Integrand size = 24, antiderivative size = 195 \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {4 b d^2 m n x}{9 e^2}-\frac {5 b d m n x^2}{36 e}+\frac {2}{27} b m n x^3-\frac {b d^2 n x \log \left (f x^m\right )}{3 e^2}+\frac {b d n x^2 \log \left (f x^m\right )}{6 e}-\frac {1}{9} b n x^3 \log \left (f x^m\right )-\frac {b d^3 m n \log (d+e x)}{9 e^3}-\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b d^3 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{3 e^3}+\frac {b d^3 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{3 e^3} \]
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
Time = 0.13 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.01 \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {6 \log \left (f x^m\right ) \left (6 a e^3 x^3+b e n x \left (-6 d^2+3 d e x-2 e^2 x^2\right )+6 b d^3 n \log (d+e x)+6 b e^3 x^3 \log \left (c (d+e x)^n\right )\right )+m \left (48 b d^2 e n x-15 b d e^2 n x^2-12 a e^3 x^3+8 b e^3 n x^3-12 b d^3 n (1+3 \log (x)) \log (d+e x)-12 b e^3 x^3 \log \left (c (d+e x)^n\right )+36 b d^3 n \log (x) \log \left (1+\frac {e x}{d}\right )\right )+36 b d^3 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{108 e^3} \]
(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)
Time = 0.46 (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 |
| \(\displaystyle -\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{3} b e n \left (-\frac {d^3 \log \left (\frac {e x}{d}+1\right ) \log \left (f x^m\right )}{e^4}-\frac {d^3 m \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}+\frac {d^2 x \log \left (f x^m\right )}{e^3}-\frac {d^2 m x}{e^3}-\frac {d x^2 \log \left (f x^m\right )}{2 e^2}+\frac {d m x^2}{4 e^2}+\frac {x^3 \log \left (f x^m\right )}{3 e}-\frac {m x^3}{9 e}\right )+\frac {1}{9} b e m n \left (-\frac {d^3 \log (d+e x)}{e^4}+\frac {d^2 x}{e^3}-\frac {d x^2}{2 e^2}+\frac {x^3}{3 e}\right )\) |
(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
3.4.59.3.1 Defintions of rubi rules used
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]
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]))
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 32.94 (sec) , antiderivative size = 1012, normalized size of antiderivative = 5.19
-1/12*I/e*n*b*d*x^2*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/18*I*n*b*x^3* Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/18*I*n*b*x^3*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*x^m)*csgn(I*f*x^m)-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*dil og(-e*x/d)-1/12*I/e*n*b*d*x^2*Pi*csgn(I*f*x^m)^3+1/6*I/e^2*n*b*d^2*x*Pi*cs gn(I*f*x^m)^3+1/18*I*n*b*x^3*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/3/e^ 2*n*b*ln(x^m)*x*d^2+49/108/e^3*n*b*m*d^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))*l n((e*x+d)^n)-1/3*m/e^3*b*d^3*n*ln(e*x+d)*ln(-e*x/d)-1/6*I/e^3*n*b*d^3*ln(e *x+d)*Pi*csgn(I*f*x^m)^3+2/27*b*m*n*x^3+1/6*I/e^3*n*b*d^3*ln(e*x+d)*Pi*csg n(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+(-1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/4*I *b*Pi*csgn(I*c)*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)^2-1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+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*l n(x^m)-2/9*m*x^3)+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+1/3/e^3*n*b*ln(x^m)*d^3*ln(e*x+d)+1/6/e*n*b*ln(x^m)*d*x^2+1/6/e* n*b*d*x^2*ln(f)-1/3/e^2*n*b*d^2*x*ln(f)-1/9*n*b*x^3*ln(f)+1/12*I/e*n*b*...
\[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{2} \log \left (f x^{m}\right ) \,d x } \]
Timed out. \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\text {Timed out} \]
Time = 0.25 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.06 \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {1}{108} \, {\left (\frac {36 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {e x + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x + d}{d}\right )\right )} b d^{3} n}{e^{3}} + \frac {12 \, b e^{3} x^{3} \log \left ({\left (e x + d\right )}^{n}\right ) + 15 \, b d e^{2} n x^{2} - 48 \, b d^{2} e n x + 12 \, b d^{3} n \log \left (e x + d\right ) + 4 \, {\left (3 \, a e^{3} - {\left (2 \, e^{3} n - 3 \, e^{3} \log \left (c\right )\right )} b\right )} x^{3}}{e^{3}}\right )} m + \frac {1}{18} \, {\left (6 \, b x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + 6 \, a x^{3} + b e n {\left (\frac {6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )}\right )} \log \left (f x^{m}\right ) \]
-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)
\[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{2} \log \left (f x^{m}\right ) \,d x } \]
Timed out. \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int x^2\,\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right ) \,d x \]