Integrand size = 24, antiderivative size = 373 \[ \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right ) \, dx=-\frac {3 a b e m n x}{2 f}+\frac {7 b^2 e m n^2 x}{4 f}-\frac {3}{8} b^2 m n^2 x^2-\frac {3 b^2 e m n x \log \left (c x^n\right )}{2 f}+\frac {1}{2} b m n x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {e m x \left (a+b \log \left (c x^n\right )\right )^2}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )^2-\frac {b^2 e^2 m n^2 \log (e+f x)}{4 f^2}+\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )+\frac {b e^2 m n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {f x}{e}\right )}{2 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x}{e}\right )}{2 f^2}+\frac {b^2 e^2 m n^2 \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{2 f^2}-\frac {b e^2 m n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{f^2}+\frac {b^2 e^2 m n^2 \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right )}{f^2} \]
-3/2*a*b*e*m*n*x/f+7/4*b^2*e*m*n^2*x/f-3/8*b^2*m*n^2*x^2-3/2*b^2*e*m*n*x*l n(c*x^n)/f+1/2*b*m*n*x^2*(a+b*ln(c*x^n))+1/2*e*m*x*(a+b*ln(c*x^n))^2/f-1/4 *m*x^2*(a+b*ln(c*x^n))^2-1/4*b^2*e^2*m*n^2*ln(f*x+e)/f^2+1/4*b^2*n^2*x^2*l n(d*(f*x+e)^m)-1/2*b*n*x^2*(a+b*ln(c*x^n))*ln(d*(f*x+e)^m)+1/2*x^2*(a+b*ln (c*x^n))^2*ln(d*(f*x+e)^m)+1/2*b*e^2*m*n*(a+b*ln(c*x^n))*ln(1+f*x/e)/f^2-1 /2*e^2*m*(a+b*ln(c*x^n))^2*ln(1+f*x/e)/f^2+1/2*b^2*e^2*m*n^2*polylog(2,-f* x/e)/f^2-b*e^2*m*n*(a+b*ln(c*x^n))*polylog(2,-f*x/e)/f^2+b^2*e^2*m*n^2*pol ylog(3,-f*x/e)/f^2
Time = 0.17 (sec) , antiderivative size = 674, normalized size of antiderivative = 1.81 \[ \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right ) \, dx=\frac {4 a^2 e f m x-12 a b e f m n x+14 b^2 e f m n^2 x-2 a^2 f^2 m x^2+4 a b f^2 m n x^2-3 b^2 f^2 m n^2 x^2+8 a b e f m x \log \left (c x^n\right )-12 b^2 e f m n x \log \left (c x^n\right )-4 a b f^2 m x^2 \log \left (c x^n\right )+4 b^2 f^2 m n x^2 \log \left (c x^n\right )+4 b^2 e f m x \log ^2\left (c x^n\right )-2 b^2 f^2 m x^2 \log ^2\left (c x^n\right )-4 a^2 e^2 m \log (e+f x)+4 a b e^2 m n \log (e+f x)-2 b^2 e^2 m n^2 \log (e+f x)+8 a b e^2 m n \log (x) \log (e+f x)-4 b^2 e^2 m n^2 \log (x) \log (e+f x)-4 b^2 e^2 m n^2 \log ^2(x) \log (e+f x)-8 a b e^2 m \log \left (c x^n\right ) \log (e+f x)+4 b^2 e^2 m n \log \left (c x^n\right ) \log (e+f x)+8 b^2 e^2 m n \log (x) \log \left (c x^n\right ) \log (e+f x)-4 b^2 e^2 m \log ^2\left (c x^n\right ) \log (e+f x)+4 a^2 f^2 x^2 \log \left (d (e+f x)^m\right )-4 a b f^2 n x^2 \log \left (d (e+f x)^m\right )+2 b^2 f^2 n^2 x^2 \log \left (d (e+f x)^m\right )+8 a b f^2 x^2 \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )-4 b^2 f^2 n x^2 \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )+4 b^2 f^2 x^2 \log ^2\left (c x^n\right ) \log \left (d (e+f x)^m\right )-8 a b e^2 m n \log (x) \log \left (1+\frac {f x}{e}\right )+4 b^2 e^2 m n^2 \log (x) \log \left (1+\frac {f x}{e}\right )+4 b^2 e^2 m n^2 \log ^2(x) \log \left (1+\frac {f x}{e}\right )-8 b^2 e^2 m n \log (x) \log \left (c x^n\right ) \log \left (1+\frac {f x}{e}\right )+4 b e^2 m n \left (-2 a+b n-2 b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )+8 b^2 e^2 m n^2 \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right )}{8 f^2} \]
(4*a^2*e*f*m*x - 12*a*b*e*f*m*n*x + 14*b^2*e*f*m*n^2*x - 2*a^2*f^2*m*x^2 + 4*a*b*f^2*m*n*x^2 - 3*b^2*f^2*m*n^2*x^2 + 8*a*b*e*f*m*x*Log[c*x^n] - 12*b ^2*e*f*m*n*x*Log[c*x^n] - 4*a*b*f^2*m*x^2*Log[c*x^n] + 4*b^2*f^2*m*n*x^2*L og[c*x^n] + 4*b^2*e*f*m*x*Log[c*x^n]^2 - 2*b^2*f^2*m*x^2*Log[c*x^n]^2 - 4* a^2*e^2*m*Log[e + f*x] + 4*a*b*e^2*m*n*Log[e + f*x] - 2*b^2*e^2*m*n^2*Log[ e + f*x] + 8*a*b*e^2*m*n*Log[x]*Log[e + f*x] - 4*b^2*e^2*m*n^2*Log[x]*Log[ e + f*x] - 4*b^2*e^2*m*n^2*Log[x]^2*Log[e + f*x] - 8*a*b*e^2*m*Log[c*x^n]* Log[e + f*x] + 4*b^2*e^2*m*n*Log[c*x^n]*Log[e + f*x] + 8*b^2*e^2*m*n*Log[x ]*Log[c*x^n]*Log[e + f*x] - 4*b^2*e^2*m*Log[c*x^n]^2*Log[e + f*x] + 4*a^2* f^2*x^2*Log[d*(e + f*x)^m] - 4*a*b*f^2*n*x^2*Log[d*(e + f*x)^m] + 2*b^2*f^ 2*n^2*x^2*Log[d*(e + f*x)^m] + 8*a*b*f^2*x^2*Log[c*x^n]*Log[d*(e + f*x)^m] - 4*b^2*f^2*n*x^2*Log[c*x^n]*Log[d*(e + f*x)^m] + 4*b^2*f^2*x^2*Log[c*x^n ]^2*Log[d*(e + f*x)^m] - 8*a*b*e^2*m*n*Log[x]*Log[1 + (f*x)/e] + 4*b^2*e^2 *m*n^2*Log[x]*Log[1 + (f*x)/e] + 4*b^2*e^2*m*n^2*Log[x]^2*Log[1 + (f*x)/e] - 8*b^2*e^2*m*n*Log[x]*Log[c*x^n]*Log[1 + (f*x)/e] + 4*b*e^2*m*n*(-2*a + b*n - 2*b*Log[c*x^n])*PolyLog[2, -((f*x)/e)] + 8*b^2*e^2*m*n^2*PolyLog[3, -((f*x)/e)])/(8*f^2)
Time = 0.72 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2825, 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.
| \(\displaystyle \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right ) \, dx\) |
| \(\Big \downarrow \) 2825 |
| \(\displaystyle -f m \int \left (\frac {\left (a+b \log \left (c x^n\right )\right )^2 x^2}{2 (e+f x)}-\frac {b n \left (a+b \log \left (c x^n\right )\right ) x^2}{2 (e+f x)}+\frac {b^2 n^2 x^2}{4 (e+f x)}\right )dx+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -f m \left (\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right ) \left (a+b \log \left (c x^n\right )\right )}{f^3}-\frac {b e^2 n \log \left (\frac {f x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 f^3}+\frac {e^2 \log \left (\frac {f x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{2 f^3}-\frac {e x \left (a+b \log \left (c x^n\right )\right )^2}{2 f^2}-\frac {b n x^2 \left (a+b \log \left (c x^n\right )\right )}{2 f}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{4 f}+\frac {3 a b e n x}{2 f^2}+\frac {3 b^2 e n x \log \left (c x^n\right )}{2 f^2}-\frac {b^2 e^2 n^2 \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{2 f^3}-\frac {b^2 e^2 n^2 \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right )}{f^3}+\frac {b^2 e^2 n^2 \log (e+f x)}{4 f^3}-\frac {7 b^2 e n^2 x}{4 f^2}+\frac {3 b^2 n^2 x^2}{8 f}\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )\) |
(b^2*n^2*x^2*Log[d*(e + f*x)^m])/4 - (b*n*x^2*(a + b*Log[c*x^n])*Log[d*(e + f*x)^m])/2 + (x^2*(a + b*Log[c*x^n])^2*Log[d*(e + f*x)^m])/2 - f*m*((3*a *b*e*n*x)/(2*f^2) - (7*b^2*e*n^2*x)/(4*f^2) + (3*b^2*n^2*x^2)/(8*f) + (3*b ^2*e*n*x*Log[c*x^n])/(2*f^2) - (b*n*x^2*(a + b*Log[c*x^n]))/(2*f) - (e*x*( a + b*Log[c*x^n])^2)/(2*f^2) + (x^2*(a + b*Log[c*x^n])^2)/(4*f) + (b^2*e^2 *n^2*Log[e + f*x])/(4*f^3) - (b*e^2*n*(a + b*Log[c*x^n])*Log[1 + (f*x)/e]) /(2*f^3) + (e^2*(a + b*Log[c*x^n])^2*Log[1 + (f*x)/e])/(2*f^3) - (b^2*e^2* n^2*PolyLog[2, -((f*x)/e)])/(2*f^3) + (b*e^2*n*(a + b*Log[c*x^n])*PolyLog[ 2, -((f*x)/e)])/f^3 - (b^2*e^2*n^2*PolyLog[3, -((f*x)/e)])/f^3)
3.1.79.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))^(p_.)*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q* (a + b*Log[c*x^n])^p, x]}, Simp[Log[d*(e + f*x^m)^r] u, x] - Simp[f*m*r Int[x^(m - 1)/(e + f*x^m) u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m , n, q}, x] && IGtQ[p, 0] && RationalQ[m] && RationalQ[q]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 77.82 (sec) , antiderivative size = 4845, normalized size of antiderivative = 12.99
1/2*m/f*e*x*ln(c)^2*b^2+m/f*b*ln(x^n)*e*x*a-5/4*m/f^2*b*n*e^2*a-1/2*m*ln(x ^n)*x^2*b^2*ln(c)+1/2*m*n*x^2*b^2*ln(c)-1/2*m*b*ln(x^n)*x^2*a+1/2*m*b*n*x^ 2*a-1/2*m*a^2*e^2/f^2*ln(f*x+e)+1/16*m*x^2*Pi^2*b^2*csgn(I*c)^2*csgn(I*c*x ^n)^4-1/8*m*x^2*Pi^2*b^2*csgn(I*c)*csgn(I*c*x^n)^5+1/16*m*x^2*Pi^2*b^2*csg n(I*x^n)^2*csgn(I*c*x^n)^4-1/8*m*x^2*Pi^2*b^2*csgn(I*x^n)*csgn(I*c*x^n)^5- m/f^2*b*ln(x^n)*e^2*ln(f*x+e)*a+m/f^2*b*n*e^2*dilog(-f*x/e)*a-m/f^2*b^2*e^ 2*ln(x)*dilog(-f*x/e)*n^2+m/f^2*b^2*n*e^2*ln(x^n)*dilog(-f*x/e)+1/2*m/f^2* b^2*e^2*n^2*ln(f*x+e)*ln(x)^2-1/2*m/f^2*b^2*e^2*n^2*ln(x)^2*ln(1+f*x/e)-m/ f^2*b^2*e^2*n^2*ln(x)*polylog(2,-f*x/e)+1/8*m/f^2*e^2*ln(f*x+e)*Pi^2*b^2*c sgn(I*c*x^n)^6+1/2*m/f^2*e^2*ln(f*x+e)*b^2*ln(c)*n-m/f^2*e^2*ln(f*x+e)*ln( c)*a*b+1/2*m/f^2*e^2*ln(f*x+e)*a*b*n+1/4*I*m*x^2*ln(c)*Pi*b^2*csgn(I*c*x^n )^3-1/4*I*m*x^2*Pi*b^2*n*csgn(I*c*x^n)^3+1/4*I*m*x^2*Pi*a*b*csgn(I*c*x^n)^ 3+1/4*I*m*ln(x^n)*x^2*b^2*Pi*csgn(I*c*x^n)^3+(1/8*I*Pi*csgn(I*(f*x+e)^m)*c sgn(I*d*(f*x+e)^m)^2-1/8*I*Pi*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)*csgn(I *d)-1/8*I*Pi*csgn(I*d*(f*x+e)^m)^3+1/8*I*Pi*csgn(I*d*(f*x+e)^m)^2*csgn(I*d )+1/4*ln(d))*(1/2*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(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)^2*x^2+2*x^2*b^2*ln(x^n)^2-2*x^2*b^2*ln(x^n)*n+b^2*n^2 *x^2+4*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(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*...
\[ \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x \log \left ({\left (f x + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right ) \, dx=\text {Timed out} \]
\[ \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x \log \left ({\left (f x + e\right )}^{m} d\right ) \,d x } \]
1/4*((2*b^2*e*f*m*x - 2*b^2*e^2*m*log(f*x + e) - (f^2*m - 2*f^2*log(d))*b^ 2*x^2)*log(x^n)^2 + (2*b^2*f^2*x^2*log(x^n)^2 + 2*(2*a*b*f^2 - (f^2*n - 2* f^2*log(c))*b^2)*x^2*log(x^n) + (2*a^2*f^2 - 2*(f^2*n - 2*f^2*log(c))*a*b + (f^2*n^2 - 2*f^2*n*log(c) + 2*f^2*log(c)^2)*b^2)*x^2)*log((f*x + e)^m))/ f^2 + integrate(-1/4*((2*(f^3*m - 2*f^3*log(d))*a^2 - 2*(f^3*m*n - 2*(f^3* m - 2*f^3*log(d))*log(c))*a*b + (f^3*m*n^2 - 2*f^3*m*n*log(c) + 2*(f^3*m - 2*f^3*log(d))*log(c)^2)*b^2)*x^3 - 4*(b^2*e*f^2*log(c)^2*log(d) + 2*a*b*e *f^2*log(c)*log(d) + a^2*e*f^2*log(d))*x^2 + 2*(2*b^2*e^2*f*m*n*x + 2*((f^ 3*m - 2*f^3*log(d))*a*b - (f^3*m*n - f^3*n*log(d) - (f^3*m - 2*f^3*log(d)) *log(c))*b^2)*x^3 - (4*a*b*e*f^2*log(d) - (e*f^2*m*n + 2*e*f^2*n*log(d) - 4*e*f^2*log(c)*log(d))*b^2)*x^2 - 2*(b^2*e^2*f*m*n*x + b^2*e^3*m*n)*log(f* x + e))*log(x^n))/(f^3*x^2 + e*f^2*x), x)
\[ \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x \log \left ({\left (f x + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right ) \, dx=\int x\,\ln \left (d\,{\left (e+f\,x\right )}^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2 \,d x \]