Integrand size = 23, antiderivative size = 398 \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=-\frac {2 a b n x}{e^4}+\frac {2 b^2 n^2 x}{e^4}-\frac {b^2 d^2 n^2}{3 e^5 (d+e x)}-\frac {b^2 d n^2 \log (x)}{3 e^5}-\frac {2 b^2 n x \log \left (c x^n\right )}{e^4}+\frac {b d^3 n \left (a+b \log \left (c x^n\right )\right )}{3 e^5 (d+e x)^2}+\frac {10 b d n x \left (a+b \log \left (c x^n\right )\right )}{3 e^4 (d+e x)}-\frac {5 d \left (a+b \log \left (c x^n\right )\right )^2}{3 e^5}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{e^4}-\frac {d^4 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^5 (d+e x)^3}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )^2}{e^5 (d+e x)^2}+\frac {6 d x \left (a+b \log \left (c x^n\right )\right )^2}{e^4 (d+e x)}-\frac {3 b^2 d n^2 \log (d+e x)}{e^5}-\frac {26 b d n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 e^5}-\frac {4 d \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^5}-\frac {26 b^2 d n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{3 e^5}-\frac {8 b d n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^5}+\frac {8 b^2 d n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^5} \]
-2*a*b*n*x/e^4+2*b^2*n^2*x/e^4-1/3*b^2*d^2*n^2/e^5/(e*x+d)-1/3*b^2*d*n^2*l n(x)/e^5-2*b^2*n*x*ln(c*x^n)/e^4+1/3*b*d^3*n*(a+b*ln(c*x^n))/e^5/(e*x+d)^2 +10/3*b*d*n*x*(a+b*ln(c*x^n))/e^4/(e*x+d)-5/3*d*(a+b*ln(c*x^n))^2/e^5+x*(a +b*ln(c*x^n))^2/e^4-1/3*d^4*(a+b*ln(c*x^n))^2/e^5/(e*x+d)^3+2*d^3*(a+b*ln( c*x^n))^2/e^5/(e*x+d)^2+6*d*x*(a+b*ln(c*x^n))^2/e^4/(e*x+d)-3*b^2*d*n^2*ln (e*x+d)/e^5-26/3*b*d*n*(a+b*ln(c*x^n))*ln(1+e*x/d)/e^5-4*d*(a+b*ln(c*x^n)) ^2*ln(1+e*x/d)/e^5-26/3*b^2*d*n^2*polylog(2,-e*x/d)/e^5-8*b*d*n*(a+b*ln(c* x^n))*polylog(2,-e*x/d)/e^5+8*b^2*d*n^2*polylog(3,-e*x/d)/e^5
Time = 0.44 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.86 \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=-\frac {-\frac {b d^3 n \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {10 b d^2 n \left (a+b \log \left (c x^n\right )\right )}{d+e x}-13 d \left (a+b \log \left (c x^n\right )\right )^2-3 e x \left (a+b \log \left (c x^n\right )\right )^2+\frac {d^4 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3}-\frac {6 d^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2}+\frac {18 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}+6 b e n x \left (a-b n+b \log \left (c x^n\right )\right )-10 b^2 d n^2 (\log (x)-\log (d+e x))+\frac {b^2 d n^2 (d+(d+e x) \log (x)-(d+e x) \log (d+e x))}{d+e x}+26 b d n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+12 d \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )+26 b^2 d n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+24 b d n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )-24 b^2 d n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{3 e^5} \]
-1/3*(-((b*d^3*n*(a + b*Log[c*x^n]))/(d + e*x)^2) + (10*b*d^2*n*(a + b*Log [c*x^n]))/(d + e*x) - 13*d*(a + b*Log[c*x^n])^2 - 3*e*x*(a + b*Log[c*x^n]) ^2 + (d^4*(a + b*Log[c*x^n])^2)/(d + e*x)^3 - (6*d^3*(a + b*Log[c*x^n])^2) /(d + e*x)^2 + (18*d^2*(a + b*Log[c*x^n])^2)/(d + e*x) + 6*b*e*n*x*(a - b* n + b*Log[c*x^n]) - 10*b^2*d*n^2*(Log[x] - Log[d + e*x]) + (b^2*d*n^2*(d + (d + e*x)*Log[x] - (d + e*x)*Log[d + e*x]))/(d + e*x) + 26*b*d*n*(a + b*L og[c*x^n])*Log[1 + (e*x)/d] + 12*d*(a + b*Log[c*x^n])^2*Log[1 + (e*x)/d] + 26*b^2*d*n^2*PolyLog[2, -((e*x)/d)] + 24*b*d*n*(a + b*Log[c*x^n])*PolyLog [2, -((e*x)/d)] - 24*b^2*d*n^2*PolyLog[3, -((e*x)/d)])/e^5
Time = 0.99 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.08, 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.
| \(\displaystyle \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \int \left (\frac {d^4 \left (a+b \log \left (c x^n\right )\right )^2}{e^4 (d+e x)^4}-\frac {4 d^3 \left (a+b \log \left (c x^n\right )\right )^2}{e^4 (d+e x)^3}+\frac {6 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{e^4 (d+e x)^2}-\frac {4 d \left (a+b \log \left (c x^n\right )\right )^2}{e^4 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{e^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^4 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^5 (d+e x)^3}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )^2}{e^5 (d+e x)^2}+\frac {b d^3 n \left (a+b \log \left (c x^n\right )\right )}{3 e^5 (d+e x)^2}-\frac {8 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^5}+\frac {10 b d n \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^5}-\frac {4 d \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e^5}-\frac {12 b d n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^5}+\frac {6 d x \left (a+b \log \left (c x^n\right )\right )^2}{e^4 (d+e x)}+\frac {10 b d n x \left (a+b \log \left (c x^n\right )\right )}{3 e^4 (d+e x)}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{e^4}-\frac {2 a b n x}{e^4}-\frac {2 b^2 n x \log \left (c x^n\right )}{e^4}-\frac {b^2 d^2 n^2}{3 e^5 (d+e x)}-\frac {10 b^2 d n^2 \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{3 e^5}-\frac {12 b^2 d n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^5}+\frac {8 b^2 d n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^5}-\frac {b^2 d n^2 \log (x)}{3 e^5}-\frac {3 b^2 d n^2 \log (d+e x)}{e^5}+\frac {2 b^2 n^2 x}{e^4}\) |
(-2*a*b*n*x)/e^4 + (2*b^2*n^2*x)/e^4 - (b^2*d^2*n^2)/(3*e^5*(d + e*x)) - ( b^2*d*n^2*Log[x])/(3*e^5) - (2*b^2*n*x*Log[c*x^n])/e^4 + (b*d^3*n*(a + b*L og[c*x^n]))/(3*e^5*(d + e*x)^2) + (10*b*d*n*x*(a + b*Log[c*x^n]))/(3*e^4*( d + e*x)) + (10*b*d*n*Log[1 + d/(e*x)]*(a + b*Log[c*x^n]))/(3*e^5) + (x*(a + b*Log[c*x^n])^2)/e^4 - (d^4*(a + b*Log[c*x^n])^2)/(3*e^5*(d + e*x)^3) + (2*d^3*(a + b*Log[c*x^n])^2)/(e^5*(d + e*x)^2) + (6*d*x*(a + b*Log[c*x^n] )^2)/(e^4*(d + e*x)) - (3*b^2*d*n^2*Log[d + e*x])/e^5 - (12*b*d*n*(a + b*L og[c*x^n])*Log[1 + (e*x)/d])/e^5 - (4*d*(a + b*Log[c*x^n])^2*Log[1 + (e*x) /d])/e^5 - (10*b^2*d*n^2*PolyLog[2, -(d/(e*x))])/(3*e^5) - (12*b^2*d*n^2*P olyLog[2, -((e*x)/d)])/e^5 - (8*b*d*n*(a + b*Log[c*x^n])*PolyLog[2, -((e*x )/d)])/e^5 + (8*b^2*d*n^2*PolyLog[3, -((e*x)/d)])/e^5
3.2.13.3.1 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 = 0.56 (sec) , antiderivative size = 943, normalized size of antiderivative = 2.37
-6*b^2*ln(x^n)^2/e^5*d^2/(e*x+d)+2*b^2*ln(x^n)^2/e^5*d^3/(e*x+d)^2-2*b^2*n *ln(x^n)*x/e^4-13/3*b^2/e^5*n^2*d*ln(x)^2+26/3*b^2/e^5*n^2*dilog(-e*x/d)*d -1/3*b^2*ln(x^n)^2*d^4/e^5/(e*x+d)^3-4*b^2*ln(x^n)^2/e^5*d*ln(e*x+d)-8*b^2 /e^5*d*n^2*ln(x)*polylog(2,-e*x/d)+1/3*b^2*n*ln(x^n)/e^5*d^3/(e*x+d)^2-26/ 3*b^2*n*ln(x^n)/e^5*d*ln(e*x+d)-10/3*b^2*n*ln(x^n)/e^5*d^2/(e*x+d)+(-I*b*P i*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)*b*(l n(x^n)*x/e^4-1/3*ln(x^n)*d^4/e^5/(e*x+d)^3-4*ln(x^n)/e^5*d*ln(e*x+d)-6*ln( x^n)/e^5*d^2/(e*x+d)+2*ln(x^n)/e^5*d^3/(e*x+d)^2-1/3*n*(1/e^5*(3*e*x+3*d-1 /2*d^3/(e*x+d)^2+13*d*ln(e*x+d)+5*d^2/(e*x+d)-13*d*ln(e*x))-12/e^5*d*(dilo g(-e*x/d)+ln(e*x+d)*ln(-e*x/d))))-1/3*b^2*d^2*n^2/e^5/(e*x+d)+3*b^2*d*n^2* ln(x)/e^5-3*b^2*d*n^2*ln(e*x+d)/e^5+8*b^2*d*n^2*polylog(3,-e*x/d)/e^5+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*(x/e^4-1/3/e^5*d^4/(e*x+d)^3-4/e^5*d*ln(e*x+d)-6/e^5*d^2/(e*x+d)+2/e^ 5*d^3/(e*x+d)^2)+b^2*ln(x^n)^2*x/e^4+26/3*b^2*n/e^5*ln(x)*ln(x^n)*d+26/3*b ^2/e^5*n^2*ln(e*x+d)*ln(-e*x/d)*d-8*b^2/e^5*d*ln(x)*dilog(-e*x/d)*n^2+8*b^ 2*n/e^5*d*ln(x^n)*dilog(-e*x/d)+4*b^2/e^5*d*n^2*ln(e*x+d)*ln(x)^2-4*b^2/e^ 5*d*n^2*ln(x)^2*ln(1+e*x/d)-8*b^2/e^5*d*ln(x)*ln(e*x+d)*ln(-e*x/d)*n^2+8*b ^2*n/e^5*d*ln(x^n)*ln(e*x+d)*ln(-e*x/d)+2*b^2*n^2*x/e^4
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{4}}{{\left (e x + d\right )}^{4}} \,d x } \]
integral((b^2*x^4*log(c*x^n)^2 + 2*a*b*x^4*log(c*x^n) + a^2*x^4)/(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)
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int \frac {x^{4} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x\right )^{4}}\, dx \]
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{4}}{{\left (e x + d\right )}^{4}} \,d x } \]
-1/3*a^2*((18*d^2*e^2*x^2 + 30*d^3*e*x + 13*d^4)/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5) - 3*x/e^4 + 12*d*log(e*x + d)/e^5) + integrate((b^2 *x^4*log(x^n)^2 + 2*(b^2*log(c) + a*b)*x^4*log(x^n) + (b^2*log(c)^2 + 2*a* b*log(c))*x^4)/(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)
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{4}}{{\left (e x + d\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int \frac {x^4\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x\right )}^4} \,d x \]