Integrand size = 23, antiderivative size = 333 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\frac {b^2 d n^2}{3 e^4 (d+e x)}+\frac {b^2 n^2 \log (x)}{3 e^4}-\frac {b d^2 n \left (a+b \log \left (c x^n\right )\right )}{3 e^4 (d+e x)^2}-\frac {7 b n x \left (a+b \log \left (c x^n\right )\right )}{3 e^3 (d+e x)}+\frac {7 \left (a+b \log \left (c x^n\right )\right )^2}{6 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^4 (d+e x)^3}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^4 (d+e x)^2}-\frac {3 x \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)}+\frac {2 b^2 n^2 \log (d+e x)}{e^4}+\frac {11 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 e^4}+\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^4}+\frac {11 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{3 e^4}+\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}-\frac {2 b^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^4} \]
1/3*b^2*d*n^2/e^4/(e*x+d)+1/3*b^2*n^2*ln(x)/e^4-1/3*b*d^2*n*(a+b*ln(c*x^n) )/e^4/(e*x+d)^2-7/3*b*n*x*(a+b*ln(c*x^n))/e^3/(e*x+d)+7/6*(a+b*ln(c*x^n))^ 2/e^4+1/3*d^3*(a+b*ln(c*x^n))^2/e^4/(e*x+d)^3-3/2*d^2*(a+b*ln(c*x^n))^2/e^ 4/(e*x+d)^2-3*x*(a+b*ln(c*x^n))^2/e^3/(e*x+d)+2*b^2*n^2*ln(e*x+d)/e^4+11/3 *b*n*(a+b*ln(c*x^n))*ln(1+e*x/d)/e^4+(a+b*ln(c*x^n))^2*ln(1+e*x/d)/e^4+11/ 3*b^2*n^2*polylog(2,-e*x/d)/e^4+2*b*n*(a+b*ln(c*x^n))*polylog(2,-e*x/d)/e^ 4-2*b^2*n^2*polylog(3,-e*x/d)/e^4
Time = 0.32 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.89 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\frac {-\frac {2 b d^2 n \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {14 b d n \left (a+b \log \left (c x^n\right )\right )}{d+e x}-11 \left (a+b \log \left (c x^n\right )\right )^2+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3}-\frac {9 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2}+\frac {18 d \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}-14 b^2 n^2 (\log (x)-\log (d+e x))+\frac {2 b^2 n^2 (d+(d+e x) \log (x)-(d+e x) \log (d+e x))}{d+e x}+22 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+6 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )+22 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+12 b n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )-12 b^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{6 e^4} \]
((-2*b*d^2*n*(a + b*Log[c*x^n]))/(d + e*x)^2 + (14*b*d*n*(a + b*Log[c*x^n] ))/(d + e*x) - 11*(a + b*Log[c*x^n])^2 + (2*d^3*(a + b*Log[c*x^n])^2)/(d + e*x)^3 - (9*d^2*(a + b*Log[c*x^n])^2)/(d + e*x)^2 + (18*d*(a + b*Log[c*x^ n])^2)/(d + e*x) - 14*b^2*n^2*(Log[x] - Log[d + e*x]) + (2*b^2*n^2*(d + (d + e*x)*Log[x] - (d + e*x)*Log[d + e*x]))/(d + e*x) + 22*b*n*(a + b*Log[c* x^n])*Log[1 + (e*x)/d] + 6*(a + b*Log[c*x^n])^2*Log[1 + (e*x)/d] + 22*b^2* n^2*PolyLog[2, -((e*x)/d)] + 12*b*n*(a + b*Log[c*x^n])*PolyLog[2, -((e*x)/ d)] - 12*b^2*n^2*PolyLog[3, -((e*x)/d)])/(6*e^4)
Time = 0.92 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.09, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2795, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \int \left (-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^4}+\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^3}-\frac {3 d \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^4 (d+e x)^3}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^4 (d+e x)^2}-\frac {b d^2 n \left (a+b \log \left (c x^n\right )\right )}{3 e^4 (d+e x)^2}+\frac {2 b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {7 b n \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {\log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e^4}+\frac {6 b n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {3 x \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)}-\frac {7 b n x \left (a+b \log \left (c x^n\right )\right )}{3 e^3 (d+e x)}+\frac {7 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{3 e^4}+\frac {6 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}-\frac {2 b^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^4}+\frac {b^2 d n^2}{3 e^4 (d+e x)}+\frac {2 b^2 n^2 \log (d+e x)}{e^4}+\frac {b^2 n^2 \log (x)}{3 e^4}\) |
(b^2*d*n^2)/(3*e^4*(d + e*x)) + (b^2*n^2*Log[x])/(3*e^4) - (b*d^2*n*(a + b *Log[c*x^n]))/(3*e^4*(d + e*x)^2) - (7*b*n*x*(a + b*Log[c*x^n]))/(3*e^3*(d + e*x)) - (7*b*n*Log[1 + d/(e*x)]*(a + b*Log[c*x^n]))/(3*e^4) + (d^3*(a + b*Log[c*x^n])^2)/(3*e^4*(d + e*x)^3) - (3*d^2*(a + b*Log[c*x^n])^2)/(2*e^ 4*(d + e*x)^2) - (3*x*(a + b*Log[c*x^n])^2)/(e^3*(d + e*x)) + (2*b^2*n^2*L og[d + e*x])/e^4 + (6*b*n*(a + b*Log[c*x^n])*Log[1 + (e*x)/d])/e^4 + ((a + b*Log[c*x^n])^2*Log[1 + (e*x)/d])/e^4 + (7*b^2*n^2*PolyLog[2, -(d/(e*x))] )/(3*e^4) + (6*b^2*n^2*PolyLog[2, -((e*x)/d)])/e^4 + (2*b*n*(a + b*Log[c*x ^n])*PolyLog[2, -((e*x)/d)])/e^4 - (2*b^2*n^2*PolyLog[3, -((e*x)/d)])/e^4
3.2.14.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[ c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b , c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0 ] && IntegerQ[m] && IntegerQ[r]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.60 (sec) , antiderivative size = 854, normalized size of antiderivative = 2.56
1/3*b^2*ln(x^n)^2/e^4*d^3/(e*x+d)^3+b^2*ln(x^n)^2/e^4*ln(e*x+d)+3*b^2*ln(x ^n)^2/e^4*d/(e*x+d)-3/2*b^2*ln(x^n)^2/e^4*d^2/(e*x+d)^2+7/3*b^2*n*ln(x^n)/ e^4*d/(e*x+d)-1/3*b^2*n*ln(x^n)/e^4*d^2/(e*x+d)^2+11/3*b^2*n*ln(x^n)/e^4*l n(e*x+d)-11/3*b^2*n/e^4*ln(x^n)*ln(x)+11/6*b^2/e^4*n^2*ln(x)^2-11/3*b^2/e^ 4*n^2*ln(e*x+d)*ln(-e*x/d)-11/3*b^2/e^4*n^2*dilog(-e*x/d)+1/3*b^2*d*n^2/e^ 4/(e*x+d)+2*b^2*n^2*ln(e*x+d)/e^4-2*b^2*n^2*ln(x)/e^4+2*b^2/e^4*ln(x)*ln(e *x+d)*ln(-e*x/d)*n^2+2*b^2/e^4*ln(x)*dilog(-e*x/d)*n^2-2*b^2*n/e^4*ln(x^n) *ln(e*x+d)*ln(-e*x/d)-2*b^2*n/e^4*ln(x^n)*dilog(-e*x/d)-b^2/e^4*n^2*ln(e*x +d)*ln(x)^2+b^2/e^4*n^2*ln(x)^2*ln(1+e*x/d)+2*b^2/e^4*n^2*ln(x)*polylog(2, -e*x/d)-2*b^2*n^2*polylog(3,-e*x/d)/e^4+(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csg n(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)*b*(1/3*ln(x^n)/e^4*d^3/(e*x+d)^ 3+ln(x^n)/e^4*ln(e*x+d)+3*ln(x^n)/e^4*d/(e*x+d)-3/2*ln(x^n)/e^4*d^2/(e*x+d )^2-1/6*n*(-7/e^4*d/(e*x+d)-11/e^4*ln(e*x+d)+1/e^4*d^2/(e*x+d)^2+11/e^4*ln (e*x)+6/e^4*ln(e*x+d)*ln(-e*x/d)+6/e^4*dilog(-e*x/d)))+1/4*(-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*b*ln(c)+2*a)^2*(1/3/e^4*d ^3/(e*x+d)^3+1/e^4*ln(e*x+d)+3/e^4*d/(e*x+d)-3/2/e^4*d^2/(e*x+d)^2)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{3}}{{\left (e x + d\right )}^{4}} \,d x } \]
integral((b^2*x^3*log(c*x^n)^2 + 2*a*b*x^3*log(c*x^n) + a^2*x^3)/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int \frac {x^{3} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x\right )^{4}}\, dx \]
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{3}}{{\left (e x + d\right )}^{4}} \,d x } \]
1/6*a^2*((18*d*e^2*x^2 + 27*d^2*e*x + 11*d^3)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d ^2*e^5*x + d^3*e^4) + 6*log(e*x + d)/e^4) + integrate((b^2*x^3*log(x^n)^2 + 2*(b^2*log(c) + a*b)*x^3*log(x^n) + (b^2*log(c)^2 + 2*a*b*log(c))*x^3)/( e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{3}}{{\left (e x + d\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int \frac {x^3\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x\right )}^4} \,d x \]