Integrand size = 23, antiderivative size = 281 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx=\frac {4 a b d n x}{e^3}-\frac {4 b^2 d n^2 x}{e^3}+\frac {b^2 n^2 x^2}{4 e^2}+\frac {4 b^2 d n x \log \left (c x^n\right )}{e^3}-\frac {b n x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {2 d x \left (a+b \log \left (c x^n\right )\right )^2}{e^3}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^2}-\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)}+\frac {2 b d^2 n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}+\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^4}+\frac {2 b^2 d^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}+\frac {6 b d^2 n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}-\frac {6 b^2 d^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^4} \]
4*a*b*d*n*x/e^3-4*b^2*d*n^2*x/e^3+1/4*b^2*n^2*x^2/e^2+4*b^2*d*n*x*ln(c*x^n )/e^3-1/2*b*n*x^2*(a+b*ln(c*x^n))/e^2-2*d*x*(a+b*ln(c*x^n))^2/e^3+1/2*x^2* (a+b*ln(c*x^n))^2/e^2-d^2*x*(a+b*ln(c*x^n))^2/e^3/(e*x+d)+2*b*d^2*n*(a+b*l n(c*x^n))*ln(1+e*x/d)/e^4+3*d^2*(a+b*ln(c*x^n))^2*ln(1+e*x/d)/e^4+2*b^2*d^ 2*n^2*polylog(2,-e*x/d)/e^4+6*b*d^2*n*(a+b*ln(c*x^n))*polylog(2,-e*x/d)/e^ 4-6*b^2*d^2*n^2*polylog(3,-e*x/d)/e^4
Time = 0.14 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.85 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx=\frac {-8 d e x \left (a+b \log \left (c x^n\right )\right )^2+2 e^2 x^2 \left (a+b \log \left (c x^n\right )\right )^2+\frac {4 d^3 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}+16 b d e n x \left (a-b n+b \log \left (c x^n\right )\right )+b e^2 n x^2 \left (b n-2 \left (a+b \log \left (c x^n\right )\right )\right )+12 d^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )+4 d^2 \left (-\left (\left (a+b \log \left (c x^n\right )\right ) \left (a+b \log \left (c x^n\right )-2 b n \log \left (1+\frac {e x}{d}\right )\right )\right )+2 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )\right )+24 b d^2 n \left (\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )-b n \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )\right )}{4 e^4} \]
(-8*d*e*x*(a + b*Log[c*x^n])^2 + 2*e^2*x^2*(a + b*Log[c*x^n])^2 + (4*d^3*( a + b*Log[c*x^n])^2)/(d + e*x) + 16*b*d*e*n*x*(a - b*n + b*Log[c*x^n]) + b *e^2*n*x^2*(b*n - 2*(a + b*Log[c*x^n])) + 12*d^2*(a + b*Log[c*x^n])^2*Log[ 1 + (e*x)/d] + 4*d^2*(-((a + b*Log[c*x^n])*(a + b*Log[c*x^n] - 2*b*n*Log[1 + (e*x)/d])) + 2*b^2*n^2*PolyLog[2, -((e*x)/d)]) + 24*b*d^2*n*((a + b*Log [c*x^n])*PolyLog[2, -((e*x)/d)] - b*n*PolyLog[3, -((e*x)/d)]))/(4*e^4)
Time = 0.52 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \int \left (-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}+\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)}-\frac {2 d \left (a+b \log \left (c x^n\right )\right )^2}{e^3}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{e^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {6 b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 b d^2 n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {3 d^2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e^4}-\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)}-\frac {2 d x \left (a+b \log \left (c x^n\right )\right )^2}{e^3}-\frac {b n x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^2}+\frac {4 a b d n x}{e^3}+\frac {4 b^2 d n x \log \left (c x^n\right )}{e^3}+\frac {2 b^2 d^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}-\frac {6 b^2 d^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^4}-\frac {4 b^2 d n^2 x}{e^3}+\frac {b^2 n^2 x^2}{4 e^2}\) |
(4*a*b*d*n*x)/e^3 - (4*b^2*d*n^2*x)/e^3 + (b^2*n^2*x^2)/(4*e^2) + (4*b^2*d *n*x*Log[c*x^n])/e^3 - (b*n*x^2*(a + b*Log[c*x^n]))/(2*e^2) - (2*d*x*(a + b*Log[c*x^n])^2)/e^3 + (x^2*(a + b*Log[c*x^n])^2)/(2*e^2) - (d^2*x*(a + b* Log[c*x^n])^2)/(e^3*(d + e*x)) + (2*b*d^2*n*(a + b*Log[c*x^n])*Log[1 + (e* x)/d])/e^4 + (3*d^2*(a + b*Log[c*x^n])^2*Log[1 + (e*x)/d])/e^4 + (2*b^2*d^ 2*n^2*PolyLog[2, -((e*x)/d)])/e^4 + (6*b*d^2*n*(a + b*Log[c*x^n])*PolyLog[ 2, -((e*x)/d)])/e^4 - (6*b^2*d^2*n^2*PolyLog[3, -((e*x)/d)])/e^4
3.1.100.3.1 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 = 0.55 (sec) , antiderivative size = 824, normalized size of antiderivative = 2.93
1/2*b^2*ln(x^n)^2/e^2*x^2-2*b^2*ln(x^n)^2/e^3*d*x+3*b^2*ln(x^n)^2/e^4*d^2* ln(e*x+d)+b^2*ln(x^n)^2*d^3/e^4/(e*x+d)+6*b^2/e^4*d^2*ln(x)*ln(e*x+d)*ln(- e*x/d)*n^2+6*b^2/e^4*d^2*ln(x)*dilog(-e*x/d)*n^2-6*b^2*n/e^4*d^2*ln(x^n)*l n(e*x+d)*ln(-e*x/d)-6*b^2*n/e^4*d^2*ln(x^n)*dilog(-e*x/d)-3*b^2/e^4*d^2*n^ 2*ln(e*x+d)*ln(x)^2+3*b^2/e^4*d^2*n^2*ln(x)^2*ln(1+e*x/d)+6*b^2/e^4*d^2*n^ 2*ln(x)*polylog(2,-e*x/d)-6*b^2*d^2*n^2*polylog(3,-e*x/d)/e^4-1/2*b^2*n*ln (x^n)/e^2*x^2+4*b^2*n*ln(x^n)/e^3*d*x+2*b^2*n*ln(x^n)/e^4*d^2*ln(e*x+d)-2* b^2*n/e^4*ln(x^n)*d^2*ln(x)+1/4*b^2*n^2*x^2/e^2-4*b^2*d*n^2*x/e^3+b^2/e^4* n^2*d^2*ln(x)^2-2*b^2/e^4*n^2*ln(-e*x/d)*ln(e*x+d)*d^2-2*b^2/e^4*n^2*dilog (-e*x/d)*d^2+(-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)/e^2*x^2-2*ln(x^n)/e^3*d*x+3*ln(x^n)/e^4*d^2 *ln(e*x+d)+ln(x^n)*d^3/e^4/(e*x+d)-n*(3/e^4*d^2*(dilog(-e*x/d)+ln(e*x+d)*l n(-e*x/d))+1/2/e^4*(1/2*(e*x+d)^2-5*d*(e*x+d)-2*d^2*ln(e*x+d)+2*d^2*ln(e*x ))))+1/4*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn(I*c)*csg n(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/e^3*(1/2*e*x^2-2*d*x)+3/e^4*d^2*ln(e*x+d)+d^3/e^4/(e*x+d) )
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{3}}{{\left (e x + d\right )}^{2}} \,d x } \]
integral((b^2*x^3*log(c*x^n)^2 + 2*a*b*x^3*log(c*x^n) + a^2*x^3)/(e^2*x^2 + 2*d*e*x + d^2), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx=\int \frac {x^{3} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x\right )^{2}}\, dx \]
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{3}}{{\left (e x + d\right )}^{2}} \,d x } \]
1/2*(2*d^3/(e^5*x + d*e^4) + 6*d^2*log(e*x + d)/e^4 + (e*x^2 - 4*d*x)/e^3) *a^2 + 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^2*x^2 + 2*d*e*x + d^2), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{3}}{{\left (e x + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx=\int \frac {x^3\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x\right )}^2} \,d x \]