Integrand size = 23, antiderivative size = 296 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=-\frac {2 a b n x}{e^3}+\frac {2 b^2 n^2 x}{e^3}-\frac {2 b^2 n x \log \left (c x^n\right )}{e^3}+\frac {b d n x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}-\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 e^4}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{e^3}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^4 (d+e x)^2}+\frac {3 d x \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)}-\frac {b^2 d n^2 \log (d+e x)}{e^4}-\frac {5 b d n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}-\frac {3 d \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^4}-\frac {5 b^2 d n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}-\frac {6 b d n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}+\frac {6 b^2 d n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^4} \]
-2*a*b*n*x/e^3+2*b^2*n^2*x/e^3-2*b^2*n*x*ln(c*x^n)/e^3+b*d*n*x*(a+b*ln(c*x ^n))/e^3/(e*x+d)-1/2*d*(a+b*ln(c*x^n))^2/e^4+x*(a+b*ln(c*x^n))^2/e^3+1/2*d ^3*(a+b*ln(c*x^n))^2/e^4/(e*x+d)^2+3*d*x*(a+b*ln(c*x^n))^2/e^3/(e*x+d)-b^2 *d*n^2*ln(e*x+d)/e^4-5*b*d*n*(a+b*ln(c*x^n))*ln(1+e*x/d)/e^4-3*d*(a+b*ln(c *x^n))^2*ln(1+e*x/d)/e^4-5*b^2*d*n^2*polylog(2,-e*x/d)/e^4-6*b*d*n*(a+b*ln (c*x^n))*polylog(2,-e*x/d)/e^4+6*b^2*d*n^2*polylog(3,-e*x/d)/e^4
Time = 0.19 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.87 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\frac {-\frac {2 b d^2 n \left (a+b \log \left (c x^n\right )\right )}{d+e x}+5 d \left (a+b \log \left (c x^n\right )\right )^2+2 e x \left (a+b \log \left (c x^n\right )\right )^2+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2}-\frac {6 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}-4 b e n x \left (a-b n+b \log \left (c x^n\right )\right )+2 b^2 d n^2 (\log (x)-\log (d+e x))-10 b d n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-6 d \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )-10 b^2 d n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )-12 b d n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+12 b^2 d n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{2 e^4} \]
((-2*b*d^2*n*(a + b*Log[c*x^n]))/(d + e*x) + 5*d*(a + b*Log[c*x^n])^2 + 2* e*x*(a + b*Log[c*x^n])^2 + (d^3*(a + b*Log[c*x^n])^2)/(d + e*x)^2 - (6*d^2 *(a + b*Log[c*x^n])^2)/(d + e*x) - 4*b*e*n*x*(a - b*n + b*Log[c*x^n]) + 2* b^2*d*n^2*(Log[x] - Log[d + e*x]) - 10*b*d*n*(a + b*Log[c*x^n])*Log[1 + (e *x)/d] - 6*d*(a + b*Log[c*x^n])^2*Log[1 + (e*x)/d] - 10*b^2*d*n^2*PolyLog[ 2, -((e*x)/d)] - 12*b*d*n*(a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)] + 12*b ^2*d*n^2*PolyLog[3, -((e*x)/d)])/(2*e^4)
Time = 0.71 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.10, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \int \left (-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^3}+\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}-\frac {3 d \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{e^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^4 (d+e x)^2}-\frac {6 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {b d n \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {3 d \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e^4}-\frac {6 b d n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {3 d x \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)}+\frac {b d n x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{e^3}-\frac {2 a b n x}{e^3}-\frac {2 b^2 n x \log \left (c x^n\right )}{e^3}-\frac {b^2 d n^2 \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{e^4}-\frac {6 b^2 d n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}+\frac {6 b^2 d n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^4}-\frac {b^2 d n^2 \log (d+e x)}{e^4}+\frac {2 b^2 n^2 x}{e^3}\) |
(-2*a*b*n*x)/e^3 + (2*b^2*n^2*x)/e^3 - (2*b^2*n*x*Log[c*x^n])/e^3 + (b*d*n *x*(a + b*Log[c*x^n]))/(e^3*(d + e*x)) + (b*d*n*Log[1 + d/(e*x)]*(a + b*Lo g[c*x^n]))/e^4 + (x*(a + b*Log[c*x^n])^2)/e^3 + (d^3*(a + b*Log[c*x^n])^2) /(2*e^4*(d + e*x)^2) + (3*d*x*(a + b*Log[c*x^n])^2)/(e^3*(d + e*x)) - (b^2 *d*n^2*Log[d + e*x])/e^4 - (6*b*d*n*(a + b*Log[c*x^n])*Log[1 + (e*x)/d])/e ^4 - (3*d*(a + b*Log[c*x^n])^2*Log[1 + (e*x)/d])/e^4 - (b^2*d*n^2*PolyLog[ 2, -(d/(e*x))])/e^4 - (6*b^2*d*n^2*PolyLog[2, -((e*x)/d)])/e^4 - (6*b*d*n* (a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)])/e^4 + (6*b^2*d*n^2*PolyLog[3, - ((e*x)/d)])/e^4
3.2.7.3.1 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 = 0.52 (sec) , antiderivative size = 827, normalized size of antiderivative = 2.79
b^2*ln(x^n)^2*x/e^3-3*b^2*ln(x^n)^2/e^4*d*ln(e*x+d)-3*b^2*ln(x^n)^2/e^4*d^ 2/(e*x+d)+1/2*b^2*ln(x^n)^2*d^3/e^4/(e*x+d)^2-2*b^2*n*ln(x^n)*x/e^3-5*b^2* n*ln(x^n)/e^4*d*ln(e*x+d)-b^2*n*ln(x^n)/e^4*d^2/(e*x+d)+5*b^2*n/e^4*ln(x)* ln(x^n)*d+2*b^2*n^2*x/e^3-b^2*d*n^2*ln(e*x+d)/e^4+b^2/e^4*n^2*d*ln(x)-5/2* b^2/e^4*n^2*d*ln(x)^2+5*b^2/e^4*n^2*ln(e*x+d)*ln(-e*x/d)*d+5*b^2/e^4*n^2*d ilog(-e*x/d)*d-6*b^2/e^4*d*ln(e*x+d)*ln(-e*x/d)*ln(x)*n^2+6*b^2*n/e^4*d*ln (x^n)*ln(e*x+d)*ln(-e*x/d)-6*b^2/e^4*d*dilog(-e*x/d)*ln(x)*n^2+6*b^2*n/e^4 *d*ln(x^n)*dilog(-e*x/d)+3*b^2/e^4*d*n^2*ln(e*x+d)*ln(x)^2-3*b^2/e^4*d*n^2 *ln(x)^2*ln(1+e*x/d)-6*b^2/e^4*d*n^2*ln(x)*polylog(2,-e*x/d)+6*b^2*d*n^2*p olylog(3,-e*x/d)/e^4+(-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*(ln(x^n)*x/e^3-3*ln(x^n)/e^4*d*ln(e*x+d)-3*ln(x^ n)/e^4*d^2/(e*x+d)+1/2*ln(x^n)*d^3/e^4/(e*x+d)^2-1/2*n*(1/e^4*(2*e*x+2*d+5 *d*ln(e*x+d)+d^2/(e*x+d)-5*d*ln(e*x))-6/e^4*d*(dilog(-e*x/d)+ln(e*x+d)*ln( -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*(x/e^3-3/e^4*d*ln(e*x+d)-3/e^4*d^2/(e*x+d)+1/2*d^3/e^4 /(e*x+d)^2)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{3}}{{\left (e x + d\right )}^{3}} \,d x } \]
integral((b^2*x^3*log(c*x^n)^2 + 2*a*b*x^3*log(c*x^n) + a^2*x^3)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int \frac {x^{3} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x\right )^{3}}\, dx \]
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{3}}{{\left (e x + d\right )}^{3}} \,d x } \]
-1/2*a^2*((6*d^2*e*x + 5*d^3)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) - 2*x/e^3 + 6*d*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^3*x^3 + 3*d*e^2*x^ 2 + 3*d^2*e*x + d^3), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{3}}{{\left (e x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int \frac {x^3\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x\right )}^3} \,d x \]