Integrand size = 23, antiderivative size = 322 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^3} \, dx=-\frac {2 b^2 n^2}{d^3 x}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {b e^2 n x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{2 d^4}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{d^3 x}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{2 d^2 (d+e x)^2}+\frac {2 e^2 x \left (a+b \log \left (c x^n\right )\right )^2}{d^4 (d+e x)}-\frac {e \left (a+b \log \left (c x^n\right )\right )^3}{b d^4 n}+\frac {b^2 e n^2 \log (d+e x)}{d^4}-\frac {5 b e n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^4}+\frac {3 e \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{d^4}-\frac {5 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^4}+\frac {6 b e n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^4}-\frac {6 b^2 e n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{d^4} \]
-2*b^2*n^2/d^3/x-2*b*n*(a+b*ln(c*x^n))/d^3/x-b*e^2*n*x*(a+b*ln(c*x^n))/d^4 /(e*x+d)+1/2*e*(a+b*ln(c*x^n))^2/d^4-(a+b*ln(c*x^n))^2/d^3/x-1/2*e*(a+b*ln (c*x^n))^2/d^2/(e*x+d)^2+2*e^2*x*(a+b*ln(c*x^n))^2/d^4/(e*x+d)-e*(a+b*ln(c *x^n))^3/b/d^4/n+b^2*e*n^2*ln(e*x+d)/d^4-5*b*e*n*(a+b*ln(c*x^n))*ln(1+e*x/ d)/d^4+3*e*(a+b*ln(c*x^n))^2*ln(1+e*x/d)/d^4-5*b^2*e*n^2*polylog(2,-e*x/d) /d^4+6*b*e*n*(a+b*ln(c*x^n))*polylog(2,-e*x/d)/d^4-6*b^2*e*n^2*polylog(3,- e*x/d)/d^4
Time = 0.29 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^3} \, dx=-\frac {\frac {4 b^2 d n^2}{x}+\frac {4 b d n \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 b d e n \left (a+b \log \left (c x^n\right )\right )}{d+e x}-5 e \left (a+b \log \left (c x^n\right )\right )^2+\frac {2 d \left (a+b \log \left (c x^n\right )\right )^2}{x}+\frac {d^2 e \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2}+\frac {4 d e \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}+\frac {2 e \left (a+b \log \left (c x^n\right )\right )^3}{b n}+2 b^2 e n^2 (\log (x)-\log (d+e x))+10 b e n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-6 e \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )+10 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )-12 b e n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+12 b^2 e n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{2 d^4} \]
-1/2*((4*b^2*d*n^2)/x + (4*b*d*n*(a + b*Log[c*x^n]))/x - (2*b*d*e*n*(a + b *Log[c*x^n]))/(d + e*x) - 5*e*(a + b*Log[c*x^n])^2 + (2*d*(a + b*Log[c*x^n ])^2)/x + (d^2*e*(a + b*Log[c*x^n])^2)/(d + e*x)^2 + (4*d*e*(a + b*Log[c*x ^n])^2)/(d + e*x) + (2*e*(a + b*Log[c*x^n])^3)/(b*n) + 2*b^2*e*n^2*(Log[x] - Log[d + e*x]) + 10*b*e*n*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] - 6*e*(a + b*Log[c*x^n])^2*Log[1 + (e*x)/d] + 10*b^2*e*n^2*PolyLog[2, -((e*x)/d)] - 12*b*e*n*(a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)] + 12*b^2*e*n^2*PolyLog[ 3, -((e*x)/d)])/d^4
Time = 0.75 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.04, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \int \left (\frac {2 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{d^3 (d+e x)^2}-\frac {3 e \left (a+b \log \left (c x^n\right )\right )^2}{d^3 x (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{d^3 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )^2}{d^2 (d+e x)^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 e^2 x \left (a+b \log \left (c x^n\right )\right )^2}{d^4 (d+e x)}-\frac {b e^2 n x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}-\frac {6 b e n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {3 e \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{d^4}-\frac {b e n \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {4 b e n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{d^3 x}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{2 d^2 (d+e x)^2}+\frac {b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4}-\frac {4 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^4}-\frac {6 b^2 e n^2 \operatorname {PolyLog}\left (3,-\frac {d}{e x}\right )}{d^4}+\frac {b^2 e n^2 \log (d+e x)}{d^4}-\frac {2 b^2 n^2}{d^3 x}\) |
(-2*b^2*n^2)/(d^3*x) - (2*b*n*(a + b*Log[c*x^n]))/(d^3*x) - (b*e^2*n*x*(a + b*Log[c*x^n]))/(d^4*(d + e*x)) - (b*e*n*Log[1 + d/(e*x)]*(a + b*Log[c*x^ n]))/d^4 - (a + b*Log[c*x^n])^2/(d^3*x) - (e*(a + b*Log[c*x^n])^2)/(2*d^2* (d + e*x)^2) + (2*e^2*x*(a + b*Log[c*x^n])^2)/(d^4*(d + e*x)) + (3*e*Log[1 + d/(e*x)]*(a + b*Log[c*x^n])^2)/d^4 + (b^2*e*n^2*Log[d + e*x])/d^4 - (4* b*e*n*(a + b*Log[c*x^n])*Log[1 + (e*x)/d])/d^4 + (b^2*e*n^2*PolyLog[2, -(d /(e*x))])/d^4 - (6*b*e*n*(a + b*Log[c*x^n])*PolyLog[2, -(d/(e*x))])/d^4 - (4*b^2*e*n^2*PolyLog[2, -((e*x)/d)])/d^4 - (6*b^2*e*n^2*PolyLog[3, -(d/(e* x))])/d^4
3.2.12.3.1 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 = 0.50 (sec) , antiderivative size = 908, normalized size of antiderivative = 2.82
-2*b^2*n*ln(x^n)/d^3/x-b^2*ln(x^n)^2/d^3/x+6*b^2/d^4*e*ln(x)*ln(e*x+d)*ln( -e*x/d)*n^2-6*b^2*n/d^4*e*ln(x^n)*ln(e*x+d)*ln(-e*x/d)-6*b^2*e*n^2*polylog (3,-e*x/d)/d^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)*b*(-1/2*ln(x^n)/d^2/(e*x+d)^2*e+3*ln(x^n)/d^4*e*ln(e*x+d )-2*ln(x^n)/d^3*e/(e*x+d)-ln(x^n)/d^3/x-3*ln(x^n)/d^4*e*ln(x)-1/2*n*(-3/d^ 4*e*ln(x)^2+6/d^4*e*(dilog(-e*x/d)+ln(e*x+d)*ln(-e*x/d))-1/d^3*e/(e*x+d)+5 /d^4*e*ln(e*x+d)+2/d^3/x-5/d^4*e*ln(x)))-2*b^2*ln(x^n)^2/d^3*e/(e*x+d)-1/2 *b^2*ln(x^n)^2/d^2/(e*x+d)^2*e-3*b^2*ln(x^n)^2/d^4*e*ln(x)-b^2/d^4*n^2*e*l n(x)-5/2*b^2/d^4*n^2*e*ln(x)^2+5*b^2/d^4*n^2*e*dilog(-e*x/d)-b^2/d^4*e*ln( x)^3*n^2+3*b^2*ln(x^n)^2/d^4*e*ln(e*x+d)+b^2*n*ln(x^n)/d^3*e/(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*(-1/2/d^2/(e*x+d)^2*e+3/d^4*e*ln(e*x+d)-2/d^3*e/(e*x+d)-1/d^3/x-3/d^4* e*ln(x))+6*b^2/d^4*e*ln(x)*dilog(-e*x/d)*n^2-6*b^2*n/d^4*e*ln(x^n)*dilog(- e*x/d)-3*b^2/d^4*e*n^2*ln(e*x+d)*ln(x)^2+3*b^2/d^4*e*n^2*ln(x)^2*ln(1+e*x/ d)+6*b^2/d^4*e*n^2*ln(x)*polylog(2,-e*x/d)-5*b^2*n*ln(x^n)/d^4*e*ln(e*x+d) +5*b^2*n*ln(x^n)/d^4*e*ln(x)+5*b^2/d^4*n^2*e*ln(e*x+d)*ln(-e*x/d)+3*b^2*n/ d^4*e*ln(x^n)*ln(x)^2+b^2*e*n^2*ln(e*x+d)/d^4-2*b^2*n^2/d^3/x
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{3} x^{2}} \,d x } \]
integral((b^2*log(c*x^n)^2 + 2*a*b*log(c*x^n) + a^2)/(e^3*x^5 + 3*d*e^2*x^ 4 + 3*d^2*e*x^3 + d^3*x^2), x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^3} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{x^{2} \left (d + e x\right )^{3}}\, dx \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{3} x^{2}} \,d x } \]
-1/2*a^2*((6*e^2*x^2 + 9*d*e*x + 2*d^2)/(d^3*e^2*x^3 + 2*d^4*e*x^2 + d^5*x ) - 6*e*log(e*x + d)/d^4 + 6*e*log(x)/d^4) + integrate((b^2*log(c)^2 + b^2 *log(x^n)^2 + 2*a*b*log(c) + 2*(b^2*log(c) + a*b)*log(x^n))/(e^3*x^5 + 3*d *e^2*x^4 + 3*d^2*e*x^3 + d^3*x^2), x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{3} x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x^2 (d+e x)^3} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{x^2\,{\left (d+e\,x\right )}^3} \,d x \]