\(\int \frac {\log (f x^m) (a+b \log (c (d+e x)^n))}{x^4} \, dx\) [365]

Optimal result

Integrand size = 24, antiderivative size = 193 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^4} \, dx=-\frac {5 b e m n}{36 d x^2}+\frac {4 b e^2 m n}{9 d^2 x}+\frac {b e^3 m n \log (x)}{9 d^3}-\frac {b e n \log \left (f x^m\right )}{6 d x^2}+\frac {b e^2 n \log \left (f x^m\right )}{3 d^2 x}-\frac {b e^3 n \log \left (1+\frac {d}{e x}\right ) \log \left (f x^m\right )}{3 d^3}-\frac {b e^3 m n \log (d+e x)}{9 d^3}-\frac {1}{9} \left (\frac {m}{x^3}+\frac {3 \log \left (f x^m\right )}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b e^3 m n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{3 d^3} \] Output:

-5/36*b*e*m*n/d/x^2+4/9*b*e^2*m*n/d^2/x+1/9*b*e^3*m*n*ln(x)/d^3-1/6*b*e*n* 
ln(f*x^m)/d/x^2+1/3*b*e^2*n*ln(f*x^m)/d^2/x-1/3*b*e^3*n*ln(1+d/e/x)*ln(f*x 
^m)/d^3-1/9*b*e^3*m*n*ln(e*x+d)/d^3-1/9*(m/x^3+3*ln(f*x^m)/x^3)*(a+b*ln(c* 
(e*x+d)^n))+1/3*b*e^3*m*n*polylog(2,-d/e/x)/d^3
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.24 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^4} \, dx=-\frac {4 a d^3 m+5 b d^2 e m n x-16 b d e^2 m n x^2+6 b e^3 m n x^3 \log ^2(x)+12 a d^3 \log \left (f x^m\right )+6 b d^2 e n x \log \left (f x^m\right )-12 b d e^2 n x^2 \log \left (f x^m\right )+4 b e^3 m n x^3 \log (d+e x)+12 b e^3 n x^3 \log \left (f x^m\right ) \log (d+e x)+4 b d^3 m \log \left (c (d+e x)^n\right )+12 b d^3 \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )-4 b e^3 n x^3 \log (x) \left (m+3 \log \left (f x^m\right )+3 m \log (d+e x)-3 m \log \left (1+\frac {e x}{d}\right )\right )+12 b e^3 m n x^3 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{36 d^3 x^3} \] Input:

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

Output:

-1/36*(4*a*d^3*m + 5*b*d^2*e*m*n*x - 16*b*d*e^2*m*n*x^2 + 6*b*e^3*m*n*x^3* 
Log[x]^2 + 12*a*d^3*Log[f*x^m] + 6*b*d^2*e*n*x*Log[f*x^m] - 12*b*d*e^2*n*x 
^2*Log[f*x^m] + 4*b*e^3*m*n*x^3*Log[d + e*x] + 12*b*e^3*n*x^3*Log[f*x^m]*L 
og[d + e*x] + 4*b*d^3*m*Log[c*(d + e*x)^n] + 12*b*d^3*Log[f*x^m]*Log[c*(d 
+ e*x)^n] - 4*b*e^3*n*x^3*Log[x]*(m + 3*Log[f*x^m] + 3*m*Log[d + e*x] - 3* 
m*Log[1 + (e*x)/d]) + 12*b*e^3*m*n*x^3*PolyLog[2, -((e*x)/d)])/(d^3*x^3)
 

Rubi [A] (verified)

Time = 1.37 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2873, 54, 2009, 2780, 2741, 2780, 2741, 2779, 2838}

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.

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

\(\Big \downarrow \) 2873

\(\displaystyle \frac {1}{3} b e n \int \frac {\log \left (f x^m\right )}{x^3 (d+e x)}dx+\frac {1}{9} b e m n \int \frac {1}{x^3 (d+e x)}dx-\frac {1}{9} \left (\frac {3 \log \left (f x^m\right )}{x^3}+\frac {m}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\)

\(\Big \downarrow \) 54

\(\displaystyle \frac {1}{9} b e m n \int \left (-\frac {e^3}{d^3 (d+e x)}+\frac {e^2}{d^3 x}-\frac {e}{d^2 x^2}+\frac {1}{d x^3}\right )dx+\frac {1}{3} b e n \int \frac {\log \left (f x^m\right )}{x^3 (d+e x)}dx-\frac {1}{9} \left (\frac {3 \log \left (f x^m\right )}{x^3}+\frac {m}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} b e n \int \frac {\log \left (f x^m\right )}{x^3 (d+e x)}dx-\frac {1}{9} \left (\frac {3 \log \left (f x^m\right )}{x^3}+\frac {m}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{9} b e m n \left (\frac {e^2 \log (x)}{d^3}-\frac {e^2 \log (d+e x)}{d^3}+\frac {e}{d^2 x}-\frac {1}{2 d x^2}\right )\)

\(\Big \downarrow \) 2780

\(\displaystyle \frac {1}{3} b e n \left (\frac {\int \frac {\log \left (f x^m\right )}{x^3}dx}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x^2 (d+e x)}dx}{d}\right )-\frac {1}{9} \left (\frac {3 \log \left (f x^m\right )}{x^3}+\frac {m}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{9} b e m n \left (\frac {e^2 \log (x)}{d^3}-\frac {e^2 \log (d+e x)}{d^3}+\frac {e}{d^2 x}-\frac {1}{2 d x^2}\right )\)

\(\Big \downarrow \) 2741

\(\displaystyle \frac {1}{3} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x^2 (d+e x)}dx}{d}\right )-\frac {1}{9} \left (\frac {3 \log \left (f x^m\right )}{x^3}+\frac {m}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{9} b e m n \left (\frac {e^2 \log (x)}{d^3}-\frac {e^2 \log (d+e x)}{d^3}+\frac {e}{d^2 x}-\frac {1}{2 d x^2}\right )\)

\(\Big \downarrow \) 2780

\(\displaystyle \frac {1}{3} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \left (\frac {\int \frac {\log \left (f x^m\right )}{x^2}dx}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x (d+e x)}dx}{d}\right )}{d}\right )-\frac {1}{9} \left (\frac {3 \log \left (f x^m\right )}{x^3}+\frac {m}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{9} b e m n \left (\frac {e^2 \log (x)}{d^3}-\frac {e^2 \log (d+e x)}{d^3}+\frac {e}{d^2 x}-\frac {1}{2 d x^2}\right )\)

\(\Big \downarrow \) 2741

\(\displaystyle \frac {1}{3} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{x}-\frac {m}{x}}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x (d+e x)}dx}{d}\right )}{d}\right )-\frac {1}{9} \left (\frac {3 \log \left (f x^m\right )}{x^3}+\frac {m}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{9} b e m n \left (\frac {e^2 \log (x)}{d^3}-\frac {e^2 \log (d+e x)}{d^3}+\frac {e}{d^2 x}-\frac {1}{2 d x^2}\right )\)

\(\Big \downarrow \) 2779

\(\displaystyle \frac {1}{3} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{x}-\frac {m}{x}}{d}-\frac {e \left (\frac {m \int \frac {\log \left (\frac {d}{e x}+1\right )}{x}dx}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \log \left (f x^m\right )}{d}\right )}{d}\right )}{d}\right )-\frac {1}{9} \left (\frac {3 \log \left (f x^m\right )}{x^3}+\frac {m}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{9} b e m n \left (\frac {e^2 \log (x)}{d^3}-\frac {e^2 \log (d+e x)}{d^3}+\frac {e}{d^2 x}-\frac {1}{2 d x^2}\right )\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {1}{9} \left (\frac {3 \log \left (f x^m\right )}{x^3}+\frac {m}{x^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{9} b e m n \left (\frac {e^2 \log (x)}{d^3}-\frac {e^2 \log (d+e x)}{d^3}+\frac {e}{d^2 x}-\frac {1}{2 d x^2}\right )+\frac {1}{3} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{x}-\frac {m}{x}}{d}-\frac {e \left (\frac {m \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \log \left (f x^m\right )}{d}\right )}{d}\right )}{d}\right )\)

Input:

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

Output:

(b*e*m*n*(-1/2*1/(d*x^2) + e/(d^2*x) + (e^2*Log[x])/d^3 - (e^2*Log[d + e*x 
])/d^3))/9 - ((m/x^3 + (3*Log[f*x^m])/x^3)*(a + b*Log[c*(d + e*x)^n]))/9 + 
 (b*e*n*((-1/4*m/x^2 - Log[f*x^m]/(2*x^2))/d - (e*((-(m/x) - Log[f*x^m]/x) 
/d - (e*(-((Log[1 + d/(e*x)]*Log[f*x^m])/d) + (m*PolyLog[2, -(d/(e*x))])/d 
))/d))/d))/3
 

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 2741

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> 
Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( 
m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
 

rule 2779

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r 
_.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) 
, x] + Simp[b*n*(p/(d*r))   Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 
 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
 

rule 2780

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)* 
(x_)^(r_.)), x_Symbol] :> Simp[1/d   Int[x^m*(a + b*Log[c*x^n])^p, x], x] - 
 Simp[e/d   Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /; Fre 
eQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 2873

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 31.69 (sec) , antiderivative size = 1070, normalized size of antiderivative = 5.54

method	result	size
risch	\(\text {Expression too large to display}\)	\(1070\)

Input:

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

Output:

(-1/3*b/x^3*ln(x^m)-1/18*(-3*I*Pi*b*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+3* 
I*Pi*b*csgn(I*f)*csgn(I*f*x^m)^2+3*I*Pi*b*csgn(I*x^m)*csgn(I*f*x^m)^2-3*I* 
Pi*b*csgn(I*f*x^m)^3+6*b*ln(f)+2*b*m)/x^3)*ln((e*x+d)^n)+1/3*e^2*n*b*ln(x^ 
m)/d^2/x-1/3*e^3*n*b*ln(x^m)/d^3*ln(e*x+d)+1/3*e^3*n*b*ln(x^m)/d^3*ln(x)+1 
/3*e^3*n*b*m/d^3*ln(e*x+d)*ln(-e*x/d)-1/6*e^3*n*b*m/d^3*ln(x)^2+1/3*e^3*n* 
b*m/d^3*dilog(-e*x/d)-1/3*e^3*n*b/d^3*ln(e*x+d)*ln(f)+1/3*e^3*n*b/d^3*ln(x 
)*ln(f)-1/6*I*e^3*n*b/d^3*ln(e*x+d)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/6*I*e 
^3*n*b/d^3*ln(x)*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/6*I*e^3*n*b/d^3*ln(x)*Pi*c 
sgn(I*x^m)*csgn(I*f*x^m)^2+1/6*I*e^2*n*b/d^2/x*Pi*csgn(I*f)*csgn(I*f*x^m)^ 
2+1/6*I*e^2*n*b/d^2/x*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/12*I*e*n*b/d/x^2*Pi 
*csgn(I*f)*csgn(I*f*x^m)^2-1/12*I*e*n*b/d/x^2*Pi*csgn(I*x^m)*csgn(I*f*x^m) 
^2+(1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/4*I*b*Pi*csgn(I*( 
e*x+d)^n)*csgn(I*c*(e*x+d)^n)*csgn(I*c)-1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+1 
/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^2*csgn(I*c)+1/2*b*ln(c)+1/2*a)*(-2/3*ln(x^m) 
/x^3-2/9*m/x^3-1/3*(-I*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+I*Pi*csgn(I* 
f)*csgn(I*f*x^m)^2+I*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-I*Pi*csgn(I*f*x^m)^3+2 
*ln(f))/x^3)+1/6*I*e^3*n*b/d^3*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f 
*x^m)-1/6*I*e^3*n*b/d^3*ln(x)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/6*I 
*e^2*n*b/d^2/x*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/12*I*e*n*b/d/x^2*P 
i*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/6*e*n*b*ln(x^m)/d/x^2+1/12*I*e*...
 

Fricas [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{4}} \,d x } \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.19 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^4} \, dx=-\frac {1}{36} \, {\left (\frac {12 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} b e^{3} n}{d^{3}} + \frac {4 \, b e^{3} n \log \left (e x + d\right )}{d^{3}} - \frac {12 \, b e^{3} n x^{3} \log \left (e x + d\right ) \log \left (x\right ) - 6 \, b e^{3} n x^{3} \log \left (x\right )^{2} + 4 \, b e^{3} n x^{3} \log \left (x\right ) + 16 \, b d e^{2} n x^{2} - 5 \, b d^{2} e n x - 4 \, b d^{3} \log \left ({\left (e x + d\right )}^{n}\right ) - 4 \, b d^{3} \log \left (c\right ) - 4 \, a d^{3}}{d^{3} x^{3}}\right )} m - \frac {1}{6} \, {\left (b e n {\left (\frac {2 \, e^{2} \log \left (e x + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac {2 \, e x - d}{d^{2} x^{2}}\right )} + \frac {2 \, b \log \left ({\left (e x + d\right )}^{n} c\right )}{x^{3}} + \frac {2 \, a}{x^{3}}\right )} \log \left (f x^{m}\right ) \] Input:

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

Output:

-1/36*(12*(log(e*x/d + 1)*log(x) + dilog(-e*x/d))*b*e^3*n/d^3 + 4*b*e^3*n* 
log(e*x + d)/d^3 - (12*b*e^3*n*x^3*log(e*x + d)*log(x) - 6*b*e^3*n*x^3*log 
(x)^2 + 4*b*e^3*n*x^3*log(x) + 16*b*d*e^2*n*x^2 - 5*b*d^2*e*n*x - 4*b*d^3* 
log((e*x + d)^n) - 4*b*d^3*log(c) - 4*a*d^3)/(d^3*x^3))*m - 1/6*(b*e*n*(2* 
e^2*log(e*x + d)/d^3 - 2*e^2*log(x)/d^3 - (2*e*x - d)/(d^2*x^2)) + 2*b*log 
((e*x + d)^n*c)/x^3 + 2*a/x^3)*log(f*x^m)
 

Giac [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{4}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^4} \, dx=\int \frac {\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{x^4} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^4} \, dx=\frac {-18 \left (\int \frac {\mathrm {log}\left (x^{m} f \right )}{e \,x^{5}+d \,x^{4}}d x \right ) b \,d^{4} n \,x^{3}-18 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) \mathrm {log}\left (x^{m} f \right ) b \,d^{3}-6 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,d^{3} m -6 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,e^{3} m \,x^{3}-18 \,\mathrm {log}\left (x^{m} f \right ) a \,d^{3}-6 \,\mathrm {log}\left (x^{m} f \right ) b \,d^{3} n +6 \,\mathrm {log}\left (x \right ) b \,e^{3} m n \,x^{3}-6 a \,d^{3} m -2 b \,d^{3} m n -3 b \,d^{2} e m n x +6 b d \,e^{2} m n \,x^{2}}{54 d^{3} x^{3}} \] Input:

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

Output:

( - 18*int(log(x**m*f)/(d*x**4 + e*x**5),x)*b*d**4*n*x**3 - 18*log((d + e* 
x)**n*c)*log(x**m*f)*b*d**3 - 6*log((d + e*x)**n*c)*b*d**3*m - 6*log((d + 
e*x)**n*c)*b*e**3*m*x**3 - 18*log(x**m*f)*a*d**3 - 6*log(x**m*f)*b*d**3*n 
+ 6*log(x)*b*e**3*m*n*x**3 - 6*a*d**3*m - 2*b*d**3*m*n - 3*b*d**2*e*m*n*x 
+ 6*b*d*e**2*m*n*x**2)/(54*d**3*x**3)