Integrand size = 24, antiderivative size = 238 \[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d x+e x^2} \, dx=\frac {\log \left (-\frac {b x}{a}\right ) \log ^3\left (c (a+b x)^n\right )}{d}-\frac {\log ^3\left (c (a+b x)^n\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}-\frac {3 n \log ^2\left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {3 n \log ^2\left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (2,1+\frac {b x}{a}\right )}{d}+\frac {6 n^2 \log \left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (3,-\frac {e (a+b x)}{b d-a e}\right )}{d}-\frac {6 n^2 \log \left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (3,1+\frac {b x}{a}\right )}{d}-\frac {6 n^3 \operatorname {PolyLog}\left (4,-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {6 n^3 \operatorname {PolyLog}\left (4,1+\frac {b x}{a}\right )}{d} \]
ln(-b*x/a)*ln(c*(b*x+a)^n)^3/d-ln(c*(b*x+a)^n)^3*ln(b*(e*x+d)/(-a*e+b*d))/ d-3*n*ln(c*(b*x+a)^n)^2*polylog(2,-e*(b*x+a)/(-a*e+b*d))/d+3*n*ln(c*(b*x+a )^n)^2*polylog(2,1+b*x/a)/d+6*n^2*ln(c*(b*x+a)^n)*polylog(3,-e*(b*x+a)/(-a *e+b*d))/d-6*n^2*ln(c*(b*x+a)^n)*polylog(3,1+b*x/a)/d-6*n^3*polylog(4,-e*( b*x+a)/(-a*e+b*d))/d+6*n^3*polylog(4,1+b*x/a)/d
Leaf count is larger than twice the leaf count of optimal. \(494\) vs. \(2(238)=476\).
Time = 0.14 (sec) , antiderivative size = 494, normalized size of antiderivative = 2.08 \[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d x+e x^2} \, dx=\frac {-\log (x) \left (n \log (a+b x)-\log \left (c (a+b x)^n\right )\right )^3+\left (n \log (a+b x)-\log \left (c (a+b x)^n\right )\right )^3 \log (d+e x)+3 n \left (-n \log (a+b x)+\log \left (c (a+b x)^n\right )\right )^2 \left (\log (x) \left (\log (a+b x)-\log \left (1+\frac {b x}{a}\right )\right )-\log (a+b x) \log \left (\frac {b (d+e x)}{b d-a e}\right )-\operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )-\operatorname {PolyLog}\left (2,\frac {e (a+b x)}{-b d+a e}\right )\right )-3 n^2 \left (n \log (a+b x)-\log \left (c (a+b x)^n\right )\right ) \left (\log \left (-\frac {b x}{a}\right ) \log ^2(a+b x)-\log ^2(a+b x) \log \left (\frac {b (d+e x)}{b d-a e}\right )-2 \log (a+b x) \operatorname {PolyLog}\left (2,\frac {e (a+b x)}{-b d+a e}\right )+2 \log (a+b x) \operatorname {PolyLog}\left (2,1+\frac {b x}{a}\right )+2 \operatorname {PolyLog}\left (3,\frac {e (a+b x)}{-b d+a e}\right )-2 \operatorname {PolyLog}\left (3,1+\frac {b x}{a}\right )\right )+n^3 \left (\log \left (-\frac {b x}{a}\right ) \log ^3(a+b x)-\log ^3(a+b x) \log \left (\frac {b (d+e x)}{b d-a e}\right )-3 \log ^2(a+b x) \operatorname {PolyLog}\left (2,\frac {e (a+b x)}{-b d+a e}\right )+3 \log ^2(a+b x) \operatorname {PolyLog}\left (2,1+\frac {b x}{a}\right )+6 \log (a+b x) \operatorname {PolyLog}\left (3,\frac {e (a+b x)}{-b d+a e}\right )-6 \log (a+b x) \operatorname {PolyLog}\left (3,1+\frac {b x}{a}\right )-6 \operatorname {PolyLog}\left (4,\frac {e (a+b x)}{-b d+a e}\right )+6 \operatorname {PolyLog}\left (4,1+\frac {b x}{a}\right )\right )}{d} \]
(-(Log[x]*(n*Log[a + b*x] - Log[c*(a + b*x)^n])^3) + (n*Log[a + b*x] - Log [c*(a + b*x)^n])^3*Log[d + e*x] + 3*n*(-(n*Log[a + b*x]) + Log[c*(a + b*x) ^n])^2*(Log[x]*(Log[a + b*x] - Log[1 + (b*x)/a]) - Log[a + b*x]*Log[(b*(d + e*x))/(b*d - a*e)] - PolyLog[2, -((b*x)/a)] - PolyLog[2, (e*(a + b*x))/( -(b*d) + a*e)]) - 3*n^2*(n*Log[a + b*x] - Log[c*(a + b*x)^n])*(Log[-((b*x) /a)]*Log[a + b*x]^2 - Log[a + b*x]^2*Log[(b*(d + e*x))/(b*d - a*e)] - 2*Lo g[a + b*x]*PolyLog[2, (e*(a + b*x))/(-(b*d) + a*e)] + 2*Log[a + b*x]*PolyL og[2, 1 + (b*x)/a] + 2*PolyLog[3, (e*(a + b*x))/(-(b*d) + a*e)] - 2*PolyLo g[3, 1 + (b*x)/a]) + n^3*(Log[-((b*x)/a)]*Log[a + b*x]^3 - Log[a + b*x]^3* Log[(b*(d + e*x))/(b*d - a*e)] - 3*Log[a + b*x]^2*PolyLog[2, (e*(a + b*x)) /(-(b*d) + a*e)] + 3*Log[a + b*x]^2*PolyLog[2, 1 + (b*x)/a] + 6*Log[a + b* x]*PolyLog[3, (e*(a + b*x))/(-(b*d) + a*e)] - 6*Log[a + b*x]*PolyLog[3, 1 + (b*x)/a] - 6*PolyLog[4, (e*(a + b*x))/(-(b*d) + a*e)] + 6*PolyLog[4, 1 + (b*x)/a]))/d
Time = 0.60 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2026, 2863, 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.
\(\displaystyle \int \frac {\log ^3\left (c (a+b x)^n\right )}{d x+e x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\log ^3\left (c (a+b x)^n\right )}{x (d+e x)}dx\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \int \left (\frac {\log ^3\left (c (a+b x)^n\right )}{d x}-\frac {e \log ^3\left (c (a+b x)^n\right )}{d (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {6 n^2 \log \left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (3,-\frac {e (a+b x)}{b d-a e}\right )}{d}-\frac {3 n \log ^2\left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d}-\frac {\log ^3\left (c (a+b x)^n\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}-\frac {6 n^2 \operatorname {PolyLog}\left (3,\frac {b x}{a}+1\right ) \log \left (c (a+b x)^n\right )}{d}+\frac {3 n \operatorname {PolyLog}\left (2,\frac {b x}{a}+1\right ) \log ^2\left (c (a+b x)^n\right )}{d}+\frac {\log \left (-\frac {b x}{a}\right ) \log ^3\left (c (a+b x)^n\right )}{d}-\frac {6 n^3 \operatorname {PolyLog}\left (4,-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {6 n^3 \operatorname {PolyLog}\left (4,\frac {b x}{a}+1\right )}{d}\) |
(Log[-((b*x)/a)]*Log[c*(a + b*x)^n]^3)/d - (Log[c*(a + b*x)^n]^3*Log[(b*(d + e*x))/(b*d - a*e)])/d - (3*n*Log[c*(a + b*x)^n]^2*PolyLog[2, -((e*(a + b*x))/(b*d - a*e))])/d + (3*n*Log[c*(a + b*x)^n]^2*PolyLog[2, 1 + (b*x)/a] )/d + (6*n^2*Log[c*(a + b*x)^n]*PolyLog[3, -((e*(a + b*x))/(b*d - a*e))])/ d - (6*n^2*Log[c*(a + b*x)^n]*PolyLog[3, 1 + (b*x)/a])/d - (6*n^3*PolyLog[ 4, -((e*(a + b*x))/(b*d - a*e))])/d + (6*n^3*PolyLog[4, 1 + (b*x)/a])/d
3.4.43.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) ^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c , d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.02 (sec) , antiderivative size = 1756, normalized size of antiderivative = 7.38
(ln((b*x+a)^n)-n*ln(b*x+a))^3/d*ln(b*x)-(ln((b*x+a)^n)-n*ln(b*x+a))^3/d*ln (e*(b*x+a)-a*e+b*d)+n^3/d*ln(b*x+a)^3*ln(1-(b*x+a)/a)+3*n^3/d*ln(b*x+a)^2* polylog(2,(b*x+a)/a)-6*n^3/d*ln(b*x+a)*polylog(3,(b*x+a)/a)+6*n^3/d*polylo g(4,(b*x+a)/a)-n^3/d*ln(b*x+a)^3*ln(1+e*(b*x+a)/(-a*e+b*d))-3*n^3/d*ln(b*x +a)^2*polylog(2,-e*(b*x+a)/(-a*e+b*d))+6*n^3/d*ln(b*x+a)*polylog(3,-e*(b*x +a)/(-a*e+b*d))-6*n^3*polylog(4,-e*(b*x+a)/(-a*e+b*d))/d+3*b*n*(ln((b*x+a) ^n)-n*ln(b*x+a))^2*(1/b/d*(dilog(-x/a*b)+ln(b*x+a)*ln(-x/a*b))-e/b/d*(dilo g((e*(b*x+a)-a*e+b*d)/(-a*e+b*d))/e+ln(b*x+a)*ln((e*(b*x+a)-a*e+b*d)/(-a*e +b*d))/e))+3*b*n^2*(ln((b*x+a)^n)-n*ln(b*x+a))*(1/b/d*(ln(b*x+a)^2*ln(1-(b *x+a)/a)+2*ln(b*x+a)*polylog(2,(b*x+a)/a)-2*polylog(3,(b*x+a)/a))-1/b/d*(l n(b*x+a)^2*ln(1+e*(b*x+a)/(-a*e+b*d))+2*ln(b*x+a)*polylog(2,-e*(b*x+a)/(-a *e+b*d))-2*polylog(3,-e*(b*x+a)/(-a*e+b*d))))+1/8*(-I*Pi*csgn(I*c*(b*x+a)^ n)^3+I*Pi*csgn(I*c*(b*x+a)^n)^2*csgn(I*(b*x+a)^n)+I*Pi*csgn(I*c*(b*x+a)^n) ^2*csgn(I*c)-I*Pi*csgn(I*c*(b*x+a)^n)*csgn(I*(b*x+a)^n)*csgn(I*c)+2*ln(c)) ^3*(-1/d*ln(e*x+d)+1/d*ln(x))+(-3/2*I*Pi*csgn(I*c*(b*x+a)^n)^3+3/2*I*Pi*cs gn(I*c*(b*x+a)^n)^2*csgn(I*(b*x+a)^n)+3/2*I*Pi*csgn(I*c*(b*x+a)^n)^2*csgn( I*c)-3/2*I*Pi*csgn(I*c*(b*x+a)^n)*csgn(I*(b*x+a)^n)*csgn(I*c)+3*ln(c))*((l n((b*x+a)^n)-n*ln(b*x+a))^2/d*ln(b*x)-(ln((b*x+a)^n)-n*ln(b*x+a))^2/d*ln(e *(b*x+a)-a*e+b*d)+b*n^2*(1/b/d*(ln(b*x+a)^2*ln(1-(b*x+a)/a)+2*ln(b*x+a)*po lylog(2,(b*x+a)/a)-2*polylog(3,(b*x+a)/a))-1/b/d*(ln(b*x+a)^2*ln(1+e*(b...
\[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d x+e x^2} \, dx=\int { \frac {\log \left ({\left (b x + a\right )}^{n} c\right )^{3}}{e x^{2} + d x} \,d x } \]
\[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d x+e x^2} \, dx=\int \frac {\log {\left (c \left (a + b x\right )^{n} \right )}^{3}}{x \left (d + e x\right )}\, dx \]
\[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d x+e x^2} \, dx=\int { \frac {\log \left ({\left (b x + a\right )}^{n} c\right )^{3}}{e x^{2} + d x} \,d x } \]
\[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d x+e x^2} \, dx=\int { \frac {\log \left ({\left (b x + a\right )}^{n} c\right )^{3}}{e x^{2} + d x} \,d x } \]
Timed out. \[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d x+e x^2} \, dx=\int \frac {{\ln \left (c\,{\left (a+b\,x\right )}^n\right )}^3}{e\,x^2+d\,x} \,d x \]