\(\int \frac {(A+B \log (\frac {e (a+b x)}{c+d x}))^2}{(a g+b g x)^2 (c i+d i x)^3} \, dx\) [105]

Optimal result

Integrand size = 42, antiderivative size = 525 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(a g+b g x)^2 (c i+d i x)^3} \, dx=-\frac {B^2 d^3 (a+b x)^2}{4 (b c-a d)^4 g^2 i^3 (c+d x)^2}-\frac {6 A b B d^2 (a+b x)}{(b c-a d)^4 g^2 i^3 (c+d x)}+\frac {6 b B^2 d^2 (a+b x)}{(b c-a d)^4 g^2 i^3 (c+d x)}-\frac {2 b^3 B^2 (c+d x)}{(b c-a d)^4 g^2 i^3 (a+b x)}-\frac {6 b B^2 d^2 (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{(b c-a d)^4 g^2 i^3 (c+d x)}+\frac {B d^3 (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{2 (b c-a d)^4 g^2 i^3 (c+d x)^2}-\frac {2 b^3 B (c+d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{(b c-a d)^4 g^2 i^3 (a+b x)}-\frac {d^3 (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{2 (b c-a d)^4 g^2 i^3 (c+d x)^2}+\frac {3 b d^2 (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(b c-a d)^4 g^2 i^3 (c+d x)}-\frac {b^3 (c+d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(b c-a d)^4 g^2 i^3 (a+b x)}-\frac {b^2 d \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^3}{B (b c-a d)^4 g^2 i^3} \] Output:

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

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 453, normalized size of antiderivative = 0.86 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(a g+b g x)^2 (c i+d i x)^3} \, dx=-\frac {\left (2 A^2-2 A B+B^2\right ) d (b c-a d)^2 (a+b x)+2 b \left (4 A^2-10 A B+11 B^2\right ) d (b c-a d) (a+b x) (c+d x)+4 b^2 \left (A^2+2 A B+2 B^2\right ) (b c-a d) (c+d x)^2+6 b^2 \left (2 A^2-2 A B+5 B^2\right ) d (a+b x) (c+d x)^2 \log (a+b x)+2 B (b c-a d) \left ((2 A-B) d (b c-a d) (a+b x)+2 b (4 A-5 B) d (a+b x) (c+d x)+4 b^2 (A+B) (c+d x)^2\right ) \log \left (\frac {e (a+b x)}{c+d x}\right )+2 B \left (a^3 B d^3-3 a^2 b B d^2 (2 c+d x)+3 a b^2 d \left (2 A (c+d x)^2-B d x (4 c+3 d x)\right )+b^3 \left (6 A d x (c+d x)^2+B \left (2 c^3+6 c^2 d x-3 d^3 x^3\right )\right )\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )+4 b^2 B^2 d (a+b x) (c+d x)^2 \log ^3\left (\frac {e (a+b x)}{c+d x}\right )-6 b^2 \left (2 A^2-2 A B+5 B^2\right ) d (a+b x) (c+d x)^2 \log (c+d x)}{4 (b c-a d)^4 g^2 i^3 (a+b x) (c+d x)^2} \] Input:

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

Output:

-1/4*((2*A^2 - 2*A*B + B^2)*d*(b*c - a*d)^2*(a + b*x) + 2*b*(4*A^2 - 10*A* 
B + 11*B^2)*d*(b*c - a*d)*(a + b*x)*(c + d*x) + 4*b^2*(A^2 + 2*A*B + 2*B^2 
)*(b*c - a*d)*(c + d*x)^2 + 6*b^2*(2*A^2 - 2*A*B + 5*B^2)*d*(a + b*x)*(c + 
 d*x)^2*Log[a + b*x] + 2*B*(b*c - a*d)*((2*A - B)*d*(b*c - a*d)*(a + b*x) 
+ 2*b*(4*A - 5*B)*d*(a + b*x)*(c + d*x) + 4*b^2*(A + B)*(c + d*x)^2)*Log[( 
e*(a + b*x))/(c + d*x)] + 2*B*(a^3*B*d^3 - 3*a^2*b*B*d^2*(2*c + d*x) + 3*a 
*b^2*d*(2*A*(c + d*x)^2 - B*d*x*(4*c + 3*d*x)) + b^3*(6*A*d*x*(c + d*x)^2 
+ B*(2*c^3 + 6*c^2*d*x - 3*d^3*x^3)))*Log[(e*(a + b*x))/(c + d*x)]^2 + 4*b 
^2*B^2*d*(a + b*x)*(c + d*x)^2*Log[(e*(a + b*x))/(c + d*x)]^3 - 6*b^2*(2*A 
^2 - 2*A*B + 5*B^2)*d*(a + b*x)*(c + d*x)^2*Log[c + d*x])/((b*c - a*d)^4*g 
^2*i^3*(a + b*x)*(c + d*x)^2)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.70, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2962, 2795, 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.

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

\(\Big \downarrow \) 2962

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

\(\Big \downarrow \) 2795

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-1/4*(B^2*d^3*(a + b*x)^2)/(c + d*x)^2 - (6*A*b*B*d^2*(a + b*x))/(c + d*x 
) + (6*b*B^2*d^2*(a + b*x))/(c + d*x) - (2*b^3*B^2*(c + d*x))/(a + b*x) - 
(6*b*B^2*d^2*(a + b*x)*Log[(e*(a + b*x))/(c + d*x)])/(c + d*x) + (B*d^3*(a 
 + b*x)^2*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(2*(c + d*x)^2) - (2*b^3*B 
*(c + d*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(a + b*x) - (d^3*(a + b*x 
)^2*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2)/(2*(c + d*x)^2) + (3*b*d^2*(a 
+ b*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2)/(c + d*x) - (b^3*(c + d*x)* 
(A + B*Log[(e*(a + b*x))/(c + d*x)])^2)/(a + b*x) - (b^2*d*(A + B*Log[(e*( 
a + b*x))/(c + d*x)])^3)/B)/((b*c - a*d)^4*g^2*i^3)
 

Defintions of rubi rules used

rule 2009

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

rule 2795

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + 
(e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[ 
c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b 
, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0 
] && IntegerQ[m] && IntegerQ[r]))
 

rule 2962

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ 
)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Sy 
mbol] :> Simp[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q   Subst[Int[x^m*((A + 
 B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; 
 FreeQ[{a, b, c, d, e, f, g, h, i, A, B, n, p}, x] && EqQ[n + mn, 0] && IGt 
Q[n, 0] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && I 
ntegersQ[m, q]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1079\) vs. \(2(519)=1038\).

Time = 3.41 (sec) , antiderivative size = 1080, normalized size of antiderivative = 2.06

method	result	size
parts	\(\text {Expression too large to display}\)	\(1080\)
derivativedivides	\(\text {Expression too large to display}\)	\(1233\)
default	\(\text {Expression too large to display}\)	\(1233\)
parallelrisch	\(\text {Expression too large to display}\)	\(1762\)
norman	\(\text {Expression too large to display}\)	\(1852\)
risch	\(\text {Expression too large to display}\)	\(2114\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 1008, normalized size of antiderivative = 1.92 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(a g+b g x)^2 (c i+d i x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/4*(4*(A^2 + 2*A*B + 2*B^2)*b^3*c^3 + 3*(2*A^2 - 10*A*B + 5*B^2)*a*b^2*c 
^2*d - 12*(A^2 - 2*A*B + 2*B^2)*a^2*b*c*d^2 + (2*A^2 - 2*A*B + B^2)*a^3*d^ 
3 + 4*(B^2*b^3*d^3*x^3 + B^2*a*b^2*c^2*d + (2*B^2*b^3*c*d^2 + B^2*a*b^2*d^ 
3)*x^2 + (B^2*b^3*c^2*d + 2*B^2*a*b^2*c*d^2)*x)*log((b*e*x + a*e)/(d*x + c 
))^3 + 6*((2*A^2 - 2*A*B + 5*B^2)*b^3*c*d^2 - (2*A^2 - 2*A*B + 5*B^2)*a*b^ 
2*d^3)*x^2 + 2*(3*(2*A*B - B^2)*b^3*d^3*x^3 + 2*B^2*b^3*c^3 + 6*A*B*a*b^2* 
c^2*d - 6*B^2*a^2*b*c*d^2 + B^2*a^3*d^3 + 3*(4*A*B*b^3*c*d^2 + (2*A*B - 3* 
B^2)*a*b^2*d^3)*x^2 - 3*(B^2*a^2*b*d^3 - 2*(A*B + B^2)*b^3*c^2*d - 4*(A*B 
- B^2)*a*b^2*c*d^2)*x)*log((b*e*x + a*e)/(d*x + c))^2 + 3*((6*A^2 - 2*A*B 
+ 13*B^2)*b^3*c^2*d - 2*(2*A^2 + 2*A*B + 3*B^2)*a*b^2*c*d^2 - (2*A^2 - 6*A 
*B + 7*B^2)*a^2*b*d^3)*x + 2*(3*(2*A^2 - 2*A*B + 5*B^2)*b^3*d^3*x^3 + 6*A^ 
2*a*b^2*c^2*d + 4*(A*B + B^2)*b^3*c^3 - 12*(A*B - B^2)*a^2*b*c*d^2 + (2*A* 
B - B^2)*a^3*d^3 + 3*(4*(A^2 + 2*B^2)*b^3*c*d^2 + (2*A^2 - 6*A*B + 7*B^2)* 
a*b^2*d^3)*x^2 + 3*(2*(A^2 + 2*A*B + 2*B^2)*b^3*c^2*d + 4*(A^2 - 2*A*B + 2 
*B^2)*a*b^2*c*d^2 - (2*A*B - 3*B^2)*a^2*b*d^3)*x)*log((b*e*x + a*e)/(d*x + 
 c)))/((b^5*c^4*d^2 - 4*a*b^4*c^3*d^3 + 6*a^2*b^3*c^2*d^4 - 4*a^3*b^2*c*d^ 
5 + a^4*b*d^6)*g^2*i^3*x^3 + (2*b^5*c^5*d - 7*a*b^4*c^4*d^2 + 8*a^2*b^3*c^ 
3*d^3 - 2*a^3*b^2*c^2*d^4 - 2*a^4*b*c*d^5 + a^5*d^6)*g^2*i^3*x^2 + (b^5*c^ 
6 - 2*a*b^4*c^5*d - 2*a^2*b^3*c^4*d^2 + 8*a^3*b^2*c^3*d^3 - 7*a^4*b*c^2*d^ 
4 + 2*a^5*c*d^5)*g^2*i^3*x + (a*b^4*c^6 - 4*a^2*b^3*c^5*d + 6*a^3*b^2*c...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2683 vs. \(2 (484) = 968\).

Time = 27.49 (sec) , antiderivative size = 2683, normalized size of antiderivative = 5.11 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(a g+b g x)^2 (c i+d i x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

-B**2*b**2*d*log(e*(a + b*x)/(c + d*x))**3/(a**4*d**4*g**2*i**3 - 4*a**3*b 
*c*d**3*g**2*i**3 + 6*a**2*b**2*c**2*d**2*g**2*i**3 - 4*a*b**3*c**3*d*g**2 
*i**3 + b**4*c**4*g**2*i**3) + 3*b**2*d*(2*A**2 - 2*A*B + 5*B**2)*log(x + 
(6*A**2*a*b**2*d**2 + 6*A**2*b**3*c*d - 6*A*B*a*b**2*d**2 - 6*A*B*b**3*c*d 
 + 15*B**2*a*b**2*d**2 + 15*B**2*b**3*c*d - 3*a**5*b**2*d**6*(2*A**2 - 2*A 
*B + 5*B**2)/(a*d - b*c)**4 + 15*a**4*b**3*c*d**5*(2*A**2 - 2*A*B + 5*B**2 
)/(a*d - b*c)**4 - 30*a**3*b**4*c**2*d**4*(2*A**2 - 2*A*B + 5*B**2)/(a*d - 
 b*c)**4 + 30*a**2*b**5*c**3*d**3*(2*A**2 - 2*A*B + 5*B**2)/(a*d - b*c)**4 
 - 15*a*b**6*c**4*d**2*(2*A**2 - 2*A*B + 5*B**2)/(a*d - b*c)**4 + 3*b**7*c 
**5*d*(2*A**2 - 2*A*B + 5*B**2)/(a*d - b*c)**4)/(12*A**2*b**3*d**2 - 12*A* 
B*b**3*d**2 + 30*B**2*b**3*d**2))/(2*g**2*i**3*(a*d - b*c)**4) - 3*b**2*d* 
(2*A**2 - 2*A*B + 5*B**2)*log(x + (6*A**2*a*b**2*d**2 + 6*A**2*b**3*c*d - 
6*A*B*a*b**2*d**2 - 6*A*B*b**3*c*d + 15*B**2*a*b**2*d**2 + 15*B**2*b**3*c* 
d + 3*a**5*b**2*d**6*(2*A**2 - 2*A*B + 5*B**2)/(a*d - b*c)**4 - 15*a**4*b* 
*3*c*d**5*(2*A**2 - 2*A*B + 5*B**2)/(a*d - b*c)**4 + 30*a**3*b**4*c**2*d** 
4*(2*A**2 - 2*A*B + 5*B**2)/(a*d - b*c)**4 - 30*a**2*b**5*c**3*d**3*(2*A** 
2 - 2*A*B + 5*B**2)/(a*d - b*c)**4 + 15*a*b**6*c**4*d**2*(2*A**2 - 2*A*B + 
 5*B**2)/(a*d - b*c)**4 - 3*b**7*c**5*d*(2*A**2 - 2*A*B + 5*B**2)/(a*d - b 
*c)**4)/(12*A**2*b**3*d**2 - 12*A*B*b**3*d**2 + 30*B**2*b**3*d**2))/(2*g** 
2*i**3*(a*d - b*c)**4) + (-2*A*B*a**2*d**2 + 10*A*B*a*b*c*d + 6*A*B*a*b...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4188 vs. \(2 (519) = 1038\).

Time = 0.56 (sec) , antiderivative size = 4188, normalized size of antiderivative = 7.98 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(a g+b g x)^2 (c i+d i x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 32.62 (sec) , antiderivative size = 1505, normalized size of antiderivative = 2.87 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(a g+b g x)^2 (c i+d i x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((4*A^2*b^2*c^2 - 2*A^2*a^2*d^2 - B^2*a^2*d^2 + 8*B^2*b^2*c^2 + 2*A*B*a^2* 
d^2 + 8*A*B*b^2*c^2 + 10*A^2*a*b*c*d + 23*B^2*a*b*c*d - 22*A*B*a*b*c*d)/(2 
*(a*d - b*c)) + (3*x^2*(2*A^2*b^2*d^2 + 5*B^2*b^2*d^2 - 2*A*B*b^2*d^2))/(a 
*d - b*c) + (3*x*(2*A^2*a*b*d^2 + 7*B^2*a*b*d^2 + 6*A^2*b^2*c*d + 13*B^2*b 
^2*c*d - 6*A*B*a*b*d^2 - 2*A*B*b^2*c*d))/(2*(a*d - b*c)))/(x*(2*b^3*c^4*g^ 
2*i^3 + 4*a^3*c*d^3*g^2*i^3 - 6*a^2*b*c^2*d^2*g^2*i^3) + x^2*(2*a^3*d^4*g^ 
2*i^3 + 4*b^3*c^3*d*g^2*i^3 - 6*a*b^2*c^2*d^2*g^2*i^3) + x^3*(2*b^3*c^2*d^ 
2*g^2*i^3 + 2*a^2*b*d^4*g^2*i^3 - 4*a*b^2*c*d^3*g^2*i^3) + 2*a^3*c^2*d^2*g 
^2*i^3 + 2*a*b^2*c^4*g^2*i^3 - 4*a^2*b*c^3*d*g^2*i^3) - log((e*(a + b*x))/ 
(c + d*x))^2*((x*((3*B^2)/(2*g^2*i^3*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) - (3 
*B^2*(a*d + b*c))/(g^2*i^3*(a*d - b*c)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))) + 
 (B^2*(a*d + 2*b*c))/(2*g^2*i^3*(a^2*b*d^3 + b^3*c^2*d - 2*a*b^2*c*d^2)) - 
 (3*B^2*a*c)/(g^2*i^3*(a*d - b*c)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) - (3*B^ 
2*b*d*x^2)/(g^2*i^3*(a*d - b*c)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/(d*x^3 + 
 (a*c^2)/(b*d) + (x^2*(a*d^2 + 2*b*c*d))/(b*d) + (x*(b*c^2 + 2*a*c*d))/(b* 
d)) + (3*B*b^2*d*(2*A - B))/(2*g^2*i^3*(a*d - b*c)^2*(a^2*d^2 + b^2*c^2 - 
2*a*b*c*d))) - (log((e*(a + b*x))/(c + d*x))*(x*((3*(B^2 + 2*A*B))/(2*g^2* 
i^3*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) - (3*B*(2*A - B)*(a*d + b*c))/(g^2*i^ 
3*(a*d - b*c)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))) + (4*B^2*b*c - B^2*a*d + 2 
*A*B*a*d + 4*A*B*b*c)/(2*g^2*i^3*(a^2*b*d^3 + b^3*c^2*d - 2*a*b^2*c*d^2...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 3298, normalized size of antiderivative = 6.28 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{(a g+b g x)^2 (c i+d i x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(i*( - 12*log(a + b*x)*a**4*b**2*c**2*d**2 - 24*log(a + b*x)*a**4*b**2*c*d 
**3*x - 12*log(a + b*x)*a**4*b**2*d**4*x**2 - 24*log(a + b*x)*a**3*b**3*c* 
*3*d - 60*log(a + b*x)*a**3*b**3*c**2*d**2*x + 36*log(a + b*x)*a**3*b**3*c 
**2*d**2 - 48*log(a + b*x)*a**3*b**3*c*d**3*x**2 + 72*log(a + b*x)*a**3*b* 
*3*c*d**3*x - 12*log(a + b*x)*a**3*b**3*d**4*x**3 + 36*log(a + b*x)*a**3*b 
**3*d**4*x**2 - 24*log(a + b*x)*a**2*b**4*c**3*d*x - 48*log(a + b*x)*a**2* 
b**4*c**2*d**2*x**2 + 36*log(a + b*x)*a**2*b**4*c**2*d**2*x - 42*log(a + b 
*x)*a**2*b**4*c**2*d**2 - 24*log(a + b*x)*a**2*b**4*c*d**3*x**3 + 72*log(a 
 + b*x)*a**2*b**4*c*d**3*x**2 - 84*log(a + b*x)*a**2*b**4*c*d**3*x + 36*lo 
g(a + b*x)*a**2*b**4*d**4*x**3 - 42*log(a + b*x)*a**2*b**4*d**4*x**2 - 48* 
log(a + b*x)*a*b**5*c**3*d - 138*log(a + b*x)*a*b**5*c**2*d**2*x - 132*log 
(a + b*x)*a*b**5*c*d**3*x**2 - 42*log(a + b*x)*a*b**5*d**4*x**3 - 48*log(a 
 + b*x)*b**6*c**3*d*x - 96*log(a + b*x)*b**6*c**2*d**2*x**2 - 48*log(a + b 
*x)*b**6*c*d**3*x**3 + 12*log(c + d*x)*a**4*b**2*c**2*d**2 + 24*log(c + d* 
x)*a**4*b**2*c*d**3*x + 12*log(c + d*x)*a**4*b**2*d**4*x**2 + 24*log(c + d 
*x)*a**3*b**3*c**3*d + 60*log(c + d*x)*a**3*b**3*c**2*d**2*x - 36*log(c + 
d*x)*a**3*b**3*c**2*d**2 + 48*log(c + d*x)*a**3*b**3*c*d**3*x**2 - 72*log( 
c + d*x)*a**3*b**3*c*d**3*x + 12*log(c + d*x)*a**3*b**3*d**4*x**3 - 36*log 
(c + d*x)*a**3*b**3*d**4*x**2 + 24*log(c + d*x)*a**2*b**4*c**3*d*x + 48*lo 
g(c + d*x)*a**2*b**4*c**2*d**2*x**2 - 36*log(c + d*x)*a**2*b**4*c**2*d*...