3.1.65 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(f+g x)^4} \, dx\) [65]

3.1.65.1 Optimal result
3.1.65.2 Mathematica [A] (verified)
3.1.65.3 Rubi [A] (verified)
3.1.65.4 Maple [B] (verified)
3.1.65.5 Fricas [F(-1)]
3.1.65.6 Sympy [F(-1)]
3.1.65.7 Maxima [B] (verification not implemented)
3.1.65.8 Giac [B] (verification not implemented)
3.1.65.9 Mupad [B] (verification not implemented)

3.1.65.1 Optimal result

Integrand size = 30, antiderivative size = 283 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^4} \, dx=-\frac {B (b c-a d) n}{6 (b f-a g) (d f-c g) (f+g x)^2}-\frac {B (b c-a d) (2 b d f-b c g-a d g) n}{3 (b f-a g)^2 (d f-c g)^2 (f+g x)}+\frac {b^3 B n \log (a+b x)}{3 g (b f-a g)^3}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{3 g (f+g x)^3}-\frac {B d^3 n \log (c+d x)}{3 g (d f-c g)^3}+\frac {B (b c-a d) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right ) n \log (f+g x)}{3 (b f-a g)^3 (d f-c g)^3} \]

output
-1/6*B*(-a*d+b*c)*n/(-a*g+b*f)/(-c*g+d*f)/(g*x+f)^2-1/3*B*(-a*d+b*c)*(-a*d 
*g-b*c*g+2*b*d*f)*n/(-a*g+b*f)^2/(-c*g+d*f)^2/(g*x+f)+1/3*b^3*B*n*ln(b*x+a 
)/g/(-a*g+b*f)^3+1/3*(-A-B*ln(e*((b*x+a)/(d*x+c))^n))/g/(g*x+f)^3-1/3*B*d^ 
3*n*ln(d*x+c)/g/(-c*g+d*f)^3+1/3*B*(-a*d+b*c)*(a^2*d^2*g^2-a*b*d*g*(-c*g+3 
*d*f)+b^2*(c^2*g^2-3*c*d*f*g+3*d^2*f^2))*n*ln(g*x+f)/(-a*g+b*f)^3/(-c*g+d* 
f)^3
 
3.1.65.2 Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.93 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^4} \, dx=\frac {-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^3}+B (b c-a d) n \left (-\frac {g}{2 (b f-a g) (d f-c g) (f+g x)^2}+\frac {g (-2 b d f+b c g+a d g)}{(b f-a g)^2 (d f-c g)^2 (f+g x)}+\frac {b^3 \log (a+b x)}{(b c-a d) (b f-a g)^3}+\frac {d^3 \log (c+d x)}{(b c-a d) (-d f+c g)^3}+\frac {g \left (a^2 d^2 g^2+a b d g (-3 d f+c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right ) \log (f+g x)}{(b f-a g)^3 (d f-c g)^3}\right )}{3 g} \]

input
Integrate[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(f + g*x)^4,x]
 
output
(-((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(f + g*x)^3) + B*(b*c - a*d)*n*( 
-1/2*g/((b*f - a*g)*(d*f - c*g)*(f + g*x)^2) + (g*(-2*b*d*f + b*c*g + a*d* 
g))/((b*f - a*g)^2*(d*f - c*g)^2*(f + g*x)) + (b^3*Log[a + b*x])/((b*c - a 
*d)*(b*f - a*g)^3) + (d^3*Log[c + d*x])/((b*c - a*d)*(-(d*f) + c*g)^3) + ( 
g*(a^2*d^2*g^2 + a*b*d*g*(-3*d*f + c*g) + b^2*(3*d^2*f^2 - 3*c*d*f*g + c^2 
*g^2))*Log[f + g*x])/((b*f - a*g)^3*(d*f - c*g)^3)))/(3*g)
 
3.1.65.3 Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.97, 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.

\(\displaystyle \int \frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{(f+g x)^4} \, dx\)

\(\Big \downarrow \) 2947

\(\displaystyle \frac {B n (b c-a d) \int \frac {1}{(a+b x) (c+d x) (f+g x)^3}dx}{3 g}-\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{3 g (f+g x)^3}\)

\(\Big \downarrow \) 93

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

\(\Big \downarrow \) 2009

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

input
Int[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(f + g*x)^4,x]
 
output
-1/3*(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(g*(f + g*x)^3) + (B*(b*c - a* 
d)*n*(-1/2*g/((b*f - a*g)*(d*f - c*g)*(f + g*x)^2) - (g*(2*b*d*f - b*c*g - 
 a*d*g))/((b*f - a*g)^2*(d*f - c*g)^2*(f + g*x)) + (b^3*Log[a + b*x])/((b* 
c - a*d)*(b*f - a*g)^3) - (d^3*Log[c + d*x])/((b*c - a*d)*(d*f - c*g)^3) + 
 (g*(a^2*d^2*g^2 - a*b*d*g*(3*d*f - c*g) + b^2*(3*d^2*f^2 - 3*c*d*f*g + c^ 
2*g^2))*Log[f + g*x])/((b*f - a*g)^3*(d*f - c*g)^3)))/(3*g)
 

3.1.65.3.1 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]
 
3.1.65.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3256\) vs. \(2(274)=548\).

Time = 41.68 (sec) , antiderivative size = 3257, normalized size of antiderivative = 11.51

method result size
parallelrisch \(\text {Expression too large to display}\) \(3257\)

input
int((A+B*ln(e*((b*x+a)/(d*x+c))^n))/(g*x+f)^4,x,method=_RETURNVERBOSE)
 
output
-1/6*(18*B*ln(e*((b*x+a)/(d*x+c))^n)*a*b^3*c*d^3*f^4*g^4*n+2*B*ln(g*x+f)*a 
^3*b*d^4*f^3*g^5*n^2-6*B*ln(g*x+f)*a^2*b^2*d^4*f^4*g^4*n^2+6*B*ln(g*x+f)*a 
*b^3*d^4*f^5*g^3*n^2-2*B*ln(g*x+f)*b^4*c^3*d*f^3*g^5*n^2+6*B*ln(g*x+f)*b^4 
*c^2*d^2*f^4*g^4*n^2-6*B*ln(g*x+f)*b^4*c*d^3*f^5*g^3*n^2-2*B*ln(b*x+a)*x^3 
*a^3*b*d^4*g^8*n^2+2*B*ln(b*x+a)*x^3*b^4*c^3*d*g^8*n^2+2*B*ln(g*x+f)*x^3*a 
^3*b*d^4*g^8*n^2-2*B*ln(g*x+f)*x^3*b^4*c^3*d*g^8*n^2-3*B*a^3*b*d^4*f^3*g^5 
*n^2+8*B*a^2*b^2*d^4*f^4*g^4*n^2-5*B*a*b^3*d^4*f^5*g^3*n^2+3*B*b^4*c^3*d*f 
^3*g^5*n^2-8*B*b^4*c^2*d^2*f^4*g^4*n^2+5*B*b^4*c*d^3*f^5*g^3*n^2+2*A*a^3*b 
*c^3*d*g^8*n-2*A*a^3*b*d^4*f^3*g^5*n+6*A*a^2*b^2*d^4*f^4*g^4*n-6*A*a*b^3*d 
^4*f^5*g^3*n-2*A*b^4*c^3*d*f^3*g^5*n+6*A*b^4*c^2*d^2*f^4*g^4*n-6*A*b^4*c*d 
^3*f^5*g^3*n+2*B*ln(e*((b*x+a)/(d*x+c))^n)*a^3*b*c^3*d*g^8*n-2*B*ln(e*((b* 
x+a)/(d*x+c))^n)*b^4*c^3*d*f^3*g^5*n+6*B*ln(e*((b*x+a)/(d*x+c))^n)*b^4*c^2 
*d^2*f^4*g^4*n-6*B*ln(e*((b*x+a)/(d*x+c))^n)*b^4*c*d^3*f^5*g^3*n-2*B*ln(b* 
x+a)*a^3*b*d^4*f^3*g^5*n^2+6*B*ln(b*x+a)*a^2*b^2*d^4*f^4*g^4*n^2-6*B*ln(b* 
x+a)*a*b^3*d^4*f^5*g^3*n^2+2*B*ln(b*x+a)*b^4*c^3*d*f^3*g^5*n^2-6*B*ln(b*x+ 
a)*b^4*c^2*d^2*f^4*g^4*n^2+6*B*ln(b*x+a)*b^4*c*d^3*f^5*g^3*n^2-6*B*ln(b*x+ 
a)*x^2*a^3*b*d^4*f*g^7*n^2+18*B*ln(b*x+a)*x^2*a^2*b^2*d^4*f^2*g^6*n^2-18*B 
*ln(b*x+a)*x^2*a*b^3*d^4*f^3*g^5*n^2+6*B*ln(b*x+a)*x^2*b^4*c^3*d*f*g^7*n^2 
-18*B*ln(b*x+a)*x^2*b^4*c^2*d^2*f^2*g^6*n^2+18*B*ln(b*x+a)*x^2*b^4*c*d^3*f 
^3*g^5*n^2+18*B*ln(g*x+f)*x^2*b^4*c^2*d^2*f^2*g^6*n^2-18*B*ln(g*x+f)*x^...
 
3.1.65.5 Fricas [F(-1)]

Timed out. \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^4} \, dx=\text {Timed out} \]

input
integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(g*x+f)^4,x, algorithm="fricas" 
)
 
output
Timed out
 
3.1.65.6 Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^4} \, dx=\text {Timed out} \]

input
integrate((A+B*ln(e*((b*x+a)/(d*x+c))**n))/(g*x+f)**4,x)
 
output
Timed out
 
3.1.65.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 852 vs. \(2 (271) = 542\).

Time = 0.24 (sec) , antiderivative size = 852, normalized size of antiderivative = 3.01 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^4} \, dx=\frac {1}{6} \, {\left (\frac {2 \, b^{3} \log \left (b x + a\right )}{b^{3} f^{3} g - 3 \, a b^{2} f^{2} g^{2} + 3 \, a^{2} b f g^{3} - a^{3} g^{4}} - \frac {2 \, d^{3} \log \left (d x + c\right )}{d^{3} f^{3} g - 3 \, c d^{2} f^{2} g^{2} + 3 \, c^{2} d f g^{3} - c^{3} g^{4}} + \frac {2 \, {\left (3 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} f^{2} - 3 \, {\left (b^{3} c^{2} d - a^{2} b d^{3}\right )} f g + {\left (b^{3} c^{3} - a^{3} d^{3}\right )} g^{2}\right )} \log \left (g x + f\right )}{b^{3} d^{3} f^{6} + a^{3} c^{3} g^{6} - 3 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} f^{5} g + 3 \, {\left (b^{3} c^{2} d + 3 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} f^{4} g^{2} - {\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} f^{3} g^{3} + 3 \, {\left (a b^{2} c^{3} + 3 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} f^{2} g^{4} - 3 \, {\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} f g^{5}} - \frac {5 \, {\left (b^{2} c d - a b d^{2}\right )} f^{2} - 3 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} f g + {\left (a b c^{2} - a^{2} c d\right )} g^{2} + 2 \, {\left (2 \, {\left (b^{2} c d - a b d^{2}\right )} f g - {\left (b^{2} c^{2} - a^{2} d^{2}\right )} g^{2}\right )} x}{b^{2} d^{2} f^{6} + a^{2} c^{2} f^{2} g^{4} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} f^{5} g + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} f^{4} g^{2} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} f^{3} g^{3} + {\left (b^{2} d^{2} f^{4} g^{2} + a^{2} c^{2} g^{6} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} f^{3} g^{3} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} f^{2} g^{4} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} f g^{5}\right )} x^{2} + 2 \, {\left (b^{2} d^{2} f^{5} g + a^{2} c^{2} f g^{5} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} f^{4} g^{2} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} f^{3} g^{3} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} f^{2} g^{4}\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 )}{3 \, {\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g\right )}} - \frac {A}{3 \, {\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g\right )}} \]

input
integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(g*x+f)^4,x, algorithm="maxima" 
)
 
output
1/6*(2*b^3*log(b*x + a)/(b^3*f^3*g - 3*a*b^2*f^2*g^2 + 3*a^2*b*f*g^3 - a^3 
*g^4) - 2*d^3*log(d*x + c)/(d^3*f^3*g - 3*c*d^2*f^2*g^2 + 3*c^2*d*f*g^3 - 
c^3*g^4) + 2*(3*(b^3*c*d^2 - a*b^2*d^3)*f^2 - 3*(b^3*c^2*d - a^2*b*d^3)*f* 
g + (b^3*c^3 - a^3*d^3)*g^2)*log(g*x + f)/(b^3*d^3*f^6 + a^3*c^3*g^6 - 3*( 
b^3*c*d^2 + a*b^2*d^3)*f^5*g + 3*(b^3*c^2*d + 3*a*b^2*c*d^2 + a^2*b*d^3)*f 
^4*g^2 - (b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*f^3*g^3 + 3*( 
a*b^2*c^3 + 3*a^2*b*c^2*d + a^3*c*d^2)*f^2*g^4 - 3*(a^2*b*c^3 + a^3*c^2*d) 
*f*g^5) - (5*(b^2*c*d - a*b*d^2)*f^2 - 3*(b^2*c^2 - a^2*d^2)*f*g + (a*b*c^ 
2 - a^2*c*d)*g^2 + 2*(2*(b^2*c*d - a*b*d^2)*f*g - (b^2*c^2 - a^2*d^2)*g^2) 
*x)/(b^2*d^2*f^6 + a^2*c^2*f^2*g^4 - 2*(b^2*c*d + a*b*d^2)*f^5*g + (b^2*c^ 
2 + 4*a*b*c*d + a^2*d^2)*f^4*g^2 - 2*(a*b*c^2 + a^2*c*d)*f^3*g^3 + (b^2*d^ 
2*f^4*g^2 + a^2*c^2*g^6 - 2*(b^2*c*d + a*b*d^2)*f^3*g^3 + (b^2*c^2 + 4*a*b 
*c*d + a^2*d^2)*f^2*g^4 - 2*(a*b*c^2 + a^2*c*d)*f*g^5)*x^2 + 2*(b^2*d^2*f^ 
5*g + a^2*c^2*f*g^5 - 2*(b^2*c*d + a*b*d^2)*f^4*g^2 + (b^2*c^2 + 4*a*b*c*d 
 + a^2*d^2)*f^3*g^3 - 2*(a*b*c^2 + a^2*c*d)*f^2*g^4)*x))*B*n - 1/3*B*log(e 
*(b*x/(d*x + c) + a/(d*x + c))^n)/(g^4*x^3 + 3*f*g^3*x^2 + 3*f^2*g^2*x + f 
^3*g) - 1/3*A/(g^4*x^3 + 3*f*g^3*x^2 + 3*f^2*g^2*x + f^3*g)
 
3.1.65.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 9692 vs. \(2 (271) = 542\).

Time = 0.85 (sec) , antiderivative size = 9692, normalized size of antiderivative = 34.25 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^4} \, dx=\text {Too large to display} \]

input
integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(g*x+f)^4,x, algorithm="giac")
 
output
1/6*(2*(3*B*b^4*c^2*d^2*f^2*n - 6*B*a*b^3*c*d^3*f^2*n + 3*B*a^2*b^2*d^4*f^ 
2*n - 3*B*b^4*c^3*d*f*g*n + 3*B*a*b^3*c^2*d^2*f*g*n + 3*B*a^2*b^2*c*d^3*f* 
g*n - 3*B*a^3*b*d^4*f*g*n + B*b^4*c^4*g^2*n - B*a*b^3*c^3*d*g^2*n - B*a^3* 
b*c*d^3*g^2*n + B*a^4*d^4*g^2*n)*log(-b*f + (b*x + a)*d*f/(d*x + c) + a*g 
- (b*x + a)*c*g/(d*x + c))/(b^3*d^3*f^6 - 3*b^3*c*d^2*f^5*g - 3*a*b^2*d^3* 
f^5*g + 3*b^3*c^2*d*f^4*g^2 + 9*a*b^2*c*d^2*f^4*g^2 + 3*a^2*b*d^3*f^4*g^2 
- b^3*c^3*f^3*g^3 - 9*a*b^2*c^2*d*f^3*g^3 - 9*a^2*b*c*d^2*f^3*g^3 - a^3*d^ 
3*f^3*g^3 + 3*a*b^2*c^3*f^2*g^4 + 9*a^2*b*c^2*d*f^2*g^4 + 3*a^3*c*d^2*f^2* 
g^4 - 3*a^2*b*c^3*f*g^5 - 3*a^3*c^2*d*f*g^5 + a^3*c^3*g^6) + 2*(3*B*b^4*c^ 
2*d^2*f^2*n - 6*B*a*b^3*c*d^3*f^2*n - 6*(b*x + a)*B*b^3*c^2*d^3*f^2*n/(d*x 
 + c) + 3*B*a^2*b^2*d^4*f^2*n + 12*(b*x + a)*B*a*b^2*c*d^4*f^2*n/(d*x + c) 
 + 3*(b*x + a)^2*B*b^2*c^2*d^4*f^2*n/(d*x + c)^2 - 6*(b*x + a)*B*a^2*b*d^5 
*f^2*n/(d*x + c) - 6*(b*x + a)^2*B*a*b*c*d^5*f^2*n/(d*x + c)^2 + 3*(b*x + 
a)^2*B*a^2*d^6*f^2*n/(d*x + c)^2 - 3*B*b^4*c^3*d*f*g*n + 3*B*a*b^3*c^2*d^2 
*f*g*n + 9*(b*x + a)*B*b^3*c^3*d^2*f*g*n/(d*x + c) + 3*B*a^2*b^2*c*d^3*f*g 
*n - 15*(b*x + a)*B*a*b^2*c^2*d^3*f*g*n/(d*x + c) - 6*(b*x + a)^2*B*b^2*c^ 
3*d^3*f*g*n/(d*x + c)^2 - 3*B*a^3*b*d^4*f*g*n + 3*(b*x + a)*B*a^2*b*c*d^4* 
f*g*n/(d*x + c) + 12*(b*x + a)^2*B*a*b*c^2*d^4*f*g*n/(d*x + c)^2 + 3*(b*x 
+ a)*B*a^3*d^5*f*g*n/(d*x + c) - 6*(b*x + a)^2*B*a^2*c*d^5*f*g*n/(d*x + c) 
^2 + B*b^4*c^4*g^2*n - B*a*b^3*c^3*d*g^2*n - 3*(b*x + a)*B*b^3*c^4*d*g^...
 
3.1.65.9 Mupad [B] (verification not implemented)

Time = 5.89 (sec) , antiderivative size = 1182, normalized size of antiderivative = 4.18 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^4} \, dx=\frac {B\,d^3\,n\,\ln \left (c+d\,x\right )}{3\,c^3\,g^4-9\,c^2\,d\,f\,g^3+9\,c\,d^2\,f^2\,g^2-3\,d^3\,f^3\,g}-\frac {\ln \left (f+g\,x\right )\,\left (g^2\,\left (B\,a^3\,d^3\,n-B\,b^3\,c^3\,n\right )-g\,\left (3\,B\,a^2\,b\,d^3\,f\,n-3\,B\,b^3\,c^2\,d\,f\,n\right )+3\,B\,a\,b^2\,d^3\,f^2\,n-3\,B\,b^3\,c\,d^2\,f^2\,n\right )}{3\,a^3\,c^3\,g^6-9\,a^3\,c^2\,d\,f\,g^5+9\,a^3\,c\,d^2\,f^2\,g^4-3\,a^3\,d^3\,f^3\,g^3-9\,a^2\,b\,c^3\,f\,g^5+27\,a^2\,b\,c^2\,d\,f^2\,g^4-27\,a^2\,b\,c\,d^2\,f^3\,g^3+9\,a^2\,b\,d^3\,f^4\,g^2+9\,a\,b^2\,c^3\,f^2\,g^4-27\,a\,b^2\,c^2\,d\,f^3\,g^3+27\,a\,b^2\,c\,d^2\,f^4\,g^2-9\,a\,b^2\,d^3\,f^5\,g-3\,b^3\,c^3\,f^3\,g^3+9\,b^3\,c^2\,d\,f^4\,g^2-9\,b^3\,c\,d^2\,f^5\,g+3\,b^3\,d^3\,f^6}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{3\,g\,\left (f^3+3\,f^2\,g\,x+3\,f\,g^2\,x^2+g^3\,x^3\right )}-\frac {B\,b^3\,n\,\ln \left (a+b\,x\right )}{3\,a^3\,g^4-9\,a^2\,b\,f\,g^3+9\,a\,b^2\,f^2\,g^2-3\,b^3\,f^3\,g}-\frac {\frac {2\,A\,a^2\,c^2\,g^4+2\,A\,b^2\,d^2\,f^4+2\,A\,a^2\,d^2\,f^2\,g^2+2\,A\,b^2\,c^2\,f^2\,g^2+3\,B\,a^2\,d^2\,f^2\,g^2\,n-3\,B\,b^2\,c^2\,f^2\,g^2\,n-4\,A\,a\,b\,c^2\,f\,g^3-4\,A\,a\,b\,d^2\,f^3\,g-4\,A\,a^2\,c\,d\,f\,g^3-4\,A\,b^2\,c\,d\,f^3\,g+8\,A\,a\,b\,c\,d\,f^2\,g^2+B\,a\,b\,c^2\,f\,g^3\,n-5\,B\,a\,b\,d^2\,f^3\,g\,n-B\,a^2\,c\,d\,f\,g^3\,n+5\,B\,b^2\,c\,d\,f^3\,g\,n}{2\,\left (a^2\,c^2\,g^4-2\,a^2\,c\,d\,f\,g^3+a^2\,d^2\,f^2\,g^2-2\,a\,b\,c^2\,f\,g^3+4\,a\,b\,c\,d\,f^2\,g^2-2\,a\,b\,d^2\,f^3\,g+b^2\,c^2\,f^2\,g^2-2\,b^2\,c\,d\,f^3\,g+b^2\,d^2\,f^4\right )}+\frac {x\,\left (-B\,n\,a^2\,c\,d\,g^4+5\,B\,n\,a^2\,d^2\,f\,g^3+B\,n\,a\,b\,c^2\,g^4-9\,B\,n\,a\,b\,d^2\,f^2\,g^2-5\,B\,n\,b^2\,c^2\,f\,g^3+9\,B\,n\,b^2\,c\,d\,f^2\,g^2\right )}{2\,\left (a^2\,c^2\,g^4-2\,a^2\,c\,d\,f\,g^3+a^2\,d^2\,f^2\,g^2-2\,a\,b\,c^2\,f\,g^3+4\,a\,b\,c\,d\,f^2\,g^2-2\,a\,b\,d^2\,f^3\,g+b^2\,c^2\,f^2\,g^2-2\,b^2\,c\,d\,f^3\,g+b^2\,d^2\,f^4\right )}+\frac {x^2\,\left (B\,n\,a^2\,d^2\,g^4-2\,B\,f\,n\,a\,b\,d^2\,g^3-B\,n\,b^2\,c^2\,g^4+2\,B\,f\,n\,b^2\,c\,d\,g^3\right )}{a^2\,c^2\,g^4-2\,a^2\,c\,d\,f\,g^3+a^2\,d^2\,f^2\,g^2-2\,a\,b\,c^2\,f\,g^3+4\,a\,b\,c\,d\,f^2\,g^2-2\,a\,b\,d^2\,f^3\,g+b^2\,c^2\,f^2\,g^2-2\,b^2\,c\,d\,f^3\,g+b^2\,d^2\,f^4}}{3\,f^3\,g+9\,f^2\,g^2\,x+9\,f\,g^3\,x^2+3\,g^4\,x^3} \]

input
int((A + B*log(e*((a + b*x)/(c + d*x))^n))/(f + g*x)^4,x)
 
output
(B*d^3*n*log(c + d*x))/(3*c^3*g^4 - 3*d^3*f^3*g + 9*c*d^2*f^2*g^2 - 9*c^2* 
d*f*g^3) - (log(f + g*x)*(g^2*(B*a^3*d^3*n - B*b^3*c^3*n) - g*(3*B*a^2*b*d 
^3*f*n - 3*B*b^3*c^2*d*f*n) + 3*B*a*b^2*d^3*f^2*n - 3*B*b^3*c*d^2*f^2*n))/ 
(3*a^3*c^3*g^6 + 3*b^3*d^3*f^6 - 3*a^3*d^3*f^3*g^3 - 3*b^3*c^3*f^3*g^3 - 9 
*a^2*b*c^3*f*g^5 - 9*a*b^2*d^3*f^5*g - 9*a^3*c^2*d*f*g^5 - 9*b^3*c*d^2*f^5 
*g + 9*a*b^2*c^3*f^2*g^4 + 9*a^2*b*d^3*f^4*g^2 + 9*a^3*c*d^2*f^2*g^4 + 9*b 
^3*c^2*d*f^4*g^2 + 27*a*b^2*c*d^2*f^4*g^2 - 27*a*b^2*c^2*d*f^3*g^3 - 27*a^ 
2*b*c*d^2*f^3*g^3 + 27*a^2*b*c^2*d*f^2*g^4) - (B*log(e*((a + b*x)/(c + d*x 
))^n))/(3*g*(f^3 + g^3*x^3 + 3*f^2*g*x + 3*f*g^2*x^2)) - (B*b^3*n*log(a + 
b*x))/(3*a^3*g^4 - 3*b^3*f^3*g + 9*a*b^2*f^2*g^2 - 9*a^2*b*f*g^3) - ((2*A* 
a^2*c^2*g^4 + 2*A*b^2*d^2*f^4 + 2*A*a^2*d^2*f^2*g^2 + 2*A*b^2*c^2*f^2*g^2 
+ 3*B*a^2*d^2*f^2*g^2*n - 3*B*b^2*c^2*f^2*g^2*n - 4*A*a*b*c^2*f*g^3 - 4*A* 
a*b*d^2*f^3*g - 4*A*a^2*c*d*f*g^3 - 4*A*b^2*c*d*f^3*g + 8*A*a*b*c*d*f^2*g^ 
2 + B*a*b*c^2*f*g^3*n - 5*B*a*b*d^2*f^3*g*n - B*a^2*c*d*f*g^3*n + 5*B*b^2* 
c*d*f^3*g*n)/(2*(a^2*c^2*g^4 + b^2*d^2*f^4 + a^2*d^2*f^2*g^2 + b^2*c^2*f^2 
*g^2 - 2*a*b*c^2*f*g^3 - 2*a*b*d^2*f^3*g - 2*a^2*c*d*f*g^3 - 2*b^2*c*d*f^3 
*g + 4*a*b*c*d*f^2*g^2)) + (x*(B*a*b*c^2*g^4*n - B*a^2*c*d*g^4*n + 5*B*a^2 
*d^2*f*g^3*n - 5*B*b^2*c^2*f*g^3*n - 9*B*a*b*d^2*f^2*g^2*n + 9*B*b^2*c*d*f 
^2*g^2*n))/(2*(a^2*c^2*g^4 + b^2*d^2*f^4 + a^2*d^2*f^2*g^2 + b^2*c^2*f^2*g 
^2 - 2*a*b*c^2*f*g^3 - 2*a*b*d^2*f^3*g - 2*a^2*c*d*f*g^3 - 2*b^2*c*d*f^...