Integrand size = 22, antiderivative size = 396 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x) \, dx=\frac {2 a b n x}{3 e^2}-\frac {26 b^2 n^2 x}{27 e^2}+\frac {19 b^2 n^2 x^2}{108 e}-\frac {2}{27} b^2 n^2 x^3+\frac {2 b^2 n x \log \left (c x^n\right )}{3 e^2}+\frac {2 b n x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}-\frac {5 b n x^2 \left (a+b \log \left (c x^n\right )\right )}{18 e}+\frac {4}{27} b n x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{3 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )^2+\frac {2 b^2 n^2 \log (1+e x)}{27 e^3}+\frac {2}{27} b^2 n^2 x^3 \log (1+e x)-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{9 e^3}-\frac {2}{9} b n x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)+\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x)}{3 e^3}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x)-\frac {2 b^2 n^2 \operatorname {PolyLog}(2,-e x)}{9 e^3}+\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,-e x)}{3 e^3}-\frac {2 b^2 n^2 \operatorname {PolyLog}(3,-e x)}{3 e^3} \]
2/3*a*b*n*x/e^2-26/27*b^2*n^2*x/e^2+19/108*b^2*n^2*x^2/e-2/27*b^2*n^2*x^3+ 2/3*b^2*n*x*ln(c*x^n)/e^2+2/9*b*n*x*(a+b*ln(c*x^n))/e^2-5/18*b*n*x^2*(a+b* ln(c*x^n))/e+4/27*b*n*x^3*(a+b*ln(c*x^n))-1/3*x*(a+b*ln(c*x^n))^2/e^2+1/6* x^2*(a+b*ln(c*x^n))^2/e-1/9*x^3*(a+b*ln(c*x^n))^2+2/27*b^2*n^2*ln(e*x+1)/e ^3+2/27*b^2*n^2*x^3*ln(e*x+1)-2/9*b*n*(a+b*ln(c*x^n))*ln(e*x+1)/e^3-2/9*b* n*x^3*(a+b*ln(c*x^n))*ln(e*x+1)+1/3*(a+b*ln(c*x^n))^2*ln(e*x+1)/e^3+1/3*x^ 3*(a+b*ln(c*x^n))^2*ln(e*x+1)-2/9*b^2*n^2*polylog(2,-e*x)/e^3+2/3*b*n*(a+b *ln(c*x^n))*polylog(2,-e*x)/e^3-2/3*b^2*n^2*polylog(3,-e*x)/e^3
Time = 0.10 (sec) , antiderivative size = 506, normalized size of antiderivative = 1.28 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x) \, dx=\frac {-36 a^2 e x+96 a b e n x-104 b^2 e n^2 x+18 a^2 e^2 x^2-30 a b e^2 n x^2+19 b^2 e^2 n^2 x^2-12 a^2 e^3 x^3+16 a b e^3 n x^3-8 b^2 e^3 n^2 x^3-72 a b e x \log \left (c x^n\right )+96 b^2 e n x \log \left (c x^n\right )+36 a b e^2 x^2 \log \left (c x^n\right )-30 b^2 e^2 n x^2 \log \left (c x^n\right )-24 a b e^3 x^3 \log \left (c x^n\right )+16 b^2 e^3 n x^3 \log \left (c x^n\right )-36 b^2 e x \log ^2\left (c x^n\right )+18 b^2 e^2 x^2 \log ^2\left (c x^n\right )-12 b^2 e^3 x^3 \log ^2\left (c x^n\right )+36 a^2 \log (1+e x)-24 a b n \log (1+e x)+8 b^2 n^2 \log (1+e x)+36 a^2 e^3 x^3 \log (1+e x)-24 a b e^3 n x^3 \log (1+e x)+8 b^2 e^3 n^2 x^3 \log (1+e x)+72 a b \log \left (c x^n\right ) \log (1+e x)-24 b^2 n \log \left (c x^n\right ) \log (1+e x)+72 a b e^3 x^3 \log \left (c x^n\right ) \log (1+e x)-24 b^2 e^3 n x^3 \log \left (c x^n\right ) \log (1+e x)+36 b^2 \log ^2\left (c x^n\right ) \log (1+e x)+36 b^2 e^3 x^3 \log ^2\left (c x^n\right ) \log (1+e x)+24 b n \left (3 a-b n+3 b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,-e x)-72 b^2 n^2 \operatorname {PolyLog}(3,-e x)}{108 e^3} \]
(-36*a^2*e*x + 96*a*b*e*n*x - 104*b^2*e*n^2*x + 18*a^2*e^2*x^2 - 30*a*b*e^ 2*n*x^2 + 19*b^2*e^2*n^2*x^2 - 12*a^2*e^3*x^3 + 16*a*b*e^3*n*x^3 - 8*b^2*e ^3*n^2*x^3 - 72*a*b*e*x*Log[c*x^n] + 96*b^2*e*n*x*Log[c*x^n] + 36*a*b*e^2* x^2*Log[c*x^n] - 30*b^2*e^2*n*x^2*Log[c*x^n] - 24*a*b*e^3*x^3*Log[c*x^n] + 16*b^2*e^3*n*x^3*Log[c*x^n] - 36*b^2*e*x*Log[c*x^n]^2 + 18*b^2*e^2*x^2*Lo g[c*x^n]^2 - 12*b^2*e^3*x^3*Log[c*x^n]^2 + 36*a^2*Log[1 + e*x] - 24*a*b*n* Log[1 + e*x] + 8*b^2*n^2*Log[1 + e*x] + 36*a^2*e^3*x^3*Log[1 + e*x] - 24*a *b*e^3*n*x^3*Log[1 + e*x] + 8*b^2*e^3*n^2*x^3*Log[1 + e*x] + 72*a*b*Log[c* x^n]*Log[1 + e*x] - 24*b^2*n*Log[c*x^n]*Log[1 + e*x] + 72*a*b*e^3*x^3*Log[ c*x^n]*Log[1 + e*x] - 24*b^2*e^3*n*x^3*Log[c*x^n]*Log[1 + e*x] + 36*b^2*Lo g[c*x^n]^2*Log[1 + e*x] + 36*b^2*e^3*x^3*Log[c*x^n]^2*Log[1 + e*x] + 24*b* n*(3*a - b*n + 3*b*Log[c*x^n])*PolyLog[2, -(e*x)] - 72*b^2*n^2*PolyLog[3, -(e*x)])/(108*e^3)
Time = 0.55 (sec) , antiderivative size = 356, normalized size of antiderivative = 0.90, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2824, 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.
| \(\displaystyle \int x^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 2824 |
| \(\displaystyle -2 b n \int \left (-\frac {1}{9} \left (a+b \log \left (c x^n\right )\right ) x^2+\frac {1}{3} \left (a+b \log \left (c x^n\right )\right ) \log (e x+1) x^2+\frac {\left (a+b \log \left (c x^n\right )\right ) x}{6 e}-\frac {a+b \log \left (c x^n\right )}{3 e^2}+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (e x+1)}{3 e^3 x}\right )dx+\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{3 e^2}+\frac {1}{3} x^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )^2+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )^2\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{3 e^2}-2 b n \left (-\frac {\operatorname {PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}+\frac {1}{9} x^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {5 x^2 \left (a+b \log \left (c x^n\right )\right )}{36 e}-\frac {2}{27} x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {a x}{3 e^2}-\frac {b x \log \left (c x^n\right )}{3 e^2}+\frac {b n \operatorname {PolyLog}(2,-e x)}{9 e^3}+\frac {b n \operatorname {PolyLog}(3,-e x)}{3 e^3}-\frac {b n \log (e x+1)}{27 e^3}+\frac {13 b n x}{27 e^2}-\frac {1}{27} b n x^3 \log (e x+1)-\frac {19 b n x^2}{216 e}+\frac {1}{27} b n x^3\right )+\frac {1}{3} x^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )^2+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )^2\) |
-1/3*(x*(a + b*Log[c*x^n])^2)/e^2 + (x^2*(a + b*Log[c*x^n])^2)/(6*e) - (x^ 3*(a + b*Log[c*x^n])^2)/9 + ((a + b*Log[c*x^n])^2*Log[1 + e*x])/(3*e^3) + (x^3*(a + b*Log[c*x^n])^2*Log[1 + e*x])/3 - 2*b*n*(-1/3*(a*x)/e^2 + (13*b* n*x)/(27*e^2) - (19*b*n*x^2)/(216*e) + (b*n*x^3)/27 - (b*x*Log[c*x^n])/(3* e^2) - (x*(a + b*Log[c*x^n]))/(9*e^2) + (5*x^2*(a + b*Log[c*x^n]))/(36*e) - (2*x^3*(a + b*Log[c*x^n]))/27 - (b*n*Log[1 + e*x])/(27*e^3) - (b*n*x^3*L og[1 + e*x])/27 + ((a + b*Log[c*x^n])*Log[1 + e*x])/(9*e^3) + (x^3*(a + b* Log[c*x^n])*Log[1 + e*x])/9 + (b*n*PolyLog[2, -(e*x)])/(9*e^3) - ((a + b*L og[c*x^n])*PolyLog[2, -(e*x)])/(3*e^3) + (b*n*PolyLog[3, -(e*x)])/(3*e^3))
3.1.11.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_ .))^(p_.)*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* (e + f*x^m)], x]}, Simp[(a + b*Log[c*x^n])^p u, x] - Simp[b*n*p Int[(a + b*Log[c*x^n])^(p - 1)/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, q}, x] && IGtQ[p, 0] && RationalQ[m] && RationalQ[q] && NeQ[q, -1] && (EqQ [p, 1] || (FractionQ[m] && IntegerQ[(q + 1)/m]) || (IGtQ[q, 0] && IntegerQ[ (q + 1)/m] && EqQ[d*e, 1]))
\[\int x^{2} {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{2} \ln \left (e x +1\right )d x\]
\[ \int x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2} \log \left (e x + 1\right ) \,d x } \]
integral(b^2*x^2*log(c*x^n)^2*log(e*x + 1) + 2*a*b*x^2*log(c*x^n)*log(e*x + 1) + a^2*x^2*log(e*x + 1), x)
Timed out. \[ \int x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x) \, dx=\text {Timed out} \]
\[ \int x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2} \log \left (e x + 1\right ) \,d x } \]
-1/18*(2*b^2*e^3*x^3 - 3*b^2*e^2*x^2 + 6*b^2*e*x - 6*(b^2*e^3*x^3 + b^2)*l og(e*x + 1))*log(x^n)^2/e^3 + 1/9*integrate((9*(b^2*e^3*log(c)^2 + 2*a*b*e ^3*log(c) + a^2*e^3)*x^3*log(e*x + 1) + (2*b^2*e^3*n*x^3 - 3*b^2*e^2*n*x^2 + 6*b^2*e*n*x + 6*((3*a*b*e^3 - (e^3*n - 3*e^3*log(c))*b^2)*x^3 - b^2*n)* log(e*x + 1))*log(x^n))/x, x)/e^3
\[ \int x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2} \log \left (e x + 1\right ) \,d x } \]
Timed out. \[ \int x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x) \, dx=\int x^2\,\ln \left (e\,x+1\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2 \,d x \]