3.1.43 \(\int \frac {A+B \log (\frac {e (a+b x)}{c+d x})}{(a g+b g x) (c i+d i x)^2} \, dx\) [43]

3.1.43.1 Optimal result
3.1.43.2 Mathematica [C] (verified)
3.1.43.3 Rubi [A] (verified)
3.1.43.4 Maple [A] (verified)
3.1.43.5 Fricas [A] (verification not implemented)
3.1.43.6 Sympy [B] (verification not implemented)
3.1.43.7 Maxima [B] (verification not implemented)
3.1.43.8 Giac [A] (verification not implemented)
3.1.43.9 Mupad [B] (verification not implemented)

3.1.43.1 Optimal result

Integrand size = 40, antiderivative size = 156 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x) (c i+d i x)^2} \, dx=-\frac {A d (a+b x)}{(b c-a d)^2 g i^2 (c+d x)}+\frac {B d (a+b x)}{(b c-a d)^2 g i^2 (c+d x)}-\frac {B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{(b c-a d)^2 g i^2 (c+d x)}+\frac {b \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{2 B (b c-a d)^2 g i^2} \]

output
-A*d*(b*x+a)/(-a*d+b*c)^2/g/i^2/(d*x+c)+B*d*(b*x+a)/(-a*d+b*c)^2/g/i^2/(d* 
x+c)-B*d*(b*x+a)*ln(e*(b*x+a)/(d*x+c))/(-a*d+b*c)^2/g/i^2/(d*x+c)+1/2*b*(A 
+B*ln(e*(b*x+a)/(d*x+c)))^2/B/(-a*d+b*c)^2/g/i^2
 
3.1.43.2 Mathematica [C] (verified)

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

Time = 0.16 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.87 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x) (c i+d i x)^2} \, dx=\frac {2 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )+2 b (c+d x) \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )-2 b (c+d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)-2 B (b c-a d+b (c+d x) \log (a+b x)-b (c+d x) \log (c+d x))-b B (c+d x) \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )+b B (c+d x) \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )}{2 (b c-a d)^2 g i^2 (c+d x)} \]

input
Integrate[(A + B*Log[(e*(a + b*x))/(c + d*x)])/((a*g + b*g*x)*(c*i + d*i*x 
)^2),x]
 
output
(2*(b*c - a*d)*(A + B*Log[(e*(a + b*x))/(c + d*x)]) + 2*b*(c + d*x)*Log[a 
+ b*x]*(A + B*Log[(e*(a + b*x))/(c + d*x)]) - 2*b*(c + d*x)*(A + B*Log[(e* 
(a + b*x))/(c + d*x)])*Log[c + d*x] - 2*B*(b*c - a*d + b*(c + d*x)*Log[a + 
 b*x] - b*(c + d*x)*Log[c + d*x]) - b*B*(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*(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)]))/(2*(b*c - 
 a*d)^2*g*i^2*(c + d*x))
 
3.1.43.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.70, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2962, 2788, 2009, 2738}

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.

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

\(\Big \downarrow \) 2962

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

\(\Big \downarrow \) 2788

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 2738

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

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

3.1.43.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2738
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Lo 
g[c*x^n])^2/(2*b*n), x] /; FreeQ[{a, b, c, n}, x]
 

rule 2788
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.)) 
/(x_), x_Symbol] :> Simp[d   Int[(d + e*x)^(q - 1)*((a + b*Log[c*x^n])^p/x) 
, x], x] + Simp[e   Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /; F 
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]
 

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]
 
3.1.43.4 Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.48

method result size
parts \(\frac {A \left (-\frac {1}{\left (a d -c b \right ) \left (d x +c \right )}-\frac {b \ln \left (d x +c \right )}{\left (a d -c b \right )^{2}}+\frac {b \ln \left (b x +a \right )}{\left (a d -c b \right )^{2}}\right )}{g \,i^{2}}-\frac {B \left (\frac {d \left (\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right ) \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )-\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}-\frac {b e}{d}\right )}{a d -c b}-\frac {b e \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}{2 \left (a d -c b \right )}\right )}{g \,i^{2} \left (a d -c b \right ) e}\) \(231\)
parallelrisch \(\frac {2 B a \,b^{2} d^{4}-2 B \,b^{3} c \,d^{3}-2 A a \,b^{2} d^{4}+2 A \,b^{3} c \,d^{3}+B x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{2} b^{3} d^{4}+2 A x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{3} d^{4}-2 B x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{3} d^{4}+B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{2} b^{3} c \,d^{3}+2 A \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{3} c \,d^{3}-2 B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) a \,b^{2} d^{4}}{2 i^{2} g \left (d x +c \right ) b^{2} d^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\) \(238\)
norman \(\frac {\frac {\left (A b c -B a d \right ) \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{g i \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {d \left (A b -B b \right ) x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{g i \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {\left (-B +A \right ) d x}{g i c \left (a d -c b \right )}+\frac {B b c \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{2}}{2 g i \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {b B d x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{2}}{2 g i \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}}{i \left (d x +c \right )}\) \(253\)
derivativedivides \(-\frac {e \left (a d -c b \right ) \left (-\frac {d^{2} A b \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{e \,i^{2} \left (a d -c b \right )^{3} g}+\frac {d^{3} A \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{e^{2} i^{2} \left (a d -c b \right )^{3} g}-\frac {d^{2} B b \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}{2 e \,i^{2} \left (a d -c b \right )^{3} g}+\frac {d^{3} B \left (\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right ) \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )-\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}-\frac {b e}{d}\right )}{e^{2} i^{2} \left (a d -c b \right )^{3} g}\right )}{d^{2}}\) \(286\)
default \(-\frac {e \left (a d -c b \right ) \left (-\frac {d^{2} A b \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{e \,i^{2} \left (a d -c b \right )^{3} g}+\frac {d^{3} A \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )}{e^{2} i^{2} \left (a d -c b \right )^{3} g}-\frac {d^{2} B b \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}{2 e \,i^{2} \left (a d -c b \right )^{3} g}+\frac {d^{3} B \left (\left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right ) \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )-\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}-\frac {b e}{d}\right )}{e^{2} i^{2} \left (a d -c b \right )^{3} g}\right )}{d^{2}}\) \(286\)
risch \(-\frac {A}{g \,i^{2} \left (a d -c b \right ) \left (d x +c \right )}-\frac {A b \ln \left (d x +c \right )}{g \,i^{2} \left (a d -c b \right )^{2}}+\frac {A b \ln \left (b x +a \right )}{g \,i^{2} \left (a d -c b \right )^{2}}-\frac {B \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right ) b}{g \,i^{2} \left (a d -c b \right )^{2}}-\frac {B d \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right ) a}{g \,i^{2} \left (a d -c b \right )^{2} \left (d x +c \right )}+\frac {B \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right ) c b}{g \,i^{2} \left (a d -c b \right )^{2} \left (d x +c \right )}+\frac {B d a}{g \,i^{2} \left (a d -c b \right )^{2} \left (d x +c \right )}-\frac {B c b}{g \,i^{2} \left (a d -c b \right )^{2} \left (d x +c \right )}+\frac {B b}{g \,i^{2} \left (a d -c b \right )^{2}}+\frac {B b \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )^{2}}{2 g \,i^{2} \left (a d -c b \right )^{2}}\) \(362\)

input
int((A+B*ln(e*(b*x+a)/(d*x+c)))/(b*g*x+a*g)/(d*i*x+c*i)^2,x,method=_RETURN 
VERBOSE)
 
output
A/g/i^2*(-1/(a*d-b*c)/(d*x+c)-b/(a*d-b*c)^2*ln(d*x+c)+b/(a*d-b*c)^2*ln(b*x 
+a))-B/g/i^2/(a*d-b*c)/e*(d/(a*d-b*c)*((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)-1/2*b*e/(a*d-b*c)* 
ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))^2)
 
3.1.43.5 Fricas [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x) (c i+d i x)^2} \, dx=\frac {2 \, {\left (A - B\right )} b c - 2 \, {\left (A - B\right )} a d + {\left (B b d x + B b c\right )} \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, {\left ({\left (A - B\right )} b d x + A b c - B a d\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{2 \, {\left ({\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} g i^{2} x + {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} g i^{2}\right )}} \]

input
integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(b*g*x+a*g)/(d*i*x+c*i)^2,x, algori 
thm="fricas")
 
output
1/2*(2*(A - B)*b*c - 2*(A - B)*a*d + (B*b*d*x + B*b*c)*log((b*e*x + a*e)/( 
d*x + c))^2 + 2*((A - B)*b*d*x + A*b*c - B*a*d)*log((b*e*x + a*e)/(d*x + c 
)))/((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*g*i^2*x + (b^2*c^3 - 2*a*b*c^2*d 
+ a^2*c*d^2)*g*i^2)
 
3.1.43.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (131) = 262\).

Time = 0.60 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.47 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x) (c i+d i x)^2} \, dx=\frac {B b \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}^{2}}{2 a^{2} d^{2} g i^{2} - 4 a b c d g i^{2} + 2 b^{2} c^{2} g i^{2}} - \frac {B \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}}{a c d g i^{2} + a d^{2} g i^{2} x - b c^{2} g i^{2} - b c d g i^{2} x} + \left (A - B\right ) \left (- \frac {b \log {\left (x + \frac {- \frac {a^{3} b d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 a^{2} b^{2} c d^{2}}{\left (a d - b c\right )^{2}} - \frac {3 a b^{3} c^{2} d}{\left (a d - b c\right )^{2}} + a b d + \frac {b^{4} c^{3}}{\left (a d - b c\right )^{2}} + b^{2} c}{2 b^{2} d} \right )}}{g i^{2} \left (a d - b c\right )^{2}} + \frac {b \log {\left (x + \frac {\frac {a^{3} b d^{3}}{\left (a d - b c\right )^{2}} - \frac {3 a^{2} b^{2} c d^{2}}{\left (a d - b c\right )^{2}} + \frac {3 a b^{3} c^{2} d}{\left (a d - b c\right )^{2}} + a b d - \frac {b^{4} c^{3}}{\left (a d - b c\right )^{2}} + b^{2} c}{2 b^{2} d} \right )}}{g i^{2} \left (a d - b c\right )^{2}} - \frac {1}{a c d g i^{2} - b c^{2} g i^{2} + x \left (a d^{2} g i^{2} - b c d g i^{2}\right )}\right ) \]

input
integrate((A+B*ln(e*(b*x+a)/(d*x+c)))/(b*g*x+a*g)/(d*i*x+c*i)**2,x)
 
output
B*b*log(e*(a + b*x)/(c + d*x))**2/(2*a**2*d**2*g*i**2 - 4*a*b*c*d*g*i**2 + 
 2*b**2*c**2*g*i**2) - B*log(e*(a + b*x)/(c + d*x))/(a*c*d*g*i**2 + a*d**2 
*g*i**2*x - b*c**2*g*i**2 - b*c*d*g*i**2*x) + (A - B)*(-b*log(x + (-a**3*b 
*d**3/(a*d - b*c)**2 + 3*a**2*b**2*c*d**2/(a*d - b*c)**2 - 3*a*b**3*c**2*d 
/(a*d - b*c)**2 + a*b*d + b**4*c**3/(a*d - b*c)**2 + b**2*c)/(2*b**2*d))/( 
g*i**2*(a*d - b*c)**2) + b*log(x + (a**3*b*d**3/(a*d - b*c)**2 - 3*a**2*b* 
*2*c*d**2/(a*d - b*c)**2 + 3*a*b**3*c**2*d/(a*d - b*c)**2 + a*b*d - b**4*c 
**3/(a*d - b*c)**2 + b**2*c)/(2*b**2*d))/(g*i**2*(a*d - b*c)**2) - 1/(a*c* 
d*g*i**2 - b*c**2*g*i**2 + x*(a*d**2*g*i**2 - b*c*d*g*i**2)))
 
3.1.43.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 421 vs. \(2 (154) = 308\).

Time = 0.22 (sec) , antiderivative size = 421, normalized size of antiderivative = 2.70 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x) (c i+d i x)^2} \, dx=B {\left (\frac {1}{{\left (b c d - a d^{2}\right )} g i^{2} x + {\left (b c^{2} - a c d\right )} g i^{2}} + \frac {b \log \left (b x + a\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} g i^{2}} - \frac {b \log \left (d x + c\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} g i^{2}}\right )} \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) + A {\left (\frac {1}{{\left (b c d - a d^{2}\right )} g i^{2} x + {\left (b c^{2} - a c d\right )} g i^{2}} + \frac {b \log \left (b x + a\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} g i^{2}} - \frac {b \log \left (d x + c\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} g i^{2}}\right )} - \frac {{\left ({\left (b d x + b c\right )} \log \left (b x + a\right )^{2} + {\left (b d x + b c\right )} \log \left (d x + c\right )^{2} + 2 \, b c - 2 \, a d + 2 \, {\left (b d x + b c\right )} \log \left (b x + a\right ) - 2 \, {\left (b d x + b c + {\left (b d x + b c\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )\right )} B}{2 \, {\left (b^{2} c^{3} g i^{2} - 2 \, a b c^{2} d g i^{2} + a^{2} c d^{2} g i^{2} + {\left (b^{2} c^{2} d g i^{2} - 2 \, a b c d^{2} g i^{2} + a^{2} d^{3} g i^{2}\right )} x\right )}} \]

input
integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(b*g*x+a*g)/(d*i*x+c*i)^2,x, algori 
thm="maxima")
 
output
B*(1/((b*c*d - a*d^2)*g*i^2*x + (b*c^2 - a*c*d)*g*i^2) + b*log(b*x + a)/(( 
b^2*c^2 - 2*a*b*c*d + a^2*d^2)*g*i^2) - b*log(d*x + c)/((b^2*c^2 - 2*a*b*c 
*d + a^2*d^2)*g*i^2))*log(b*e*x/(d*x + c) + a*e/(d*x + c)) + A*(1/((b*c*d 
- a*d^2)*g*i^2*x + (b*c^2 - a*c*d)*g*i^2) + b*log(b*x + a)/((b^2*c^2 - 2*a 
*b*c*d + a^2*d^2)*g*i^2) - b*log(d*x + c)/((b^2*c^2 - 2*a*b*c*d + a^2*d^2) 
*g*i^2)) - 1/2*((b*d*x + b*c)*log(b*x + a)^2 + (b*d*x + b*c)*log(d*x + c)^ 
2 + 2*b*c - 2*a*d + 2*(b*d*x + b*c)*log(b*x + a) - 2*(b*d*x + b*c + (b*d*x 
 + b*c)*log(b*x + a))*log(d*x + c))*B/(b^2*c^3*g*i^2 - 2*a*b*c^2*d*g*i^2 + 
 a^2*c*d^2*g*i^2 + (b^2*c^2*d*g*i^2 - 2*a*b*c*d^2*g*i^2 + a^2*d^3*g*i^2)*x 
)
 
3.1.43.8 Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.49 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x) (c i+d i x)^2} \, dx=\frac {1}{2} \, {\left (\frac {B b e \log \left (\frac {b e x + a e}{d x + c}\right )^{2}}{b c g i^{2} - a d g i^{2}} + \frac {2 \, A b e \log \left (\frac {b e x + a e}{d x + c}\right )}{b c g i^{2} - a d g i^{2}} - \frac {2 \, {\left (b e x + a e\right )} B d \log \left (\frac {b e x + a e}{d x + c}\right )}{{\left (b c g i^{2} - a d g i^{2}\right )} {\left (d x + c\right )}} - \frac {2 \, {\left (b e x + a e\right )} {\left (A d - B d\right )}}{{\left (b c g i^{2} - a d g i^{2}\right )} {\left (d x + c\right )}}\right )} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )} \]

input
integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(b*g*x+a*g)/(d*i*x+c*i)^2,x, algori 
thm="giac")
 
output
1/2*(B*b*e*log((b*e*x + a*e)/(d*x + c))^2/(b*c*g*i^2 - a*d*g*i^2) + 2*A*b* 
e*log((b*e*x + a*e)/(d*x + c))/(b*c*g*i^2 - a*d*g*i^2) - 2*(b*e*x + a*e)*B 
*d*log((b*e*x + a*e)/(d*x + c))/((b*c*g*i^2 - a*d*g*i^2)*(d*x + c)) - 2*(b 
*e*x + a*e)*(A*d - B*d)/((b*c*g*i^2 - a*d*g*i^2)*(d*x + c)))*(b*c/((b*c*e 
- a*d*e)*(b*c - a*d)) - a*d/((b*c*e - a*d*e)*(b*c - a*d)))
 
3.1.43.9 Mupad [B] (verification not implemented)

Time = 2.50 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.58 \[ \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a g+b g x) (c i+d i x)^2} \, dx=\frac {B\,b\,{\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}^2}{2\,g\,i^2\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}-\frac {A-B}{\left (a\,d-b\,c\right )\,\left (c\,g\,i^2+d\,g\,i^2\,x\right )}-\frac {B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\,\left (a\,d-b\,c\right )}{b\,d\,g\,i^2\,\left (\frac {x}{b}+\frac {c}{b\,d}\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}-\frac {b\,\mathrm {atan}\left (\frac {\left (2\,b\,d\,x+\frac {a^2\,d^2\,g\,i^2-b^2\,c^2\,g\,i^2}{g\,i^2\,\left (a\,d-b\,c\right )}\right )\,1{}\mathrm {i}}{a\,d-b\,c}\right )\,\left (A-B\right )\,2{}\mathrm {i}}{g\,i^2\,{\left (a\,d-b\,c\right )}^2} \]

input
int((A + B*log((e*(a + b*x))/(c + d*x)))/((a*g + b*g*x)*(c*i + d*i*x)^2),x 
)
 
output
(B*b*log((e*(a + b*x))/(c + d*x))^2)/(2*g*i^2*(a^2*d^2 + b^2*c^2 - 2*a*b*c 
*d)) - (b*atan(((2*b*d*x + (a^2*d^2*g*i^2 - b^2*c^2*g*i^2)/(g*i^2*(a*d - b 
*c)))*1i)/(a*d - b*c))*(A - B)*2i)/(g*i^2*(a*d - b*c)^2) - (A - B)/((a*d - 
 b*c)*(c*g*i^2 + d*g*i^2*x)) - (B*log((e*(a + b*x))/(c + d*x))*(a*d - b*c) 
)/(b*d*g*i^2*(x/b + c/(b*d))*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))