Integrand size = 24, antiderivative size = 232 \[ \int x^3 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {5 b d^3 m n x}{16 e^3}+\frac {3 b d^2 m n x^2}{32 e^2}-\frac {7 b d m n x^3}{144 e}+\frac {1}{32} b m n x^4+\frac {b d^3 n x \log \left (f x^m\right )}{4 e^3}-\frac {b d^2 n x^2 \log \left (f x^m\right )}{8 e^2}+\frac {b d n x^3 \log \left (f x^m\right )}{12 e}-\frac {1}{16} b n x^4 \log \left (f x^m\right )+\frac {b d^4 m n \log (d+e x)}{16 e^4}-\frac {1}{16} \left (m x^4-4 x^4 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^4 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{4 e^4}-\frac {b d^4 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 e^4} \]
-5/16*b*d^3*m*n*x/e^3+3/32*b*d^2*m*n*x^2/e^2-7/144*b*d*m*n*x^3/e+1/32*b*m* n*x^4+1/4*b*d^3*n*x*ln(f*x^m)/e^3-1/8*b*d^2*n*x^2*ln(f*x^m)/e^2+1/12*b*d*n *x^3*ln(f*x^m)/e-1/16*b*n*x^4*ln(f*x^m)+1/16*b*d^4*m*n*ln(e*x+d)/e^4-1/16* (m*x^4-4*x^4*ln(f*x^m))*(a+b*ln(c*(e*x+d)^n))-1/4*b*d^4*n*ln(f*x^m)*ln(1+e *x/d)/e^4-1/4*b*d^4*m*n*polylog(2,-e*x/d)/e^4
Time = 0.14 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.95 \[ \int x^3 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {-6 \log \left (f x^m\right ) \left (-12 a e^4 x^4+b e n x \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )+12 b d^4 n \log (d+e x)-12 b e^4 x^4 \log \left (c (d+e x)^n\right )\right )+m \left (-90 b d^3 e n x+27 b d^2 e^2 n x^2-14 b d e^3 n x^3-18 a e^4 x^4+9 b e^4 n x^4+18 b d^4 n (1+4 \log (x)) \log (d+e x)-18 b e^4 x^4 \log \left (c (d+e x)^n\right )-72 b d^4 n \log (x) \log \left (1+\frac {e x}{d}\right )\right )-72 b d^4 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{288 e^4} \]
(-6*Log[f*x^m]*(-12*a*e^4*x^4 + b*e*n*x*(-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2 + 3*e^3*x^3) + 12*b*d^4*n*Log[d + e*x] - 12*b*e^4*x^4*Log[c*(d + e*x)^n]) + m*(-90*b*d^3*e*n*x + 27*b*d^2*e^2*n*x^2 - 14*b*d*e^3*n*x^3 - 18*a*e^4*x ^4 + 9*b*e^4*n*x^4 + 18*b*d^4*n*(1 + 4*Log[x])*Log[d + e*x] - 18*b*e^4*x^4 *Log[c*(d + e*x)^n] - 72*b*d^4*n*Log[x]*Log[1 + (e*x)/d]) - 72*b*d^4*m*n*P olyLog[2, -((e*x)/d)])/(288*e^4)
Time = 0.52 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.12, 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.
| \(\displaystyle \int x^3 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2873 |
| \(\displaystyle -\frac {1}{4} b e n \int \frac {x^4 \log \left (f x^m\right )}{d+e x}dx+\frac {1}{16} b e m n \int \frac {x^4}{d+e x}dx-\frac {1}{16} \left (m x^4-4 x^4 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {1}{16} b e m n \int \left (\frac {d^4}{e^4 (d+e x)}-\frac {d^3}{e^4}+\frac {x d^2}{e^3}-\frac {x^2 d}{e^2}+\frac {x^3}{e}\right )dx-\frac {1}{4} b e n \int \frac {x^4 \log \left (f x^m\right )}{d+e x}dx-\frac {1}{16} \left (m x^4-4 x^4 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{4} b e n \int \frac {x^4 \log \left (f x^m\right )}{d+e x}dx-\frac {1}{16} \left (m x^4-4 x^4 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (\frac {d^4 \log (d+e x)}{e^5}-\frac {d^3 x}{e^4}+\frac {d^2 x^2}{2 e^3}-\frac {d x^3}{3 e^2}+\frac {x^4}{4 e}\right )\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle -\frac {1}{4} b e n \int \left (\frac {\log \left (f x^m\right ) d^4}{e^4 (d+e x)}-\frac {\log \left (f x^m\right ) d^3}{e^4}+\frac {x \log \left (f x^m\right ) d^2}{e^3}-\frac {x^2 \log \left (f x^m\right ) d}{e^2}+\frac {x^3 \log \left (f x^m\right )}{e}\right )dx-\frac {1}{16} \left (m x^4-4 x^4 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (\frac {d^4 \log (d+e x)}{e^5}-\frac {d^3 x}{e^4}+\frac {d^2 x^2}{2 e^3}-\frac {d x^3}{3 e^2}+\frac {x^4}{4 e}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{16} \left (m x^4-4 x^4 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{4} b e n \left (\frac {d^4 \log \left (\frac {e x}{d}+1\right ) \log \left (f x^m\right )}{e^5}+\frac {d^4 m \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^5}-\frac {d^3 x \log \left (f x^m\right )}{e^4}+\frac {d^3 m x}{e^4}+\frac {d^2 x^2 \log \left (f x^m\right )}{2 e^3}-\frac {d^2 m x^2}{4 e^3}-\frac {d x^3 \log \left (f x^m\right )}{3 e^2}+\frac {d m x^3}{9 e^2}+\frac {x^4 \log \left (f x^m\right )}{4 e}-\frac {m x^4}{16 e}\right )+\frac {1}{16} b e m n \left (\frac {d^4 \log (d+e x)}{e^5}-\frac {d^3 x}{e^4}+\frac {d^2 x^2}{2 e^3}-\frac {d x^3}{3 e^2}+\frac {x^4}{4 e}\right )\) |
(b*e*m*n*(-((d^3*x)/e^4) + (d^2*x^2)/(2*e^3) - (d*x^3)/(3*e^2) + x^4/(4*e) + (d^4*Log[d + e*x])/e^5))/16 - ((m*x^4 - 4*x^4*Log[f*x^m])*(a + b*Log[c* (d + e*x)^n]))/16 - (b*e*n*((d^3*m*x)/e^4 - (d^2*m*x^2)/(4*e^3) + (d*m*x^3 )/(9*e^2) - (m*x^4)/(16*e) - (d^3*x*Log[f*x^m])/e^4 + (d^2*x^2*Log[f*x^m]) /(2*e^3) - (d*x^3*Log[f*x^m])/(3*e^2) + (x^4*Log[f*x^m])/(4*e) + (d^4*Log[ f*x^m]*Log[1 + (e*x)/d])/e^5 + (d^4*m*PolyLog[2, -((e*x)/d)])/e^5))/4
3.4.58.3.1 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 = 78.15 (sec) , antiderivative size = 1180, normalized size of antiderivative = 5.09
-205/576*b*d^4*m*n/e^4-1/32*I*n*b*x^4*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/32*I* n*b*x^4*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)*(1/4*(-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^4+1/2*x^4*ln(x^m)-1/8*m*x^4)-1/8/e^2*n*b*ln(x^m)*x^2*d ^2+1/4/e^3*n*b*ln(x^m)*x*d^3-1/4/e^4*n*b*ln(x^m)*d^4*ln(e*x+d)+1/12/e*n*b* d*x^3*ln(f)-1/8/e^2*n*b*d^2*x^2*ln(f)+1/4/e^3*n*b*d^3*x*ln(f)-1/4/e^4*n*b* d^4*ln(e*x+d)*ln(f)+1/32*I*n*b*x^4*Pi*csgn(I*f*x^m)^3+1/4*m/e^4*b*d^4*n*di log(-e*x/d)-1/16*n*b*ln(x^m)*x^4-1/16*n*b*x^4*ln(f)-1/16*I/e^2*n*b*d^2*x^2 *Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/8*I/e^3*n*b*d^3*x*Pi*csgn(I*f)*csgn(I*f* x^m)^2+1/8*I/e^3*n*b*d^3*x*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/8*I/e^4*n*b*d^ 4*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2+(1/4*b*x^4*ln(x^m)+1/16*b*x^4*(-2 *I*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+2*I*Pi*csgn(I*f)*csgn(I*f*x^m)^2 +2*I*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-2*I*Pi*csgn(I*f*x^m)^3+4*ln(f)-m))*ln( (e*x+d)^n)-1/24*I/e*n*b*d*x^3*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/16* I/e^2*n*b*d^2*x^2*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/8*I/e^3*n*b*d^3 *x*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/8*I/e^4*n*b*d^4*ln(e*x+d)*Pi*c sgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/8*I/e^4*n*b*d^4*ln(e*x+d)*Pi*csgn(...
\[ \int x^3 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3} \log \left (f x^{m}\right ) \,d x } \]
Timed out. \[ \int x^3 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.00 \[ \int x^3 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {1}{288} \, {\left (\frac {72 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {e x + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x + d}{d}\right )\right )} b d^{4} n}{e^{4}} - \frac {18 \, b e^{4} x^{4} \log \left ({\left (e x + d\right )}^{n}\right ) + 14 \, b d e^{3} n x^{3} - 27 \, b d^{2} e^{2} n x^{2} + 90 \, b d^{3} e n x - 18 \, b d^{4} n \log \left (e x + d\right ) + 9 \, {\left (2 \, a e^{4} - {\left (e^{4} n - 2 \, e^{4} \log \left (c\right )\right )} b\right )} x^{4}}{e^{4}}\right )} m + \frac {1}{48} \, {\left (12 \, b x^{4} \log \left ({\left (e x + d\right )}^{n} c\right ) + 12 \, a x^{4} - b e n {\left (\frac {12 \, d^{4} \log \left (e x + d\right )}{e^{5}} + \frac {3 \, e^{3} x^{4} - 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} - 12 \, d^{3} x}{e^{4}}\right )}\right )} \log \left (f x^{m}\right ) \]
1/288*(72*(log(e*x + d)*log(-(e*x + d)/d + 1) + dilog((e*x + d)/d))*b*d^4* n/e^4 - (18*b*e^4*x^4*log((e*x + d)^n) + 14*b*d*e^3*n*x^3 - 27*b*d^2*e^2*n *x^2 + 90*b*d^3*e*n*x - 18*b*d^4*n*log(e*x + d) + 9*(2*a*e^4 - (e^4*n - 2* e^4*log(c))*b)*x^4)/e^4)*m + 1/48*(12*b*x^4*log((e*x + d)^n*c) + 12*a*x^4 - b*e*n*(12*d^4*log(e*x + d)/e^5 + (3*e^3*x^4 - 4*d*e^2*x^3 + 6*d^2*e*x^2 - 12*d^3*x)/e^4))*log(f*x^m)
\[ \int x^3 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3} \log \left (f x^{m}\right ) \,d x } \]
Timed out. \[ \int x^3 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int x^3\,\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right ) \,d x \]