\(\int x \log (f x^m) (a+b \log (c (d+e x)^n)) \, dx\) [360]

Optimal result

Integrand size = 22, antiderivative size = 158 \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {3 b d m n x}{4 e}+\frac {1}{4} b m n x^2+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 e^2}-\frac {b d^2 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 e^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.04 \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {\log \left (f x^m\right ) \left (-2 b d^2 n \log (d+e x)+e x \left (2 b d n+2 a e x-b e n x+2 b e x \log \left (c (d+e x)^n\right )\right )\right )+m \left (-3 b d e n x-a e^2 x^2+b e^2 n x^2+b d^2 n (1+2 \log (x)) \log (d+e x)-b e^2 x^2 \log \left (c (d+e x)^n\right )-2 b d^2 n \log (x) \log \left (1+\frac {e x}{d}\right )\right )-2 b d^2 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 e^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2873, 49, 2009, 2793, 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.

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

\(\Big \downarrow \) 2873

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 2793

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 49

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

rule 2009

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

rule 2793

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* 
(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[a + b*Log[c*x^n], 
 (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, 
 f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && Integer 
Q[r]))
 

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 = 31.29 (sec) , antiderivative size = 843, normalized size of antiderivative = 5.34

method	result	size
risch	\(\text {Expression too large to display}\)	\(843\)

Input:

int(x*ln(f*x^m)*(a+b*ln(c*(e*x+d)^n)),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)*(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+x^2*ln(x^m)-1/2*m* 
x^2)+(1/2*b*x^2*ln(x^m)+1/4*b*x^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*cs 
gn(I*f*x^m)^3+2*ln(f)-m))*ln((e*x+d)^n)+1/2/e*n*b*d*x*ln(f)+1/2/e^2*n*b*m* 
d^2*ln(e*x+d)*ln(-e*x/d)-1/8*I*n*b*x^2*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/8*I* 
n*b*x^2*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-5/8/e^2*n*b*m*d^2+1/2/e*n*b*ln(x^m) 
*d*x+1/4*I/e*n*b*d*x*Pi*csgn(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^2*n*b*d^2*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*f*x 
^m)^2-1/4*I/e^2*n*b*d^2*ln(e*x+d)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/2/e^2*n 
*b*ln(x^m)*d^2*ln(e*x+d)+1/2/e^2*n*b*m*d^2*dilog(-e*x/d)-1/4*I/e*n*b*d*x*P 
i*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+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*n*b*d*x*Pi*csgn(I*f*x^m)^3-1/2/e^2* 
n*b*d^2*ln(e*x+d)*ln(f)+1/8*I*n*b*x^2*Pi*csgn(I*f*x^m)^3+1/4*I/e^2*n*b*d^2 
*ln(e*x+d)*Pi*csgn(I*f*x^m)^3+1/8*I*n*b*x^2*Pi*csgn(I*f)*csgn(I*x^m)*csgn( 
I*f*x^m)+1/4*b*m*n*x^2-1/4*n*b*ln(x^m)*x^2-1/4*n*b*x^2*ln(f)-3/4*b*d*m*n*x 
/e+1/4*b*d^2*m*n*ln(e*x+d)/e^2
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.13 \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {1}{4} \, {\left (\frac {2 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {e x + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x + d}{d}\right )\right )} b d^{2} n}{e^{2}} - \frac {b e^{2} x^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + 3 \, b d e n x - b d^{2} n \log \left (e x + d\right ) + {\left (a e^{2} - {\left (e^{2} n - e^{2} \log \left (c\right )\right )} b\right )} x^{2}}{e^{2}}\right )} m - \frac {1}{4} \, {\left (b e n {\left (\frac {2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac {e x^{2} - 2 \, d x}{e^{2}}\right )} - 2 \, b x^{2} \log \left ({\left (e x + d\right )}^{n} c\right ) - 2 \, a x^{2}\right )} \log \left (f x^{m}\right ) \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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