/12*B*a*b*d^2*g^2*i^2*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)) + 1/6*B*b^2*c^ 
2*g^2*i^2*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)) + 2/3*B*a*b*c*d*g^2...
                                                                                    
                                                                                    
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3352 vs. \(2 (334) = 668\).

Time = 1.13 (sec) , antiderivative size = 3352, normalized size of antiderivative = 9.52 \[ \int (a g+b g x)^2 (c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/60*(2*(B*b^8*c^6*g^2*i^2*n - 6*B*a*b^7*c^5*d*g^2*i^2*n - 5*(b*x + a)*B*b 
^7*c^6*d*g^2*i^2*n/(d*x + c) + 15*B*a^2*b^6*c^4*d^2*g^2*i^2*n + 30*(b*x + 
a)*B*a*b^6*c^5*d^2*g^2*i^2*n/(d*x + c) + 10*(b*x + a)^2*B*b^6*c^6*d^2*g^2* 
i^2*n/(d*x + c)^2 - 20*B*a^3*b^5*c^3*d^3*g^2*i^2*n - 75*(b*x + a)*B*a^2*b^ 
5*c^4*d^3*g^2*i^2*n/(d*x + c) - 60*(b*x + a)^2*B*a*b^5*c^5*d^3*g^2*i^2*n/( 
d*x + c)^2 + 15*B*a^4*b^4*c^2*d^4*g^2*i^2*n + 100*(b*x + a)*B*a^3*b^4*c^3* 
d^4*g^2*i^2*n/(d*x + c) + 150*(b*x + a)^2*B*a^2*b^4*c^4*d^4*g^2*i^2*n/(d*x 
 + c)^2 - 6*B*a^5*b^3*c*d^5*g^2*i^2*n - 75*(b*x + a)*B*a^4*b^3*c^2*d^5*g^2 
*i^2*n/(d*x + c) - 200*(b*x + a)^2*B*a^3*b^3*c^3*d^5*g^2*i^2*n/(d*x + c)^2 
 + B*a^6*b^2*d^6*g^2*i^2*n + 30*(b*x + a)*B*a^5*b^2*c*d^6*g^2*i^2*n/(d*x + 
 c) + 150*(b*x + a)^2*B*a^4*b^2*c^2*d^6*g^2*i^2*n/(d*x + c)^2 - 5*(b*x + a 
)*B*a^6*b*d^7*g^2*i^2*n/(d*x + c) - 60*(b*x + a)^2*B*a^5*b*c*d^7*g^2*i^2*n 
/(d*x + c)^2 + 10*(b*x + a)^2*B*a^6*d^8*g^2*i^2*n/(d*x + c)^2)*log((b*x + 
a)/(d*x + c))/(b^5*d^3 - 5*(b*x + a)*b^4*d^4/(d*x + c) + 10*(b*x + a)^2*b^ 
3*d^5/(d*x + c)^2 - 10*(b*x + a)^3*b^2*d^6/(d*x + c)^3 + 5*(b*x + a)^4*b*d 
^7/(d*x + c)^4 - (b*x + a)^5*d^8/(d*x + c)^5) + (2*(b*x + a)*B*b^9*c^6*d*g 
^2*i^2*n/(d*x + c) - 12*(b*x + a)*B*a*b^8*c^5*d^2*g^2*i^2*n/(d*x + c) - 9* 
(b*x + a)^2*B*b^8*c^6*d^2*g^2*i^2*n/(d*x + c)^2 + 30*(b*x + a)*B*a^2*b^7*c 
^4*d^3*g^2*i^2*n/(d*x + c) + 54*(b*x + a)^2*B*a*b^7*c^5*d^3*g^2*i^2*n/(d*x 
 + c)^2 + 9*(b*x + a)^3*B*b^7*c^6*d^3*g^2*i^2*n/(d*x + c)^3 - 40*(b*x +...
 

Mupad [B] (verification not implemented)

Time = 26.55 (sec) , antiderivative size = 1328, normalized size of antiderivative = 3.77 \[ \int (a g+b g x)^2 (c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

x^2*(((30*a*d + 30*b*c)*((((b*d*g^2*i^2*(15*A*a*d + 15*A*b*c + B*a*d*n - B 
*b*c*n))/5 - (A*b*d*g^2*i^2*(30*a*d + 30*b*c))/30)*(30*a*d + 30*b*c))/(30* 
b*d) - (g^2*i^2*(6*A*a^2*d^2 + 6*A*b^2*c^2 + B*a^2*d^2*n - B*b^2*c^2*n + 1 
8*A*a*b*c*d))/2 + A*a*b*c*d*g^2*i^2))/(60*b*d) - (a*c*((b*d*g^2*i^2*(15*A* 
a*d + 15*A*b*c + B*a*d*n - B*b*c*n))/5 - (A*b*d*g^2*i^2*(30*a*d + 30*b*c)) 
/30))/(2*b*d) + (g^2*i^2*(3*A*a^3*d^3 + 3*A*b^3*c^3 + B*a^3*d^3*n - B*b^3* 
c^3*n + 27*A*a*b^2*c^2*d + 27*A*a^2*b*c*d^2 - 3*B*a*b^2*c^2*d*n + 3*B*a^2* 
b*c*d^2*n))/(6*b*d)) + log(e*((a + b*x)/(c + d*x))^n)*((B*g^2*i^2*x^3*(a^2 
*d^2 + b^2*c^2 + 4*a*b*c*d))/3 + B*a^2*c^2*g^2*i^2*x + (B*b^2*d^2*g^2*i^2* 
x^5)/5 + B*a*c*g^2*i^2*x^2*(a*d + b*c) + (B*b*d*g^2*i^2*x^4*(a*d + b*c))/2 
) - x^3*((((b*d*g^2*i^2*(15*A*a*d + 15*A*b*c + B*a*d*n - B*b*c*n))/5 - (A* 
b*d*g^2*i^2*(30*a*d + 30*b*c))/30)*(30*a*d + 30*b*c))/(90*b*d) - (g^2*i^2* 
(6*A*a^2*d^2 + 6*A*b^2*c^2 + B*a^2*d^2*n - B*b^2*c^2*n + 18*A*a*b*c*d))/6 
+ (A*a*b*c*d*g^2*i^2)/3) + x*((a*c*((((b*d*g^2*i^2*(15*A*a*d + 15*A*b*c + 
B*a*d*n - B*b*c*n))/5 - (A*b*d*g^2*i^2*(30*a*d + 30*b*c))/30)*(30*a*d + 30 
*b*c))/(30*b*d) - (g^2*i^2*(6*A*a^2*d^2 + 6*A*b^2*c^2 + B*a^2*d^2*n - B*b^ 
2*c^2*n + 18*A*a*b*c*d))/2 + A*a*b*c*d*g^2*i^2))/(b*d) - ((30*a*d + 30*b*c 
)*(((30*a*d + 30*b*c)*((((b*d*g^2*i^2*(15*A*a*d + 15*A*b*c + B*a*d*n - B*b 
*c*n))/5 - (A*b*d*g^2*i^2*(30*a*d + 30*b*c))/30)*(30*a*d + 30*b*c))/(30*b* 
d) - (g^2*i^2*(6*A*a^2*d^2 + 6*A*b^2*c^2 + B*a^2*d^2*n - B*b^2*c^2*n + ...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 796, normalized size of antiderivative = 2.26 \[ \int (a g+b g x)^2 (c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

(g**2*( - 2*log(c + d*x)*a**5*d**5*n + 10*log(c + d*x)*a**4*b*c*d**4*n - 2 
0*log(c + d*x)*a**3*b**2*c**2*d**3*n + 20*log(c + d*x)*a**2*b**3*c**3*d**2 
*n - 10*log(c + d*x)*a*b**4*c**4*d*n + 2*log(c + d*x)*b**5*c**5*n - 2*log( 
((a + b*x)**n*e)/(c + d*x)**n)*a**5*d**5 + 10*log(((a + b*x)**n*e)/(c + d* 
x)**n)*a**4*b*c*d**4 - 20*log(((a + b*x)**n*e)/(c + d*x)**n)*a**3*b**2*c** 
2*d**3 - 60*log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b**3*c**2*d**3*x - 60* 
log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b**3*c*d**4*x**2 - 20*log(((a + b* 
x)**n*e)/(c + d*x)**n)*a**2*b**3*d**5*x**3 - 60*log(((a + b*x)**n*e)/(c + 
d*x)**n)*a*b**4*c**2*d**3*x**2 - 80*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b 
**4*c*d**4*x**3 - 30*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**4*d**5*x**4 - 
 20*log(((a + b*x)**n*e)/(c + d*x)**n)*b**5*c**2*d**3*x**3 - 30*log(((a + 
b*x)**n*e)/(c + d*x)**n)*b**5*c*d**4*x**4 - 12*log(((a + b*x)**n*e)/(c + d 
*x)**n)*b**5*d**5*x**5 + 2*a**4*b*d**5*n*x - 60*a**3*b**2*c**2*d**3*x - 10 
*a**3*b**2*c*d**4*n*x - 60*a**3*b**2*c*d**4*x**2 - a**3*b**2*d**5*n*x**2 - 
 20*a**3*b**2*d**5*x**3 - 60*a**2*b**3*c**2*d**3*x**2 - 15*a**2*b**3*c*d** 
4*n*x**2 - 80*a**2*b**3*c*d**4*x**3 - 6*a**2*b**3*d**5*n*x**3 - 30*a**2*b* 
*3*d**5*x**4 + 10*a*b**4*c**3*d**2*n*x + 15*a*b**4*c**2*d**3*n*x**2 - 20*a 
*b**4*c**2*d**3*x**3 - 30*a*b**4*c*d**4*x**4 - 3*a*b**4*d**5*n*x**4 - 12*a 
*b**4*d**5*x**5 - 2*b**5*c**4*d*n*x + b**5*c**3*d**2*n*x**2 + 6*b**5*c**2* 
d**3*n*x**3 + 3*b**5*c*d**4*n*x**4))/(60*b**2*d**3)