Integrand size = 23, antiderivative size = 200 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx=\frac {2 a b d n x}{e^2}-\frac {2 b^2 d n^2 x}{e^2}+\frac {b^2 n^2 x^2}{4 e}+\frac {2 b^2 d n x \log \left (c x^n\right )}{e^2}-\frac {b n x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {d x \left (a+b \log \left (c x^n\right )\right )^2}{e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {2 b d^2 n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3}-\frac {2 b^2 d^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^3} \]
2*a*b*d*n*x/e^2-2*b^2*d*n^2*x/e^2+1/4*b^2*n^2*x^2/e+2*b^2*d*n*x*ln(c*x^n)/ e^2-1/2*b*n*x^2*(a+b*ln(c*x^n))/e-d*x*(a+b*ln(c*x^n))^2/e^2+1/2*x^2*(a+b*l n(c*x^n))^2/e+d^2*(a+b*ln(c*x^n))^2*ln(1+e*x/d)/e^3+2*b*d^2*n*(a+b*ln(c*x^ n))*polylog(2,-e*x/d)/e^3-2*b^2*d^2*n^2*polylog(3,-e*x/d)/e^3
Time = 0.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.79 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx=\frac {-4 d e x \left (a+b \log \left (c x^n\right )\right )^2+2 e^2 x^2 \left (a+b \log \left (c x^n\right )\right )^2+8 b d e n x \left (a-b n+b \log \left (c x^n\right )\right )+b e^2 n x^2 \left (b n-2 \left (a+b \log \left (c x^n\right )\right )\right )+4 d^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )+8 b d^2 n \left (\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )-b n \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )\right )}{4 e^3} \]
(-4*d*e*x*(a + b*Log[c*x^n])^2 + 2*e^2*x^2*(a + b*Log[c*x^n])^2 + 8*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])) + 4*d^2*(a + b*Log[c*x^n])^2*Log[1 + (e*x)/d] + 8*b*d^2*n*((a + b*Log[c*x^n] )*PolyLog[2, -((e*x)/d)] - b*n*PolyLog[3, -((e*x)/d)]))/(4*e^3)
Time = 0.44 (sec) , antiderivative size = 200, 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.
| \(\displaystyle \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \int \left (\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}-\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{e^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{e}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {d^2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e^3}-\frac {d x \left (a+b \log \left (c x^n\right )\right )^2}{e^2}-\frac {b n x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e}+\frac {2 a b d n x}{e^2}+\frac {2 b^2 d n x \log \left (c x^n\right )}{e^2}-\frac {2 b^2 d^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^3}-\frac {2 b^2 d n^2 x}{e^2}+\frac {b^2 n^2 x^2}{4 e}\) |
(2*a*b*d*n*x)/e^2 - (2*b^2*d*n^2*x)/e^2 + (b^2*n^2*x^2)/(4*e) + (2*b^2*d*n *x*Log[c*x^n])/e^2 - (b*n*x^2*(a + b*Log[c*x^n]))/(2*e) - (d*x*(a + b*Log[ c*x^n])^2)/e^2 + (x^2*(a + b*Log[c*x^n])^2)/(2*e) + (d^2*(a + b*Log[c*x^n] )^2*Log[1 + (e*x)/d])/e^3 + (2*b*d^2*n*(a + b*Log[c*x^n])*PolyLog[2, -((e* x)/d)])/e^3 - (2*b^2*d^2*n^2*PolyLog[3, -((e*x)/d)])/e^3
3.1.93.3.1 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 = 0.49 (sec) , antiderivative size = 637, normalized size of antiderivative = 3.18
| method | result | size |
| risch | \(\frac {b^{2} \ln \left (x^{n}\right )^{2} x^{2}}{2 e}-\frac {b^{2} \ln \left (x^{n}\right )^{2} d x}{e^{2}}+\frac {b^{2} \ln \left (x^{n}\right )^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}+\frac {2 b^{2} d^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right ) \ln \left (x \right ) n^{2}}{e^{3}}-\frac {2 b^{2} n \,d^{2} \ln \left (x^{n}\right ) \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{3}}+\frac {2 b^{2} d^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right ) \ln \left (x \right ) n^{2}}{e^{3}}-\frac {2 b^{2} n \,d^{2} \ln \left (x^{n}\right ) \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{3}}-\frac {b^{2} d^{2} n^{2} \ln \left (e x +d \right ) \ln \left (x \right )^{2}}{e^{3}}+\frac {b^{2} d^{2} n^{2} \ln \left (x \right )^{2} \ln \left (1+\frac {e x}{d}\right )}{e^{3}}+\frac {2 b^{2} d^{2} n^{2} \ln \left (x \right ) \operatorname {Li}_{2}\left (-\frac {e x}{d}\right )}{e^{3}}-\frac {2 b^{2} d^{2} n^{2} \operatorname {Li}_{3}\left (-\frac {e x}{d}\right )}{e^{3}}-\frac {b^{2} n \ln \left (x^{n}\right ) x^{2}}{2 e}+\frac {2 b^{2} n \ln \left (x^{n}\right ) d x}{e^{2}}+\frac {b^{2} n^{2} x^{2}}{4 e}-\frac {2 b^{2} d \,n^{2} x}{e^{2}}+\left (-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 a \right ) b \left (\frac {\ln \left (x^{n}\right ) x^{2}}{2 e}-\frac {\ln \left (x^{n}\right ) d x}{e^{2}}+\frac {\ln \left (x^{n}\right ) d^{2} \ln \left (e x +d \right )}{e^{3}}-n \left (\frac {d^{2} \left (\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )\right )}{e^{3}}+\frac {\frac {\left (e x +d \right )^{2}}{2}-3 d \left (e x +d \right )}{2 e^{3}}\right )\right )+\frac {{\left (-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 a \right )}^{2} \left (\frac {\frac {1}{2} e \,x^{2}-d x}{e^{2}}+\frac {d^{2} \ln \left (e x +d \right )}{e^{3}}\right )}{4}\) | \(637\) |
1/2*b^2*ln(x^n)^2/e*x^2-b^2*ln(x^n)^2/e^2*d*x+b^2*ln(x^n)^2*d^2/e^3*ln(e*x +d)+2*b^2*d^2/e^3*ln(e*x+d)*ln(-e*x/d)*ln(x)*n^2-2*b^2*n*d^2/e^3*ln(x^n)*l n(e*x+d)*ln(-e*x/d)+2*b^2*d^2/e^3*dilog(-e*x/d)*ln(x)*n^2-2*b^2*n*d^2/e^3* ln(x^n)*dilog(-e*x/d)-b^2*d^2/e^3*n^2*ln(e*x+d)*ln(x)^2+b^2*d^2/e^3*n^2*ln (x)^2*ln(1+e*x/d)+2*b^2*d^2/e^3*n^2*ln(x)*polylog(2,-e*x/d)-2*b^2*d^2*n^2* polylog(3,-e*x/d)/e^3-1/2*b^2*n*ln(x^n)/e*x^2+2*b^2*n*ln(x^n)/e^2*d*x+1/4* b^2*n^2*x^2/e-2*b^2*d*n^2*x/e^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*x^2-ln(x^n)/e^2*d*x+ln (x^n)*d^2/e^3*ln(e*x+d)-n*(d^2/e^3*(dilog(-e*x/d)+ln(e*x+d)*ln(-e*x/d))+1/ 2/e^3*(1/2*(e*x+d)^2-3*d*(e*x+d))))+1/4*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csg n(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/e^2*(1/2*e*x^2-d*x)+d^2/e^ 3*ln(e*x+d))
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2}}{e x + d} \,d x } \]
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx=\int \frac {x^{2} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{d + e x}\, dx \]
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2}}{e x + d} \,d x } \]
1/2*a^2*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + integrate((b^2*x^ 2*log(x^n)^2 + 2*(b^2*log(c) + a*b)*x^2*log(x^n) + (b^2*log(c)^2 + 2*a*b*l og(c))*x^2)/(e*x + d), x)
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2}}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx=\int \frac {x^2\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{d+e\,x} \,d x \]