\(\int x^2 \log (f x^m) (a+b \log (c (d+e x)^n)) \, dx\) [359]

Optimal result

Integrand size = 24, antiderivative size = 195 \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {4 b d^2 m n x}{9 e^2}-\frac {5 b d m n x^2}{36 e}+\frac {2}{27} b m n x^3-\frac {b d^2 n x \log \left (f x^m\right )}{3 e^2}+\frac {b d n x^2 \log \left (f x^m\right )}{6 e}-\frac {1}{9} b n x^3 \log \left (f x^m\right )-\frac {b d^3 m n \log (d+e x)}{9 e^3}-\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b d^3 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{3 e^3}+\frac {b d^3 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{3 e^3} \] Output:

4/9*b*d^2*m*n*x/e^2-5/36*b*d*m*n*x^2/e+2/27*b*m*n*x^3-1/3*b*d^2*n*x*ln(f*x 
^m)/e^2+1/6*b*d*n*x^2*ln(f*x^m)/e-1/9*b*n*x^3*ln(f*x^m)-1/9*b*d^3*m*n*ln(e 
*x+d)/e^3-1/9*(m*x^3-3*x^3*ln(f*x^m))*(a+b*ln(c*(e*x+d)^n))+1/3*b*d^3*n*ln 
(f*x^m)*ln(1+e*x/d)/e^3+1/3*b*d^3*m*n*polylog(2,-e*x/d)/e^3
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.01 \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {6 \log \left (f x^m\right ) \left (6 a e^3 x^3+b e n x \left (-6 d^2+3 d e x-2 e^2 x^2\right )+6 b d^3 n \log (d+e x)+6 b e^3 x^3 \log \left (c (d+e x)^n\right )\right )+m \left (48 b d^2 e n x-15 b d e^2 n x^2-12 a e^3 x^3+8 b e^3 n x^3-12 b d^3 n (1+3 \log (x)) \log (d+e x)-12 b e^3 x^3 \log \left (c (d+e x)^n\right )+36 b d^3 n \log (x) \log \left (1+\frac {e x}{d}\right )\right )+36 b d^3 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{108 e^3} \] Input:

Integrate[x^2*Log[f*x^m]*(a + b*Log[c*(d + e*x)^n]),x]
 

Output:

(6*Log[f*x^m]*(6*a*e^3*x^3 + b*e*n*x*(-6*d^2 + 3*d*e*x - 2*e^2*x^2) + 6*b* 
d^3*n*Log[d + e*x] + 6*b*e^3*x^3*Log[c*(d + e*x)^n]) + m*(48*b*d^2*e*n*x - 
 15*b*d*e^2*n*x^2 - 12*a*e^3*x^3 + 8*b*e^3*n*x^3 - 12*b*d^3*n*(1 + 3*Log[x 
])*Log[d + e*x] - 12*b*e^3*x^3*Log[c*(d + e*x)^n] + 36*b*d^3*n*Log[x]*Log[ 
1 + (e*x)/d]) + 36*b*d^3*m*n*PolyLog[2, -((e*x)/d)])/(108*e^3)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, 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.

\(\displaystyle \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx\)

\(\Big \downarrow \) 2873

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 2793

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^2*Log[f*x^m]*(a + b*Log[c*(d + e*x)^n]),x]
 

Output:

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

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 = 76.57 (sec) , antiderivative size = 1012, normalized size of antiderivative = 5.19

method	result	size
risch	\(\text {Expression too large to display}\)	\(1012\)

Input:

int(x^2*ln(f*x^m)*(a+b*ln(c*(e*x+d)^n)),x,method=_RETURNVERBOSE)
 

Output:

-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)-1/3*m/ 
e^3*b*d^3*n*dilog(-e*x/d)+1/3/e^3*n*b*ln(x^m)*d^3*ln(e*x+d)+1/3/e^3*n*b*d^ 
3*ln(e*x+d)*ln(f)+1/18*I*n*b*x^3*Pi*csgn(I*f*x^m)^3+(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)*(1/3*(-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+2/3*x^3*ln(x^m)-2/9*m*x^3)+(1/3*b*x^3*ln 
(x^m)+1/18*b*x^3*(-3*I*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+3*I*Pi*csgn( 
I*f)*csgn(I*f*x^m)^2+3*I*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-3*I*Pi*csgn(I*f*x^ 
m)^3+6*ln(f)-2*m))*ln((e*x+d)^n)-1/9*n*b*ln(x^m)*x^3-1/18*I*n*b*x^3*Pi*csg 
n(I*f)*csgn(I*f*x^m)^2+1/12*I/e*n*b*x^2*d*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/1 
2*I/e*n*b*x^2*d*Pi*csgn(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 
/6*I/e^3*n*b*d^3*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/6*I/e^3*n*b*d^3* 
ln(e*x+d)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+2/27*b*m*n*x^3-1/12*I/e*n*b*x^2*d 
*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)*c 
sgn(I*x^m)*csgn(I*f*x^m)-1/18*I*n*b*x^3*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/1 
8*I*n*b*x^3*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/12*I/e*n*b*x^2*d*Pi*c 
sgn(I*f*x^m)^3+1/6*I/e^2*n*b*d^2*x*Pi*csgn(I*f*x^m)^3-1/6*I/e^3*n*b*d^3...
 

Fricas [F]

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

integrate(x^2*log(f*x^m)*(a+b*log(c*(e*x+d)^n)),x, algorithm="fricas")
 

Output:

integral(b*x^2*log((e*x + d)^n*c)*log(f*x^m) + a*x^2*log(f*x^m), x)
 

Sympy [F(-1)]

Timed out. \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\text {Timed out} \] Input:

integrate(x**2*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 = 207, normalized size of antiderivative = 1.06 \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {1}{108} \, {\left (\frac {36 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {e x + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x + d}{d}\right )\right )} b d^{3} n}{e^{3}} + \frac {12 \, b e^{3} x^{3} \log \left ({\left (e x + d\right )}^{n}\right ) + 15 \, b d e^{2} n x^{2} - 48 \, b d^{2} e n x + 12 \, b d^{3} n \log \left (e x + d\right ) + 4 \, {\left (3 \, a e^{3} - {\left (2 \, e^{3} n - 3 \, e^{3} \log \left (c\right )\right )} b\right )} x^{3}}{e^{3}}\right )} m + \frac {1}{18} \, {\left (6 \, b x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + 6 \, a x^{3} + b e n {\left (\frac {6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )}\right )} \log \left (f x^{m}\right ) \] Input:

integrate(x^2*log(f*x^m)*(a+b*log(c*(e*x+d)^n)),x, algorithm="maxima")
 

Output:

-1/108*(36*(log(e*x + d)*log(-(e*x + d)/d + 1) + dilog((e*x + d)/d))*b*d^3 
*n/e^3 + (12*b*e^3*x^3*log((e*x + d)^n) + 15*b*d*e^2*n*x^2 - 48*b*d^2*e*n* 
x + 12*b*d^3*n*log(e*x + d) + 4*(3*a*e^3 - (2*e^3*n - 3*e^3*log(c))*b)*x^3 
)/e^3)*m + 1/18*(6*b*x^3*log((e*x + d)^n*c) + 6*a*x^3 + b*e*n*(6*d^3*log(e 
*x + d)/e^4 - (2*e^2*x^3 - 3*d*e*x^2 + 6*d^2*x)/e^3))*log(f*x^m)
 

Giac [F]

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

integrate(x^2*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^2*log(f*x^m), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int x^2\,\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right ) \,d x \] Input:

int(x^2*log(f*x^m)*(a + b*log(c*(d + e*x)^n)),x)
 

Output:

int(x^2*log(f*x^m)*(a + b*log(c*(d + e*x)^n)), x)
 

Reduce [F]

\[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {-36 \left (\int \frac {\mathrm {log}\left (x^{m} f \right )}{e \,x^{2}+d x}d x \right ) b \,d^{4} m n +36 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) \mathrm {log}\left (x^{m} f \right ) b \,e^{3} m \,x^{3}-12 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,d^{3} m^{2}-12 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,e^{3} m^{2} x^{3}+18 \mathrm {log}\left (x^{m} f \right )^{2} b \,d^{3} n +36 \,\mathrm {log}\left (x^{m} f \right ) a \,e^{3} m \,x^{3}-36 \,\mathrm {log}\left (x^{m} f \right ) b \,d^{2} e m n x +18 \,\mathrm {log}\left (x^{m} f \right ) b d \,e^{2} m n \,x^{2}-12 \,\mathrm {log}\left (x^{m} f \right ) b \,e^{3} m n \,x^{3}-12 a \,e^{3} m^{2} x^{3}+48 b \,d^{2} e \,m^{2} n x -15 b d \,e^{2} m^{2} n \,x^{2}+8 b \,e^{3} m^{2} n \,x^{3}}{108 e^{3} m} \] Input:

int(x^2*log(f*x^m)*(a+b*log(c*(e*x+d)^n)),x)
 

Output:

( - 36*int(log(x**m*f)/(d*x + e*x**2),x)*b*d**4*m*n + 36*log((d + e*x)**n* 
c)*log(x**m*f)*b*e**3*m*x**3 - 12*log((d + e*x)**n*c)*b*d**3*m**2 - 12*log 
((d + e*x)**n*c)*b*e**3*m**2*x**3 + 18*log(x**m*f)**2*b*d**3*n + 36*log(x* 
*m*f)*a*e**3*m*x**3 - 36*log(x**m*f)*b*d**2*e*m*n*x + 18*log(x**m*f)*b*d*e 
**2*m*n*x**2 - 12*log(x**m*f)*b*e**3*m*n*x**3 - 12*a*e**3*m**2*x**3 + 48*b 
*d**2*e*m**2*n*x - 15*b*d*e**2*m**2*n*x**2 + 8*b*e**3*m**2*n*x**3)/(108*e* 
*3*m)