Integrand size = 22, antiderivative size = 158 \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {3 b d m n x}{4 e}+\frac {1}{4} b m n x^2+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 e^2}-\frac {b d^2 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 e^2} \]
-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
Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.04 \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {\log \left (f x^m\right ) \left (-2 b d^2 n \log (d+e x)+e x \left (2 b d n+2 a e x-b e n x+2 b e x \log \left (c (d+e x)^n\right )\right )\right )+m \left (-3 b d e n x-a e^2 x^2+b e^2 n x^2+b d^2 n (1+2 \log (x)) \log (d+e x)-b e^2 x^2 \log \left (c (d+e x)^n\right )-2 b d^2 n \log (x) \log \left (1+\frac {e x}{d}\right )\right )-2 b d^2 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 e^2} \]
(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)
Time = 0.41 (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 |
| \(\displaystyle -\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{2} b e n \left (\frac {d^2 \log \left (\frac {e x}{d}+1\right ) \log \left (f x^m\right )}{e^3}+\frac {d^2 m \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3}-\frac {d x \log \left (f x^m\right )}{e^2}+\frac {d m x}{e^2}+\frac {x^2 \log \left (f x^m\right )}{2 e}-\frac {m x^2}{4 e}\right )+\frac {1}{4} b e m n \left (\frac {d^2 \log (d+e x)}{e^3}-\frac {d x}{e^2}+\frac {x^2}{2 e}\right )\) |
(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
3.4.60.3.1 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 = 13.04 (sec) , antiderivative size = 843, normalized size of antiderivative = 5.34
-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*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)-1/2/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)-1/2/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)*P i*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)*csg n(I*f*x^m)^2-1/8*I*n*b*x^2*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)+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)*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+1/2/e^2*n*b*m*d^2*ln(e*x +d)*ln(-e*x/d)+1/4*b*m*n*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*csgn(I*f*x^m)^3+2*ln(f)-m))*ln((e*x+d)^n)-1/8*I*n*b* x^2*Pi*csgn(I*f)*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)*(1/2*(-I*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+I*Pi*csg n(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)-1/4*n*b*x^2*ln(f)-1/4*n*b*ln(x^m)*x ^2-5/8/e^2*n*b*m*d^2+1/2/e*n*b*d*x*ln(f)+1/2/e*n*b*ln(x^m)*d*x-3/4*b*d*m*n *x/e+1/4*b*d^2*m*n*ln(e*x+d)/e^2
\[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x \log \left (f x^{m}\right ) \,d x } \]
Timed out. \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\text {Timed out} \]
Time = 0.24 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.13 \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {1}{4} \, {\left (\frac {2 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {e x + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x + d}{d}\right )\right )} b d^{2} n}{e^{2}} - \frac {b e^{2} x^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + 3 \, b d e n x - b d^{2} n \log \left (e x + d\right ) + {\left (a e^{2} - {\left (e^{2} n - e^{2} \log \left (c\right )\right )} b\right )} x^{2}}{e^{2}}\right )} m - \frac {1}{4} \, {\left (b e n {\left (\frac {2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac {e x^{2} - 2 \, d x}{e^{2}}\right )} - 2 \, b x^{2} \log \left ({\left (e x + d\right )}^{n} c\right ) - 2 \, a x^{2}\right )} \log \left (f x^{m}\right ) \]
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)
\[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x \log \left (f x^{m}\right ) \,d x } \]
Timed out. \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int x\,\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right ) \,d x \]