Integrand size = 24, antiderivative size = 158 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g}+\frac {3 b n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g}-\frac {6 b^2 n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \operatorname {PolyLog}\left (3,-\frac {g (d+e x)}{e f-d g}\right )}{g}+\frac {6 b^3 n^3 \operatorname {PolyLog}\left (4,-\frac {g (d+e x)}{e f-d g}\right )}{g} \]
(a+b*ln(c*(e*x+d)^n))^3*ln(e*(g*x+f)/(-d*g+e*f))/g+3*b*n*(a+b*ln(c*(e*x+d) ^n))^2*polylog(2,-g*(e*x+d)/(-d*g+e*f))/g-6*b^2*n^2*(a+b*ln(c*(e*x+d)^n))* polylog(3,-g*(e*x+d)/(-d*g+e*f))/g+6*b^3*n^3*polylog(4,-g*(e*x+d)/(-d*g+e* f))/g
Leaf count is larger than twice the leaf count of optimal. \(335\) vs. \(2(158)=316\).
Time = 0.07 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.12 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\frac {\left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^3 \log (f+g x)+3 b n \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2 \left (\log (d+e x) \log \left (\frac {e (f+g x)}{e f-d g}\right )+\operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )\right )+6 b^2 n^2 \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \left (\frac {1}{2} \log ^2(d+e x) \log \left (\frac {e (f+g x)}{e f-d g}\right )+\log (d+e x) \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )-\operatorname {PolyLog}\left (3,\frac {g (d+e x)}{-e f+d g}\right )\right )+b^3 n^3 \left (\log ^3(d+e x) \log \left (\frac {e (f+g x)}{e f-d g}\right )+3 \log ^2(d+e x) \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )-6 \log (d+e x) \operatorname {PolyLog}\left (3,\frac {g (d+e x)}{-e f+d g}\right )+6 \operatorname {PolyLog}\left (4,\frac {g (d+e x)}{-e f+d g}\right )\right )}{g} \]
((a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^3*Log[f + g*x] + 3*b*n*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2*(Log[d + e*x]*Log[(e*(f + g*x) )/(e*f - d*g)] + PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)]) + 6*b^2*n^2*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])*((Log[d + e*x]^2*Log[(e*(f + g* x))/(e*f - d*g)])/2 + Log[d + e*x]*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g) ] - PolyLog[3, (g*(d + e*x))/(-(e*f) + d*g)]) + b^3*n^3*(Log[d + e*x]^3*Lo g[(e*(f + g*x))/(e*f - d*g)] + 3*Log[d + e*x]^2*PolyLog[2, (g*(d + e*x))/( -(e*f) + d*g)] - 6*Log[d + e*x]*PolyLog[3, (g*(d + e*x))/(-(e*f) + d*g)] + 6*PolyLog[4, (g*(d + e*x))/(-(e*f) + d*g)]))/g
Time = 0.56 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2843, 2881, 2821, 2830, 7143}
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.
\(\displaystyle \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx\) |
\(\Big \downarrow \) 2843 |
\(\displaystyle \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{g}-\frac {3 b e n \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{d+e x}dx}{g}\) |
\(\Big \downarrow \) 2881 |
\(\displaystyle \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{g}-\frac {3 b n \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac {e \left (f-\frac {d g}{e}\right )+g (d+e x)}{e f-d g}\right )}{d+e x}d(d+e x)}{g}\) |
\(\Big \downarrow \) 2821 |
\(\displaystyle \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{g}-\frac {3 b n \left (2 b n \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{d+e x}d(d+e x)-\operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2\right )}{g}\) |
\(\Big \downarrow \) 2830 |
\(\displaystyle \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{g}-\frac {3 b n \left (2 b n \left (\operatorname {PolyLog}\left (3,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-b n \int \frac {\operatorname {PolyLog}\left (3,-\frac {g (d+e x)}{e f-d g}\right )}{d+e x}d(d+e x)\right )-\operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2\right )}{g}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{g}-\frac {3 b n \left (2 b n \left (\operatorname {PolyLog}\left (3,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-b n \operatorname {PolyLog}\left (4,-\frac {g (d+e x)}{e f-d g}\right )\right )-\operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2\right )}{g}\) |
((a + b*Log[c*(d + e*x)^n])^3*Log[(e*(f + g*x))/(e*f - d*g)])/g - (3*b*n*( -((a + b*Log[c*(d + e*x)^n])^2*PolyLog[2, -((g*(d + e*x))/(e*f - d*g))]) + 2*b*n*((a + b*Log[c*(d + e*x)^n])*PolyLog[3, -((g*(d + e*x))/(e*f - d*g)) ] - b*n*PolyLog[4, -((g*(d + e*x))/(e*f - d*g))])))/g
3.3.31.3.1 Defintions of rubi rules used
Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b _.))^(p_.))/(x_), x_Symbol] :> Simp[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c *x^n])^p/m), x] + Simp[b*n*(p/m) Int[PolyLog[2, (-d)*f*x^m]*((a + b*Log[c *x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_ .)])/(x_), x_Symbol] :> Simp[PolyLog[k + 1, e*x^q]*((a + b*Log[c*x^n])^p/q) , x] - Simp[b*n*(p/q) Int[PolyLog[k + 1, e*x^q]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, e, k, n, q}, x] && GtQ[p, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_. )*(x_)), x_Symbol] :> Simp[Log[e*((f + g*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])^p/g), x] - Simp[b*e*n*(p/g) Int[Log[(e*(f + g*x))/(e*f - d*g)] *((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*f - d*g, 0] && IGtQ[p, 1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log [(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Sym bol] :> Simp[1/e Subst[Int[(k*(x/d))^r*(a + b*Log[c*x^n])^p*(f + g*Log[h* ((e*i - d*j)/e + j*(x/e))^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r}, x] && EqQ[e*k - d*l, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.06 (sec) , antiderivative size = 1396, normalized size of antiderivative = 8.84
-b^3*ln(g*(e*x+d)-d*g+e*f)/g*ln(e*x+d)^3*n^3+3*b^3*ln(g*(e*x+d)-d*g+e*f)/g *ln((e*x+d)^n)*ln(e*x+d)^2*n^2-3*b^3*ln(g*(e*x+d)-d*g+e*f)/g*ln((e*x+d)^n) ^2*ln(e*x+d)*n+b^3*ln(g*(e*x+d)-d*g+e*f)/g*ln((e*x+d)^n)^3-2*b^3*n^3/g*ln( e*x+d)^3*ln(1-g*(e*x+d)/(d*g-e*f))-3*b^3*n^3/g*ln(e*x+d)^2*polylog(2,g*(e* x+d)/(d*g-e*f))+6*b^3*n^3/g*polylog(4,g*(e*x+d)/(d*g-e*f))+3*b^3*n^3*dilog ((g*(e*x+d)-d*g+e*f)/(-d*g+e*f))/g*ln(e*x+d)^2-6*b^3*n^2*dilog((g*(e*x+d)- d*g+e*f)/(-d*g+e*f))/g*ln((e*x+d)^n)*ln(e*x+d)+3*b^3*n*dilog((g*(e*x+d)-d* g+e*f)/(-d*g+e*f))/g*ln((e*x+d)^n)^2+3*b^3*n^3*ln(e*x+d)^3*ln((g*(e*x+d)-d *g+e*f)/(-d*g+e*f))/g-6*b^3*n^2*ln(e*x+d)^2*ln((g*(e*x+d)-d*g+e*f)/(-d*g+e *f))/g*ln((e*x+d)^n)+3*b^3*n*ln(e*x+d)*ln((g*(e*x+d)-d*g+e*f)/(-d*g+e*f))/ g*ln((e*x+d)^n)^2+3*b^3*n^2/g*ln((e*x+d)^n)*ln(e*x+d)^2*ln(1-g*(e*x+d)/(d* g-e*f))+6*b^3*n^2/g*ln((e*x+d)^n)*ln(e*x+d)*polylog(2,g*(e*x+d)/(d*g-e*f)) -6*b^3*n^2/g*ln((e*x+d)^n)*polylog(3,g*(e*x+d)/(d*g-e*f))+1/8*(-I*b*Pi*csg n(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+ d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e* x+d)^n)^3*b+2*b*ln(c)+2*a)^3*ln(g*x+f)/g+3/2*(-I*b*Pi*csgn(I*c*(e*x+d)^n)* csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*cs gn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*b*l n(c)+2*a)*b^2*((ln((e*x+d)^n)-n*ln(e*x+d))^2*ln(g*(e*x+d)-d*g+e*f)/g+n^2/g *ln(e*x+d)^2*ln(1-g*(e*x+d)/(d*g-e*f))+2*n^2/g*ln(e*x+d)*polylog(2,g*(e...
\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{3}}{g x + f} \,d x } \]
integral((b^3*log((e*x + d)^n*c)^3 + 3*a*b^2*log((e*x + d)^n*c)^2 + 3*a^2* b*log((e*x + d)^n*c) + a^3)/(g*x + f), x)
\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\int \frac {\left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{3}}{f + g x}\, dx \]
\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{3}}{g x + f} \,d x } \]
a^3*log(g*x + f)/g + integrate((b^3*log((e*x + d)^n)^3 + b^3*log(c)^3 + 3* a*b^2*log(c)^2 + 3*a^2*b*log(c) + 3*(b^3*log(c) + a*b^2)*log((e*x + d)^n)^ 2 + 3*(b^3*log(c)^2 + 2*a*b^2*log(c) + a^2*b)*log((e*x + d)^n))/(g*x + f), x)
\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{3}}{g x + f} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^3}{f+g\,x} \,d x \]