Integrand size = 23, antiderivative size = 420 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^4} \, dx=-\frac {2 b^2 n^2}{d^4 x}-\frac {b^2 e n^2}{3 d^4 (d+e x)}-\frac {b^2 e n^2 \log (x)}{3 d^5}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right )}{d^4 x}+\frac {b e n \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)^2}-\frac {8 b e^2 n x \left (a+b \log \left (c x^n\right )\right )}{3 d^5 (d+e x)}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )^2}{3 d^5}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{d^4 x}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{3 d^2 (d+e x)^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{d^3 (d+e x)^2}+\frac {3 e^2 x \left (a+b \log \left (c x^n\right )\right )^2}{d^5 (d+e x)}-\frac {4 e \left (a+b \log \left (c x^n\right )\right )^3}{3 b d^5 n}+\frac {3 b^2 e n^2 \log (d+e x)}{d^5}-\frac {26 b e n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 d^5}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{d^5}-\frac {26 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{3 d^5}+\frac {8 b e n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^5}-\frac {8 b^2 e n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{d^5} \]
-2*b^2*n^2/d^4/x-1/3*b^2*e*n^2/d^4/(e*x+d)-1/3*b^2*e*n^2*ln(x)/d^5-2*b*n*( a+b*ln(c*x^n))/d^4/x+1/3*b*e*n*(a+b*ln(c*x^n))/d^3/(e*x+d)^2-8/3*b*e^2*n*x *(a+b*ln(c*x^n))/d^5/(e*x+d)+4/3*e*(a+b*ln(c*x^n))^2/d^5-(a+b*ln(c*x^n))^2 /d^4/x-1/3*e*(a+b*ln(c*x^n))^2/d^2/(e*x+d)^3-e*(a+b*ln(c*x^n))^2/d^3/(e*x+ d)^2+3*e^2*x*(a+b*ln(c*x^n))^2/d^5/(e*x+d)-4/3*e*(a+b*ln(c*x^n))^3/b/d^5/n +3*b^2*e*n^2*ln(e*x+d)/d^5-26/3*b*e*n*(a+b*ln(c*x^n))*ln(1+e*x/d)/d^5+4*e* (a+b*ln(c*x^n))^2*ln(1+e*x/d)/d^5-26/3*b^2*e*n^2*polylog(2,-e*x/d)/d^5+8*b *e*n*(a+b*ln(c*x^n))*polylog(2,-e*x/d)/d^5-8*b^2*e*n^2*polylog(3,-e*x/d)/d ^5
Time = 0.45 (sec) , antiderivative size = 378, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^4} \, dx=-\frac {\frac {6 b^2 d n^2}{x}+\frac {6 b d n \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {b d^2 e n \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac {8 b d e n \left (a+b \log \left (c x^n\right )\right )}{d+e x}-13 e \left (a+b \log \left (c x^n\right )\right )^2+\frac {3 d \left (a+b \log \left (c x^n\right )\right )^2}{x}+\frac {d^3 e \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3}+\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2}+\frac {9 d e \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )^3}{b n}+8 b^2 e n^2 (\log (x)-\log (d+e x))+\frac {b^2 e n^2 (d+(d+e x) \log (x)-(d+e x) \log (d+e x))}{d+e x}+26 b e n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-12 e \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )+26 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )-24 b e n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+24 b^2 e n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{3 d^5} \]
-1/3*((6*b^2*d*n^2)/x + (6*b*d*n*(a + b*Log[c*x^n]))/x - (b*d^2*e*n*(a + b *Log[c*x^n]))/(d + e*x)^2 - (8*b*d*e*n*(a + b*Log[c*x^n]))/(d + e*x) - 13* e*(a + b*Log[c*x^n])^2 + (3*d*(a + b*Log[c*x^n])^2)/x + (d^3*e*(a + b*Log[ c*x^n])^2)/(d + e*x)^3 + (3*d^2*e*(a + b*Log[c*x^n])^2)/(d + e*x)^2 + (9*d *e*(a + b*Log[c*x^n])^2)/(d + e*x) + (4*e*(a + b*Log[c*x^n])^3)/(b*n) + 8* b^2*e*n^2*(Log[x] - Log[d + e*x]) + (b^2*e*n^2*(d + (d + e*x)*Log[x] - (d + e*x)*Log[d + e*x]))/(d + e*x) + 26*b*e*n*(a + b*Log[c*x^n])*Log[1 + (e*x )/d] - 12*e*(a + b*Log[c*x^n])^2*Log[1 + (e*x)/d] + 26*b^2*e*n^2*PolyLog[2 , -((e*x)/d)] - 24*b*e*n*(a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)] + 24*b^ 2*e*n^2*PolyLog[3, -((e*x)/d)])/d^5
Time = 1.06 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.03, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \int \left (\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{d^4 (d+e x)^2}-\frac {4 e \left (a+b \log \left (c x^n\right )\right )^2}{d^4 x (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{d^4 x^2}+\frac {2 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{d^3 (d+e x)^3}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )^2}{d^2 (d+e x)^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 e^2 x \left (a+b \log \left (c x^n\right )\right )^2}{d^5 (d+e x)}-\frac {8 b e^2 n x \left (a+b \log \left (c x^n\right )\right )}{3 d^5 (d+e x)}-\frac {8 b e n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^5}+\frac {4 e \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{d^5}-\frac {8 b e n \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^5}-\frac {6 b e n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^5}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{d^4 x}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right )}{d^4 x}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{d^3 (d+e x)^2}+\frac {b e n \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)^2}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{3 d^2 (d+e x)^3}+\frac {8 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{3 d^5}-\frac {6 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^5}-\frac {8 b^2 e n^2 \operatorname {PolyLog}\left (3,-\frac {d}{e x}\right )}{d^5}-\frac {b^2 e n^2 \log (x)}{3 d^5}+\frac {3 b^2 e n^2 \log (d+e x)}{d^5}-\frac {b^2 e n^2}{3 d^4 (d+e x)}-\frac {2 b^2 n^2}{d^4 x}\) |
(-2*b^2*n^2)/(d^4*x) - (b^2*e*n^2)/(3*d^4*(d + e*x)) - (b^2*e*n^2*Log[x])/ (3*d^5) - (2*b*n*(a + b*Log[c*x^n]))/(d^4*x) + (b*e*n*(a + b*Log[c*x^n]))/ (3*d^3*(d + e*x)^2) - (8*b*e^2*n*x*(a + b*Log[c*x^n]))/(3*d^5*(d + e*x)) - (8*b*e*n*Log[1 + d/(e*x)]*(a + b*Log[c*x^n]))/(3*d^5) - (a + b*Log[c*x^n] )^2/(d^4*x) - (e*(a + b*Log[c*x^n])^2)/(3*d^2*(d + e*x)^3) - (e*(a + b*Log [c*x^n])^2)/(d^3*(d + e*x)^2) + (3*e^2*x*(a + b*Log[c*x^n])^2)/(d^5*(d + e *x)) + (4*e*Log[1 + d/(e*x)]*(a + b*Log[c*x^n])^2)/d^5 + (3*b^2*e*n^2*Log[ d + e*x])/d^5 - (6*b*e*n*(a + b*Log[c*x^n])*Log[1 + (e*x)/d])/d^5 + (8*b^2 *e*n^2*PolyLog[2, -(d/(e*x))])/(3*d^5) - (8*b*e*n*(a + b*Log[c*x^n])*PolyL og[2, -(d/(e*x))])/d^5 - (6*b^2*e*n^2*PolyLog[2, -((e*x)/d)])/d^5 - (8*b^2 *e*n^2*PolyLog[3, -(d/(e*x))])/d^5
3.2.19.3.1 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 = 0.66 (sec) , antiderivative size = 1015, normalized size of antiderivative = 2.42
-2*b^2*n*ln(x^n)/d^4/x-3*b^2*ln(x^n)^2/d^4*e/(e*x+d)-b^2*ln(x^n)^2/d^3/(e* x+d)^2*e-1/3*b^2*ln(x^n)^2/d^2/(e*x+d)^3*e+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/d^2/(e*x+d)^3*e+ 4/d^5*e*ln(e*x+d)-3/d^4*e/(e*x+d)-1/d^3/(e*x+d)^2*e-1/d^4/x-4/d^5*e*ln(x)) +4*b^2*ln(x^n)^2/d^5*e*ln(e*x+d)-26/3*b^2*n*ln(x^n)/d^5*e*ln(e*x+d)+26/3*b ^2*n*ln(x^n)/d^5*e*ln(x)+26/3*b^2/d^5*n^2*e*ln(e*x+d)*ln(-e*x/d)+4*b^2*n/d ^5*e*ln(x^n)*ln(x)^2+8*b^2/d^5*e*ln(x)*dilog(-e*x/d)*n^2-8*b^2*n/d^5*e*ln( x^n)*dilog(-e*x/d)-4*b^2/d^5*e*n^2*ln(e*x+d)*ln(x)^2+4*b^2/d^5*e*n^2*ln(x) ^2*ln(1+e*x/d)+8*b^2/d^5*e*n^2*ln(x)*polylog(2,-e*x/d)-4*b^2*ln(x^n)^2/d^5 *e*ln(x)-13/3*b^2/d^5*n^2*e*ln(x)^2+26/3*b^2/d^5*n^2*e*dilog(-e*x/d)-4/3*b ^2/d^5*e*ln(x)^3*n^2+(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*c sgn(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)/d^2/(e*x+d)^3*e+4*ln(x^n)/d^5*e*ln (e*x+d)-3*ln(x^n)/d^4*e/(e*x+d)-ln(x^n)/d^3/(e*x+d)^2*e-ln(x^n)/d^4/x-4*ln (x^n)/d^5*e*ln(x)-1/3*n*(-6/d^5*e*ln(x)^2+12/d^5*e*(dilog(-e*x/d)+ln(e*x+d )*ln(-e*x/d))-4/d^4*e/(e*x+d)+13/d^5*e*ln(e*x+d)-1/2/d^3/(e*x+d)^2*e+3/d^4 /x-13/d^5*e*ln(x)))+8/3*b^2*n*ln(x^n)/d^4*e/(e*x+d)+1/3*b^2*n*ln(x^n)/d^3/ (e*x+d)^2*e+8*b^2/d^5*e*ln(x)*ln(e*x+d)*ln(-e*x/d)*n^2-8*b^2*n/d^5*e*ln(x^ n)*ln(e*x+d)*ln(-e*x/d)-1/3*b^2*e*n^2/d^4/(e*x+d)-3*b^2*e*n^2*ln(x)/d^5...
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{4} x^{2}} \,d x } \]
integral((b^2*log(c*x^n)^2 + 2*a*b*log(c*x^n) + a^2)/(e^4*x^6 + 4*d*e^3*x^ 5 + 6*d^2*e^2*x^4 + 4*d^3*e*x^3 + d^4*x^2), x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^4} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{x^{2} \left (d + e x\right )^{4}}\, dx \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{4} x^{2}} \,d x } \]
-1/3*a^2*((12*e^3*x^3 + 30*d*e^2*x^2 + 22*d^2*e*x + 3*d^3)/(d^4*e^3*x^4 + 3*d^5*e^2*x^3 + 3*d^6*e*x^2 + d^7*x) - 12*e*log(e*x + d)/d^5 + 12*e*log(x) /d^5) + integrate((b^2*log(c)^2 + b^2*log(x^n)^2 + 2*a*b*log(c) + 2*(b^2*l og(c) + a*b)*log(x^n))/(e^4*x^6 + 4*d*e^3*x^5 + 6*d^2*e^2*x^4 + 4*d^3*e*x^ 3 + d^4*x^2), x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{4} x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^4} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{x^2\,{\left (d+e\,x\right )}^4} \,d x \]