\(\int \frac {a+b \log (c (d+e x)^n)}{(f+g x)^4} \, dx\) [43]

Optimal result

Integrand size = 22, antiderivative size = 141 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^4} \, dx=\frac {b e n}{6 g (e f-d g) (f+g x)^2}+\frac {b e^2 n}{3 g (e f-d g)^2 (f+g x)}+\frac {b e^3 n \log (d+e x)}{3 g (e f-d g)^3}-\frac {a+b \log \left (c (d+e x)^n\right )}{3 g (f+g x)^3}-\frac {b e^3 n \log (f+g x)}{3 g (e f-d g)^3} \] Output:

1/6*b*e*n/g/(-d*g+e*f)/(g*x+f)^2+1/3*b*e^2*n/g/(-d*g+e*f)^2/(g*x+f)+1/3*b* 
e^3*n*ln(e*x+d)/g/(-d*g+e*f)^3-1/3*(a+b*ln(c*(e*x+d)^n))/g/(g*x+f)^3-1/3*b 
*e^3*n*ln(g*x+f)/g/(-d*g+e*f)^3
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.78 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^4} \, dx=\frac {-2 \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b e n (f+g x) \left ((e f-d g) (3 e f-d g+2 e g x)+2 e^2 (f+g x)^2 \log (d+e x)-2 e^2 (f+g x)^2 \log (f+g x)\right )}{(e f-d g)^3}}{6 g (f+g x)^3} \] Input:

Integrate[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^4,x]
 

Output:

(-2*(a + b*Log[c*(d + e*x)^n]) + (b*e*n*(f + g*x)*((e*f - d*g)*(3*e*f - d* 
g + 2*e*g*x) + 2*e^2*(f + g*x)^2*Log[d + e*x] - 2*e^2*(f + g*x)^2*Log[f + 
g*x]))/(e*f - d*g)^3)/(6*g*(f + g*x)^3)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2842, 54, 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.

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

\(\Big \downarrow \) 2842

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

\(\Big \downarrow \) 54

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b e n \left (\frac {e^2 \log (d+e x)}{(e f-d g)^3}-\frac {e^2 \log (f+g x)}{(e f-d g)^3}+\frac {e}{(f+g x) (e f-d g)^2}+\frac {1}{2 (f+g x)^2 (e f-d g)}\right )}{3 g}-\frac {a+b \log \left (c (d+e x)^n\right )}{3 g (f+g x)^3}\)

Input:

Int[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^4,x]
 

Output:

-1/3*(a + b*Log[c*(d + e*x)^n])/(g*(f + g*x)^3) + (b*e*n*(1/(2*(e*f - d*g) 
*(f + g*x)^2) + e/((e*f - d*g)^2*(f + g*x)) + (e^2*Log[d + e*x])/(e*f - d* 
g)^3 - (e^2*Log[f + g*x])/(e*f - d*g)^3))/(3*g)
 

Defintions of rubi rules used

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 2009

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

rule 2842

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_ 
))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/( 
g*(q + 1))), x] - Simp[b*e*(n/(g*(q + 1)))   Int[(f + g*x)^(q + 1)/(d + e*x 
), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && 
NeQ[q, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(454\) vs. \(2(131)=262\).

Time = 3.09 (sec) , antiderivative size = 455, normalized size of antiderivative = 3.23

method	result	size
parallelrisch	\(-\frac {3 b \,e^{4} f^{3} g^{2} n -6 a \,d^{2} e^{2} f \,g^{4}+6 a d \,e^{3} f^{2} g^{3}+2 \ln \left (c \left (e x +d \right )^{n}\right ) b \,d^{3} e \,g^{5}-2 \ln \left (c \left (e x +d \right )^{n}\right ) b \,e^{4} f^{3} g^{2}+2 \ln \left (e x +d \right ) b \,e^{4} f^{3} g^{2} n -2 \ln \left (g x +f \right ) b \,e^{4} f^{3} g^{2} n -2 x^{2} b d \,e^{3} g^{5} n +2 x^{2} b \,e^{4} f \,g^{4} n -2 \ln \left (g x +f \right ) x^{3} b \,e^{4} g^{5} n +x b \,d^{2} e^{2} g^{5} n +5 x b \,e^{4} f^{2} g^{3} n -6 \ln \left (c \left (e x +d \right )^{n}\right ) b \,d^{2} e^{2} f \,g^{4}+6 \ln \left (c \left (e x +d \right )^{n}\right ) b d \,e^{3} f^{2} g^{3}+2 \ln \left (e x +d \right ) x^{3} b \,e^{4} g^{5} n +2 a \,d^{3} e \,g^{5}-2 a \,e^{4} f^{3} g^{2}-6 \ln \left (g x +f \right ) x b \,e^{4} f^{2} g^{3} n +b \,d^{2} e^{2} f \,g^{4} n -4 b d \,e^{3} f^{2} g^{3} n +6 \ln \left (e x +d \right ) x^{2} b \,e^{4} f \,g^{4} n -6 \ln \left (g x +f \right ) x^{2} b \,e^{4} f \,g^{4} n +6 \ln \left (e x +d \right ) x b \,e^{4} f^{2} g^{3} n -6 x b d \,e^{3} f \,g^{4} n}{6 \left (d^{3} g^{3}-3 d^{2} e f \,g^{2}+3 d \,e^{2} f^{2} g -e^{3} f^{3}\right ) \left (g x +f \right )^{3} g^{3} e}\)	\(455\)
risch	\(\text {Expression too large to display}\)	\(950\)

Input:

int((a+b*ln(c*(e*x+d)^n))/(g*x+f)^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 507 vs. \(2 (131) = 262\).

Time = 0.09 (sec) , antiderivative size = 507, normalized size of antiderivative = 3.60 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^4} \, dx=-\frac {2 \, a e^{3} f^{3} - 6 \, a d e^{2} f^{2} g + 6 \, a d^{2} e f g^{2} - 2 \, a d^{3} g^{3} - 2 \, {\left (b e^{3} f g^{2} - b d e^{2} g^{3}\right )} n x^{2} - {\left (5 \, b e^{3} f^{2} g - 6 \, b d e^{2} f g^{2} + b d^{2} e g^{3}\right )} n x - {\left (3 \, b e^{3} f^{3} - 4 \, b d e^{2} f^{2} g + b d^{2} e f g^{2}\right )} n - 2 \, {\left (b e^{3} g^{3} n x^{3} + 3 \, b e^{3} f g^{2} n x^{2} + 3 \, b e^{3} f^{2} g n x + {\left (3 \, b d e^{2} f^{2} g - 3 \, b d^{2} e f g^{2} + b d^{3} g^{3}\right )} n\right )} \log \left (e x + d\right ) + 2 \, {\left (b e^{3} g^{3} n x^{3} + 3 \, b e^{3} f g^{2} n x^{2} + 3 \, b e^{3} f^{2} g n x + b e^{3} f^{3} n\right )} \log \left (g x + f\right ) + 2 \, {\left (b e^{3} f^{3} - 3 \, b d e^{2} f^{2} g + 3 \, b d^{2} e f g^{2} - b d^{3} g^{3}\right )} \log \left (c\right )}{6 \, {\left (e^{3} f^{6} g - 3 \, d e^{2} f^{5} g^{2} + 3 \, d^{2} e f^{4} g^{3} - d^{3} f^{3} g^{4} + {\left (e^{3} f^{3} g^{4} - 3 \, d e^{2} f^{2} g^{5} + 3 \, d^{2} e f g^{6} - d^{3} g^{7}\right )} x^{3} + 3 \, {\left (e^{3} f^{4} g^{3} - 3 \, d e^{2} f^{3} g^{4} + 3 \, d^{2} e f^{2} g^{5} - d^{3} f g^{6}\right )} x^{2} + 3 \, {\left (e^{3} f^{5} g^{2} - 3 \, d e^{2} f^{4} g^{3} + 3 \, d^{2} e f^{3} g^{4} - d^{3} f^{2} g^{5}\right )} x\right )}} \] Input:

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^4,x, algorithm="fricas")
 

Output:

-1/6*(2*a*e^3*f^3 - 6*a*d*e^2*f^2*g + 6*a*d^2*e*f*g^2 - 2*a*d^3*g^3 - 2*(b 
*e^3*f*g^2 - b*d*e^2*g^3)*n*x^2 - (5*b*e^3*f^2*g - 6*b*d*e^2*f*g^2 + b*d^2 
*e*g^3)*n*x - (3*b*e^3*f^3 - 4*b*d*e^2*f^2*g + b*d^2*e*f*g^2)*n - 2*(b*e^3 
*g^3*n*x^3 + 3*b*e^3*f*g^2*n*x^2 + 3*b*e^3*f^2*g*n*x + (3*b*d*e^2*f^2*g - 
3*b*d^2*e*f*g^2 + b*d^3*g^3)*n)*log(e*x + d) + 2*(b*e^3*g^3*n*x^3 + 3*b*e^ 
3*f*g^2*n*x^2 + 3*b*e^3*f^2*g*n*x + b*e^3*f^3*n)*log(g*x + f) + 2*(b*e^3*f 
^3 - 3*b*d*e^2*f^2*g + 3*b*d^2*e*f*g^2 - b*d^3*g^3)*log(c))/(e^3*f^6*g - 3 
*d*e^2*f^5*g^2 + 3*d^2*e*f^4*g^3 - d^3*f^3*g^4 + (e^3*f^3*g^4 - 3*d*e^2*f^ 
2*g^5 + 3*d^2*e*f*g^6 - d^3*g^7)*x^3 + 3*(e^3*f^4*g^3 - 3*d*e^2*f^3*g^4 + 
3*d^2*e*f^2*g^5 - d^3*f*g^6)*x^2 + 3*(e^3*f^5*g^2 - 3*d*e^2*f^4*g^3 + 3*d^ 
2*e*f^3*g^4 - d^3*f^2*g^5)*x)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^4} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate((a+b*ln(c*(e*x+d)**n))/(g*x+f)**4,x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (131) = 262\).

Time = 0.04 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.13 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^4} \, dx=\frac {1}{6} \, {\left (\frac {2 \, e^{2} \log \left (e x + d\right )}{e^{3} f^{3} g - 3 \, d e^{2} f^{2} g^{2} + 3 \, d^{2} e f g^{3} - d^{3} g^{4}} - \frac {2 \, e^{2} \log \left (g x + f\right )}{e^{3} f^{3} g - 3 \, d e^{2} f^{2} g^{2} + 3 \, d^{2} e f g^{3} - d^{3} g^{4}} + \frac {2 \, e g x + 3 \, e f - d g}{e^{2} f^{4} g - 2 \, d e f^{3} g^{2} + d^{2} f^{2} g^{3} + {\left (e^{2} f^{2} g^{3} - 2 \, d e f g^{4} + d^{2} g^{5}\right )} x^{2} + 2 \, {\left (e^{2} f^{3} g^{2} - 2 \, d e f^{2} g^{3} + d^{2} f g^{4}\right )} x}\right )} b e n - \frac {b \log \left ({\left (e x + d\right )}^{n} c\right )}{3 \, {\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g\right )}} - \frac {a}{3 \, {\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g\right )}} \] Input:

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^4,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 401 vs. \(2 (131) = 262\).

Time = 0.12 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.84 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^4} \, dx=\frac {b e^{3} n \log \left (e x + d\right )}{3 \, {\left (e^{3} f^{3} g - 3 \, d e^{2} f^{2} g^{2} + 3 \, d^{2} e f g^{3} - d^{3} g^{4}\right )}} - \frac {b e^{3} n \log \left (g x + f\right )}{3 \, {\left (e^{3} f^{3} g - 3 \, d e^{2} f^{2} g^{2} + 3 \, d^{2} e f g^{3} - d^{3} g^{4}\right )}} - \frac {b n \log \left (e x + d\right )}{3 \, {\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g\right )}} + \frac {2 \, b e^{2} g^{2} n x^{2} + 5 \, b e^{2} f g n x - b d e g^{2} n x + 3 \, b e^{2} f^{2} n - b d e f g n - 2 \, b e^{2} f^{2} \log \left (c\right ) + 4 \, b d e f g \log \left (c\right ) - 2 \, b d^{2} g^{2} \log \left (c\right ) - 2 \, a e^{2} f^{2} + 4 \, a d e f g - 2 \, a d^{2} g^{2}}{6 \, {\left (e^{2} f^{2} g^{4} x^{3} - 2 \, d e f g^{5} x^{3} + d^{2} g^{6} x^{3} + 3 \, e^{2} f^{3} g^{3} x^{2} - 6 \, d e f^{2} g^{4} x^{2} + 3 \, d^{2} f g^{5} x^{2} + 3 \, e^{2} f^{4} g^{2} x - 6 \, d e f^{3} g^{3} x + 3 \, d^{2} f^{2} g^{4} x + e^{2} f^{5} g - 2 \, d e f^{4} g^{2} + d^{2} f^{3} g^{3}\right )}} \] Input:

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^4,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 26.23 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.01 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^4} \, dx=\frac {2\,a\,d\,e\,f}{3\,{\left (f+g\,x\right )}^3\,{\left (d\,g-e\,f\right )}^2}-\frac {a\,d^2\,g}{3\,{\left (f+g\,x\right )}^3\,{\left (d\,g-e\,f\right )}^2}-\frac {b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{3\,g\,{\left (f+g\,x\right )}^3}-\frac {a\,e^2\,f^2}{3\,g\,{\left (f+g\,x\right )}^3\,{\left (d\,g-e\,f\right )}^2}+\frac {5\,b\,e^2\,f\,n\,x}{6\,{\left (f+g\,x\right )}^3\,{\left (d\,g-e\,f\right )}^2}+\frac {b\,e^2\,g\,n\,x^2}{3\,{\left (f+g\,x\right )}^3\,{\left (d\,g-e\,f\right )}^2}-\frac {b\,d\,e\,f\,n}{6\,{\left (f+g\,x\right )}^3\,{\left (d\,g-e\,f\right )}^2}+\frac {b\,e^2\,f^2\,n}{2\,g\,{\left (f+g\,x\right )}^3\,{\left (d\,g-e\,f\right )}^2}-\frac {b\,d\,e\,g\,n\,x}{6\,{\left (f+g\,x\right )}^3\,{\left (d\,g-e\,f\right )}^2}+\frac {b\,e^3\,n\,\mathrm {atan}\left (\frac {d\,g\,1{}\mathrm {i}+e\,f\,1{}\mathrm {i}+e\,g\,x\,2{}\mathrm {i}}{d\,g-e\,f}\right )\,2{}\mathrm {i}}{3\,g\,{\left (d\,g-e\,f\right )}^3} \] Input:

int((a + b*log(c*(d + e*x)^n))/(f + g*x)^4,x)
 

Output:

(2*a*d*e*f)/(3*(f + g*x)^3*(d*g - e*f)^2) - (a*d^2*g)/(3*(f + g*x)^3*(d*g 
- e*f)^2) - (b*log(c*(d + e*x)^n))/(3*g*(f + g*x)^3) - (a*e^2*f^2)/(3*g*(f 
 + g*x)^3*(d*g - e*f)^2) + (b*e^3*n*atan((d*g*1i + e*f*1i + e*g*x*2i)/(d*g 
 - e*f))*2i)/(3*g*(d*g - e*f)^3) + (5*b*e^2*f*n*x)/(6*(f + g*x)^3*(d*g - e 
*f)^2) + (b*e^2*g*n*x^2)/(3*(f + g*x)^3*(d*g - e*f)^2) - (b*d*e*f*n)/(6*(f 
 + g*x)^3*(d*g - e*f)^2) + (b*e^2*f^2*n)/(2*g*(f + g*x)^3*(d*g - e*f)^2) - 
 (b*d*e*g*n*x)/(6*(f + g*x)^3*(d*g - e*f)^2)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 979, normalized size of antiderivative = 6.94 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^4} \, dx =\text {Too large to display} \] Input:

int((a+b*log(c*(e*x+d)^n))/(g*x+f)^4,x)
 

Output:

( - 6*log(d + e*x)*b*d**3*f**3*g**3*n - 18*log(d + e*x)*b*d**3*f**2*g**4*n 
*x - 18*log(d + e*x)*b*d**3*f*g**5*n*x**2 - 6*log(d + e*x)*b*d**3*g**6*n*x 
**3 + 18*log(d + e*x)*b*d**2*e*f**4*g**2*n + 54*log(d + e*x)*b*d**2*e*f**3 
*g**3*n*x + 54*log(d + e*x)*b*d**2*e*f**2*g**4*n*x**2 + 18*log(d + e*x)*b* 
d**2*e*f*g**5*n*x**3 - 18*log(d + e*x)*b*d*e**2*f**5*g*n - 54*log(d + e*x) 
*b*d*e**2*f**4*g**2*n*x - 54*log(d + e*x)*b*d*e**2*f**3*g**3*n*x**2 - 18*l 
og(d + e*x)*b*d*e**2*f**2*g**4*n*x**3 + 6*log(f + g*x)*b*e**3*f**6*n + 18* 
log(f + g*x)*b*e**3*f**5*g*n*x + 18*log(f + g*x)*b*e**3*f**4*g**2*n*x**2 + 
 6*log(f + g*x)*b*e**3*f**3*g**3*n*x**3 + 18*log((d + e*x)**n*c)*b*d**3*f* 
*2*g**4*x + 18*log((d + e*x)**n*c)*b*d**3*f*g**5*x**2 + 6*log((d + e*x)**n 
*c)*b*d**3*g**6*x**3 - 54*log((d + e*x)**n*c)*b*d**2*e*f**3*g**3*x - 54*lo 
g((d + e*x)**n*c)*b*d**2*e*f**2*g**4*x**2 - 18*log((d + e*x)**n*c)*b*d**2* 
e*f*g**5*x**3 + 54*log((d + e*x)**n*c)*b*d*e**2*f**4*g**2*x + 54*log((d + 
e*x)**n*c)*b*d*e**2*f**3*g**3*x**2 + 18*log((d + e*x)**n*c)*b*d*e**2*f**2* 
g**4*x**3 - 18*log((d + e*x)**n*c)*b*e**3*f**5*g*x - 18*log((d + e*x)**n*c 
)*b*e**3*f**4*g**2*x**2 - 6*log((d + e*x)**n*c)*b*e**3*f**3*g**3*x**3 - 6* 
a*d**3*f**3*g**3 + 18*a*d**2*e*f**4*g**2 - 18*a*d*e**2*f**5*g + 6*a*e**3*f 
**6 - 3*b*d**2*e*f**4*g**2*n - 3*b*d**2*e*f**3*g**3*n*x + 10*b*d*e**2*f**5 
*g*n + 12*b*d*e**2*f**4*g**2*n*x - 2*b*d*e**2*f**2*g**4*n*x**3 - 7*b*e**3* 
f**6*n - 9*b*e**3*f**5*g*n*x + 2*b*e**3*f**3*g**3*n*x**3)/(18*f**3*g*(d...