\(\int \frac {\log (f x^m) (a+b \log (c (d+e x)^n))}{x^3} \, dx\) [364]

Optimal result

Integrand size = 24, antiderivative size = 156 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=-\frac {3 b e m n}{4 d x}-\frac {b e^2 m n \log (x)}{4 d^2}-\frac {b e n \log \left (f x^m\right )}{2 d x}+\frac {b e^2 n \log \left (1+\frac {d}{e x}\right ) \log \left (f x^m\right )}{2 d^2}+\frac {b e^2 m n \log (d+e x)}{4 d^2}-\frac {1}{4} \left (\frac {m}{x^2}+\frac {2 \log \left (f x^m\right )}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b e^2 m n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{2 d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.31 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=-\frac {a d^2 m+3 b d e m n x-b e^2 m n x^2 \log ^2(x)+2 a d^2 \log \left (f x^m\right )+2 b d e n x \log \left (f x^m\right )-b e^2 m n x^2 \log (d+e x)-2 b e^2 n x^2 \log \left (f x^m\right ) \log (d+e x)+b d^2 m \log \left (c (d+e x)^n\right )+2 b d^2 \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )+b e^2 n x^2 \log (x) \left (m+2 \log \left (f x^m\right )+2 m \log (d+e x)-2 m \log \left (1+\frac {e x}{d}\right )\right )-2 b e^2 m n x^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 d^2 x^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2873, 54, 2009, 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.

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

\(\Big \downarrow \) 2873

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

\(\Big \downarrow \) 54

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 2780

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

\(\Big \downarrow \) 2741

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

\(\Big \downarrow \) 2779

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

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 = 11.24 (sec) , antiderivative size = 901, normalized size of antiderivative = 5.78

method	result	size
risch	\(\text {Expression too large to display}\)	\(901\)

Input:

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

Output:

(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)*(-ln(x^m)/x^2-1/ 
2*m/x^2-1/2*(-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^2)+1/4*I*e*n*b/d/x*Pi*csgn(I*f*x^m)^3-1/4*I*e^2*n*b/d^2*ln(e*x+d)*Pi*cs 
gn(I*f*x^m)^3+1/4*I*e^2*n*b/d^2*ln(x)*Pi*csgn(I*f*x^m)^3-1/2*e*n*b/d/x*ln( 
f)-1/2*e*n*b*ln(x^m)/d/x-1/4*I*e^2*n*b/d^2*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*x 
^m)*csgn(I*f*x^m)+1/4*I*e^2*n*b/d^2*ln(x)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I* 
f*x^m)+(-1/2*b/x^2*ln(x^m)-1/4*(-I*Pi*b*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m 
)+I*Pi*b*csgn(I*f)*csgn(I*f*x^m)^2+I*Pi*b*csgn(I*x^m)*csgn(I*f*x^m)^2-I*Pi 
*b*csgn(I*f*x^m)^3+2*b*ln(f)+b*m)/x^2)*ln((e*x+d)^n)+1/2*e^2*n*b/d^2*ln(e* 
x+d)*ln(f)-1/2*e^2*n*b/d^2*ln(x)*ln(f)-1/2*e^2*n*b*ln(x^m)/d^2*ln(x)-1/2*e 
^2*n*b*m/d^2*dilog(-e*x/d)+1/4*e^2*n*b*m/d^2*ln(x)^2+1/2*e^2*n*b*ln(x^m)/d 
^2*ln(e*x+d)+1/4*I*e^2*n*b/d^2*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/2* 
e^2*n*b*m/d^2*ln(e*x+d)*ln(-e*x/d)+1/4*I*e^2*n*b/d^2*ln(e*x+d)*Pi*csgn(I*x 
^m)*csgn(I*f*x^m)^2-1/4*I*e^2*n*b/d^2*ln(x)*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1 
/4*I*e^2*n*b/d^2*ln(x)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/4*I*e*n*b/d/x*Pi*c 
sgn(I*f)*csgn(I*f*x^m)^2-1/4*I*e*n*b/d/x*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/ 
4*I*e*n*b/d/x*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-3/4*b*e*m*n/d/x-1/...
 

Fricas [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{3}} \,d x } \] Input:

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

Output:

integral((b*log((e*x + d)^n*c)*log(f*x^m) + a*log(f*x^m))/x^3, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.27 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\frac {1}{4} \, {\left (\frac {2 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} b e^{2} n}{d^{2}} + \frac {b e^{2} n \log \left (e x + d\right )}{d^{2}} - \frac {2 \, b e^{2} n x^{2} \log \left (e x + d\right ) \log \left (x\right ) - b e^{2} n x^{2} \log \left (x\right )^{2} + b e^{2} n x^{2} \log \left (x\right ) + 3 \, b d e n x + b d^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + b d^{2} \log \left (c\right ) + a d^{2}}{d^{2} x^{2}}\right )} m + \frac {1}{2} \, {\left (b e n {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \left (x\right )}{d^{2}} - \frac {1}{d x}\right )} - \frac {b \log \left ({\left (e x + d\right )}^{n} c\right )}{x^{2}} - \frac {a}{x^{2}}\right )} \log \left (f x^{m}\right ) \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{3}} \,d x } \] Input:

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

Output:

integrate((b*log((e*x + d)^n*c) + a)*log(f*x^m)/x^3, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\frac {-4 \left (\int \frac {\mathrm {log}\left (x^{m} f \right )}{e \,x^{4}+d \,x^{3}}d x \right ) b \,d^{3} n \,x^{2}-4 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) \mathrm {log}\left (x^{m} f \right ) b \,d^{2}-2 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,d^{2} m +2 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,e^{2} m \,x^{2}-4 \,\mathrm {log}\left (x^{m} f \right ) a \,d^{2}-2 \,\mathrm {log}\left (x^{m} f \right ) b \,d^{2} n -2 \,\mathrm {log}\left (x \right ) b \,e^{2} m n \,x^{2}-2 a \,d^{2} m -b \,d^{2} m n -2 b d e m n x}{8 d^{2} x^{2}} \] Input:

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

Output:

( - 4*int(log(x**m*f)/(d*x**3 + e*x**4),x)*b*d**3*n*x**2 - 4*log((d + e*x) 
**n*c)*log(x**m*f)*b*d**2 - 2*log((d + e*x)**n*c)*b*d**2*m + 2*log((d + e* 
x)**n*c)*b*e**2*m*x**2 - 4*log(x**m*f)*a*d**2 - 2*log(x**m*f)*b*d**2*n - 2 
*log(x)*b*e**2*m*n*x**2 - 2*a*d**2*m - b*d**2*m*n - 2*b*d*e*m*n*x)/(8*d**2 
*x**2)