\(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(f+g x)^3} \, dx\) [64]

Optimal result

Integrand size = 30, antiderivative size = 190 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^3} \, dx=-\frac {B (b c-a d) n}{2 (b f-a g) (d f-c g) (f+g x)}+\frac {b^2 B n \log (a+b x)}{2 g (b f-a g)^2}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 g (f+g x)^2}-\frac {B d^2 n \log (c+d x)}{2 g (d f-c g)^2}+\frac {B (b c-a d) (2 b d f-b c g-a d g) n \log (f+g x)}{2 (b f-a g)^2 (d f-c g)^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^3} \, dx=\frac {-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^2}+B (b c-a d) n \left (\frac {b^2 \log (a+b x)}{(b c-a d) (b f-a g)^2}+\frac {\frac {g (-d f+c g)}{(b f-a g) (f+g x)}+\frac {d^2 \log (c+d x)}{-b c+a d}-\frac {g (-2 b d f+b c g+a d g) \log (f+g x)}{(b f-a g)^2}}{(d f-c g)^2}\right )}{2 g} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2947, 93, 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.

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

\(\Big \downarrow \) 2947

\(\displaystyle \frac {B n (b c-a d) \int \frac {1}{(a+b x) (c+d x) (f+g x)^2}dx}{2 g}-\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{2 g (f+g x)^2}\)

\(\Big \downarrow \) 93

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 93

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre 
eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
 

rule 2009

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

rule 2947

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*((A + 
 B*Log[e*((a + b*x)/(c + d*x))^n])/(g*(m + 1))), x] - Simp[B*n*((b*c - a*d) 
/(g*(m + 1)))   Int[(f + g*x)^(m + 1)/((a + b*x)*(c + d*x)), x], x] /; Free 
Q[{a, b, c, d, e, f, g, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] 
&& NeQ[m, -2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1375\) vs. \(2(180)=360\).

Time = 9.43 (sec) , antiderivative size = 1376, normalized size of antiderivative = 7.24

method	result	size
parallelrisch	\(\text {Expression too large to display}\)	\(1376\)

Input:

int((A+B*ln(e*((b*x+a)/(d*x+c))^n))/(g*x+f)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1175 vs. \(2 (180) = 360\).

Time = 41.22 (sec) , antiderivative size = 1175, normalized size of antiderivative = 6.18 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.87 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^3} \, dx=\frac {1}{2} \, {\left (\frac {b^{2} \log \left (b x + a\right )}{b^{2} f^{2} g - 2 \, a b f g^{2} + a^{2} g^{3}} - \frac {d^{2} \log \left (d x + c\right )}{d^{2} f^{2} g - 2 \, c d f g^{2} + c^{2} g^{3}} + \frac {{\left (2 \, {\left (b^{2} c d - a b d^{2}\right )} f - {\left (b^{2} c^{2} - a^{2} d^{2}\right )} g\right )} \log \left (g x + f\right )}{b^{2} d^{2} f^{4} + a^{2} c^{2} g^{4} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} f^{3} g + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} f^{2} g^{2} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} f g^{3}} - \frac {b c - a d}{b d f^{3} + a c f g^{2} - {\left (b c + a d\right )} f^{2} g + {\left (b d f^{2} g + a c g^{3} - {\left (b c + a d\right )} f g^{2}\right )} x}\right )} B n - \frac {B \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{2 \, {\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} - \frac {A}{2 \, {\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2994 vs. \(2 (180) = 360\).

Time = 0.61 (sec) , antiderivative size = 2994, normalized size of antiderivative = 15.76 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*((2*B*b^3*c^2*d*f*n - 4*B*a*b^2*c*d^2*f*n + 2*B*a^2*b*d^3*f*n - B*b^3* 
c^3*g*n + B*a*b^2*c^2*d*g*n + B*a^2*b*c*d^2*g*n - B*a^3*d^3*g*n)*log(-b*f 
+ (b*x + a)*d*f/(d*x + c) + a*g - (b*x + a)*c*g/(d*x + c))/(b^2*d^2*f^4 - 
2*b^2*c*d*f^3*g - 2*a*b*d^2*f^3*g + b^2*c^2*f^2*g^2 + 4*a*b*c*d*f^2*g^2 + 
a^2*d^2*f^2*g^2 - 2*a*b*c^2*f*g^3 - 2*a^2*c*d*f*g^3 + a^2*c^2*g^4) + (2*B* 
b^3*c^2*d*f*n - 4*B*a*b^2*c*d^2*f*n - 2*(b*x + a)*B*b^2*c^2*d^2*f*n/(d*x + 
 c) + 2*B*a^2*b*d^3*f*n + 4*(b*x + a)*B*a*b*c*d^3*f*n/(d*x + c) - 2*(b*x + 
 a)*B*a^2*d^4*f*n/(d*x + c) - B*b^3*c^3*g*n + B*a*b^2*c^2*d*g*n + 2*(b*x + 
 a)*B*b^2*c^3*d*g*n/(d*x + c) + B*a^2*b*c*d^2*g*n - 4*(b*x + a)*B*a*b*c^2* 
d^2*g*n/(d*x + c) - B*a^3*d^3*g*n + 2*(b*x + a)*B*a^2*c*d^3*g*n/(d*x + c)) 
*log((b*x + a)/(d*x + c))/(b^2*d^2*f^4 - 2*(b*x + a)*b*d^3*f^4/(d*x + c) + 
 (b*x + a)^2*d^4*f^4/(d*x + c)^2 - 2*b^2*c*d*f^3*g - 2*a*b*d^2*f^3*g + 6*( 
b*x + a)*b*c*d^2*f^3*g/(d*x + c) + 2*(b*x + a)*a*d^3*f^3*g/(d*x + c) - 4*( 
b*x + a)^2*c*d^3*f^3*g/(d*x + c)^2 + b^2*c^2*f^2*g^2 + 4*a*b*c*d*f^2*g^2 - 
 6*(b*x + a)*b*c^2*d*f^2*g^2/(d*x + c) + a^2*d^2*f^2*g^2 - 6*(b*x + a)*a*c 
*d^2*f^2*g^2/(d*x + c) + 6*(b*x + a)^2*c^2*d^2*f^2*g^2/(d*x + c)^2 - 2*a*b 
*c^2*f*g^3 + 2*(b*x + a)*b*c^3*f*g^3/(d*x + c) - 2*a^2*c*d*f*g^3 + 6*(b*x 
+ a)*a*c^2*d*f*g^3/(d*x + c) - 4*(b*x + a)^2*c^3*d*f*g^3/(d*x + c)^2 + a^2 
*c^2*g^4 - 2*(b*x + a)*a*c^3*g^4/(d*x + c) + (b*x + a)^2*c^4*g^4/(d*x + c) 
^2) - (2*B*b^3*c^2*d*f*n - 4*B*a*b^2*c*d^2*f*n + 2*B*a^2*b*d^3*f*n - B*...
 

Mupad [B] (verification not implemented)

Time = 27.97 (sec) , antiderivative size = 430, normalized size of antiderivative = 2.26 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^3} \, dx=\frac {\ln \left (f+g\,x\right )\,\left (g\,\left (B\,a^2\,d^2\,n-B\,b^2\,c^2\,n\right )-2\,B\,a\,b\,d^2\,f\,n+2\,B\,b^2\,c\,d\,f\,n\right )}{2\,a^2\,c^2\,g^4-4\,a^2\,c\,d\,f\,g^3+2\,a^2\,d^2\,f^2\,g^2-4\,a\,b\,c^2\,f\,g^3+8\,a\,b\,c\,d\,f^2\,g^2-4\,a\,b\,d^2\,f^3\,g+2\,b^2\,c^2\,f^2\,g^2-4\,b^2\,c\,d\,f^3\,g+2\,b^2\,d^2\,f^4}-\frac {\frac {A\,a\,c\,g^2+A\,b\,d\,f^2-A\,a\,d\,f\,g-A\,b\,c\,f\,g-B\,a\,d\,f\,g\,n+B\,b\,c\,f\,g\,n}{a\,c\,g^2+b\,d\,f^2-a\,d\,f\,g-b\,c\,f\,g}-\frac {x\,\left (B\,a\,d\,g^2\,n-B\,b\,c\,g^2\,n\right )}{a\,c\,g^2+b\,d\,f^2-a\,d\,f\,g-b\,c\,f\,g}}{2\,f^2\,g+4\,f\,g^2\,x+2\,g^3\,x^2}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{2\,g\,\left (f^2+2\,f\,g\,x+g^2\,x^2\right )}+\frac {B\,b^2\,n\,\ln \left (a+b\,x\right )}{2\,a^2\,g^3-4\,a\,b\,f\,g^2+2\,b^2\,f^2\,g}-\frac {B\,d^2\,n\,\ln \left (c+d\,x\right )}{2\,c^2\,g^3-4\,c\,d\,f\,g^2+2\,d^2\,f^2\,g} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 2381, normalized size of antiderivative = 12.53 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*log(a + b*x)*a**2*b*c**2*f**2*g**4*n - 4*log(a + b*x)*a**2*b*c**2*f* 
g**5*n*x - 2*log(a + b*x)*a**2*b*c**2*g**6*n*x**2 + 4*log(a + b*x)*a**2*b* 
c*d*f**3*g**3*n + 8*log(a + b*x)*a**2*b*c*d*f**2*g**4*n*x + 4*log(a + b*x) 
*a**2*b*c*d*f*g**5*n*x**2 - 2*log(a + b*x)*a**2*b*d**2*f**4*g**2*n - 4*log 
(a + b*x)*a**2*b*d**2*f**3*g**3*n*x - 2*log(a + b*x)*a**2*b*d**2*f**2*g**4 
*n*x**2 + 4*log(a + b*x)*a*b**2*c**2*f**3*g**3*n + 8*log(a + b*x)*a*b**2*c 
**2*f**2*g**4*n*x + 4*log(a + b*x)*a*b**2*c**2*f*g**5*n*x**2 - 8*log(a + b 
*x)*a*b**2*c*d*f**4*g**2*n - 16*log(a + b*x)*a*b**2*c*d*f**3*g**3*n*x - 8* 
log(a + b*x)*a*b**2*c*d*f**2*g**4*n*x**2 + 4*log(a + b*x)*a*b**2*d**2*f**5 
*g*n + 8*log(a + b*x)*a*b**2*d**2*f**4*g**2*n*x + 4*log(a + b*x)*a*b**2*d* 
*2*f**3*g**3*n*x**2 + 2*log(c + d*x)*a**2*b*c**2*f**2*g**4*n + 4*log(c + d 
*x)*a**2*b*c**2*f*g**5*n*x + 2*log(c + d*x)*a**2*b*c**2*g**6*n*x**2 - 4*lo 
g(c + d*x)*a**2*b*c*d*f**3*g**3*n - 8*log(c + d*x)*a**2*b*c*d*f**2*g**4*n* 
x - 4*log(c + d*x)*a**2*b*c*d*f*g**5*n*x**2 - 4*log(c + d*x)*a*b**2*c**2*f 
**3*g**3*n - 8*log(c + d*x)*a*b**2*c**2*f**2*g**4*n*x - 4*log(c + d*x)*a*b 
**2*c**2*f*g**5*n*x**2 + 8*log(c + d*x)*a*b**2*c*d*f**4*g**2*n + 16*log(c 
+ d*x)*a*b**2*c*d*f**3*g**3*n*x + 8*log(c + d*x)*a*b**2*c*d*f**2*g**4*n*x* 
*2 + 2*log(c + d*x)*b**3*c**2*f**4*g**2*n + 4*log(c + d*x)*b**3*c**2*f**3* 
g**3*n*x + 2*log(c + d*x)*b**3*c**2*f**2*g**4*n*x**2 - 4*log(c + d*x)*b**3 
*c*d*f**5*g*n - 8*log(c + d*x)*b**3*c*d*f**4*g**2*n*x - 4*log(c + d*x)*...