Integrand size = 22, antiderivative size = 203 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=-\frac {3 b e m n x}{4 f}+\frac {1}{4} b m n x^2+\frac {e m x \left (a+b \log \left (c x^n\right )\right )}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b e^2 m n \log (e+f x)}{4 f^2}+\frac {b e^2 m n \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{2 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right ) \log (e+f x)}{2 f^2}-\frac {1}{4} b n x^2 \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {b e^2 m n \operatorname {PolyLog}\left (2,1+\frac {f x}{e}\right )}{2 f^2} \]
-3/4*b*e*m*n*x/f+1/4*b*m*n*x^2+1/2*e*m*x*(a+b*ln(c*x^n))/f-1/4*m*x^2*(a+b* ln(c*x^n))+1/4*b*e^2*m*n*ln(f*x+e)/f^2+1/2*b*e^2*m*n*ln(-f*x/e)*ln(f*x+e)/ f^2-1/2*e^2*m*(a+b*ln(c*x^n))*ln(f*x+e)/f^2-1/4*b*n*x^2*ln(d*(f*x+e)^m)+1/ 2*x^2*(a+b*ln(c*x^n))*ln(d*(f*x+e)^m)+1/2*b*e^2*m*n*polylog(2,1+f*x/e)/f^2
Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.02 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=\frac {2 a e f m x-3 b e f m n x-a f^2 m x^2+b f^2 m n x^2-2 a e^2 m \log (e+f x)+b e^2 m n \log (e+f x)+2 b e^2 m n \log (x) \log (e+f x)+2 a f^2 x^2 \log \left (d (e+f x)^m\right )-b f^2 n x^2 \log \left (d (e+f x)^m\right )+b \log \left (c x^n\right ) \left (-2 e^2 m \log (e+f x)+f x \left (2 e m-f m x+2 f x \log \left (d (e+f x)^m\right )\right )\right )-2 b e^2 m n \log (x) \log \left (1+\frac {f x}{e}\right )-2 b e^2 m n \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{4 f^2} \]
(2*a*e*f*m*x - 3*b*e*f*m*n*x - a*f^2*m*x^2 + b*f^2*m*n*x^2 - 2*a*e^2*m*Log [e + f*x] + b*e^2*m*n*Log[e + f*x] + 2*b*e^2*m*n*Log[x]*Log[e + f*x] + 2*a *f^2*x^2*Log[d*(e + f*x)^m] - b*f^2*n*x^2*Log[d*(e + f*x)^m] + b*Log[c*x^n ]*(-2*e^2*m*Log[e + f*x] + f*x*(2*e*m - f*m*x + 2*f*x*Log[d*(e + f*x)^m])) - 2*b*e^2*m*n*Log[x]*Log[1 + (f*x)/e] - 2*b*e^2*m*n*PolyLog[2, -((f*x)/e) ])/(4*f^2)
Time = 0.35 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2823, 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.
\(\displaystyle \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx\) |
\(\Big \downarrow \) 2823 |
\(\displaystyle -b n \int \left (-\frac {m \log (e+f x) e^2}{2 f^2 x}+\frac {m e}{2 f}-\frac {m x}{4}+\frac {1}{2} x \log \left (d (e+f x)^m\right )\right )dx+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )-\frac {e^2 m \log (e+f x) \left (a+b \log \left (c x^n\right )\right )}{2 f^2}+\frac {e m x \left (a+b \log \left (c x^n\right )\right )}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )-\frac {e^2 m \log (e+f x) \left (a+b \log \left (c x^n\right )\right )}{2 f^2}+\frac {e m x \left (a+b \log \left (c x^n\right )\right )}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {1}{4} x^2 \log \left (d (e+f x)^m\right )-\frac {e^2 m \operatorname {PolyLog}\left (2,\frac {f x}{e}+1\right )}{2 f^2}-\frac {e^2 m \log (e+f x)}{4 f^2}-\frac {e^2 m \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{2 f^2}+\frac {3 e m x}{4 f}-\frac {m x^2}{4}\right )\) |
(e*m*x*(a + b*Log[c*x^n]))/(2*f) - (m*x^2*(a + b*Log[c*x^n]))/4 - (e^2*m*( a + b*Log[c*x^n])*Log[e + f*x])/(2*f^2) + (x^2*(a + b*Log[c*x^n])*Log[d*(e + f*x)^m])/2 - b*n*((3*e*m*x)/(4*f) - (m*x^2)/4 - (e^2*m*Log[e + f*x])/(4 *f^2) - (e^2*m*Log[-((f*x)/e)]*Log[e + f*x])/(2*f^2) + (x^2*Log[d*(e + f*x )^m])/4 - (e^2*m*PolyLog[2, 1 + (f*x)/e])/(2*f^2))
3.1.72.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* (e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[1/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 15.08 (sec) , antiderivative size = 885, normalized size of antiderivative = 4.36
1/2*m/f*e*x*b*ln(c)-1/4*I*m/f*e*x*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n) +1/4*I*m/f^2*e^2*ln(f*x+e)*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+(1/4*I *Pi*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2-1/4*I*Pi*csgn(I*(f*x+e)^m)*csg n(I*d*(f*x+e)^m)*csgn(I*d)-1/4*I*Pi*csgn(I*d*(f*x+e)^m)^3+1/4*I*Pi*csgn(I* d*(f*x+e)^m)^2*csgn(I*d)+1/2*ln(d))*(1/2*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*cs gn(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x ^n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)+2*a)*x^2+b*x^2*ln(x^n)-1/2*b*n*x^2) -1/2*m*a*e^2/f^2*ln(f*x+e)+1/8*I*m*x^2*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c *x^n)-1/4*I*m/f*e*x*b*Pi*csgn(I*c*x^n)^3+1/4*I*m/f^2*e^2*ln(f*x+e)*b*Pi*cs gn(I*c*x^n)^3+(1/2*b*x^2*ln(x^n)+1/4*x^2*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*cs gn(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x ^n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)-b*n+2*a))*ln((f*x+e)^m)-1/2*m/f^2*e ^2*ln(f*x+e)*b*ln(c)-1/8*I*m*x^2*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/8*I*m*x^ 2*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/2*m/f*e*x*a+1/2*m/f*b*ln(x^n)*e*x-1/4 *x^2*a*m-5/8*m/f^2*b*n*e^2+1/4*I*m/f*e*x*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/ 4*I*m/f*e*x*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*m/f^2*e^2*ln(f*x+e)*b*P i*csgn(I*c)*csgn(I*c*x^n)^2-1/4*I*m/f^2*e^2*ln(f*x+e)*b*Pi*csgn(I*x^n)*csg n(I*c*x^n)^2+1/2*b*e^2*m*n*ln(-f*x/e)*ln(f*x+e)/f^2+1/4*b*m*n*x^2-3/4*b*e* m*n*x/f+1/4*b*e^2*m*n*ln(f*x+e)/f^2+1/2*n*b/f^2*e^2*m*dilog(-f*x/e)-1/2*m/ f^2*b*ln(x^n)*e^2*ln(f*x+e)+1/8*I*m*x^2*b*Pi*csgn(I*c*x^n)^3-1/4*m*x^2*...
\[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x \log \left ({\left (f x + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.33 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=-\frac {{\left (\log \left (\frac {f x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {f x}{e}\right )\right )} b e^{2} m n}{2 \, f^{2}} - \frac {{\left (2 \, a e^{2} m - {\left (e^{2} m n - 2 \, e^{2} m \log \left (c\right )\right )} b\right )} \log \left (f x + e\right )}{4 \, f^{2}} + \frac {2 \, b e^{2} m n \log \left (f x + e\right ) \log \left (x\right ) - {\left ({\left (f^{2} m - 2 \, f^{2} \log \left (d\right )\right )} a - {\left (f^{2} m n - f^{2} n \log \left (d\right ) - {\left (f^{2} m - 2 \, f^{2} \log \left (d\right )\right )} \log \left (c\right )\right )} b\right )} x^{2} + {\left (2 \, a e f m - {\left (3 \, e f m n - 2 \, e f m \log \left (c\right )\right )} b\right )} x + {\left (2 \, b f^{2} x^{2} \log \left (x^{n}\right ) + {\left (2 \, a f^{2} - {\left (f^{2} n - 2 \, f^{2} \log \left (c\right )\right )} b\right )} x^{2}\right )} \log \left ({\left (f x + e\right )}^{m}\right ) + {\left (2 \, b e f m x - 2 \, b e^{2} m \log \left (f x + e\right ) - {\left (f^{2} m - 2 \, f^{2} \log \left (d\right )\right )} b x^{2}\right )} \log \left (x^{n}\right )}{4 \, f^{2}} \]
-1/2*(log(f*x/e + 1)*log(x) + dilog(-f*x/e))*b*e^2*m*n/f^2 - 1/4*(2*a*e^2* m - (e^2*m*n - 2*e^2*m*log(c))*b)*log(f*x + e)/f^2 + 1/4*(2*b*e^2*m*n*log( f*x + e)*log(x) - ((f^2*m - 2*f^2*log(d))*a - (f^2*m*n - f^2*n*log(d) - (f ^2*m - 2*f^2*log(d))*log(c))*b)*x^2 + (2*a*e*f*m - (3*e*f*m*n - 2*e*f*m*lo g(c))*b)*x + (2*b*f^2*x^2*log(x^n) + (2*a*f^2 - (f^2*n - 2*f^2*log(c))*b)* x^2)*log((f*x + e)^m) + (2*b*e*f*m*x - 2*b*e^2*m*log(f*x + e) - (f^2*m - 2 *f^2*log(d))*b*x^2)*log(x^n))/f^2
\[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x \log \left ({\left (f x + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=\int x\,\ln \left (d\,{\left (e+f\,x\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]