Integrand size = 26, antiderivative size = 555 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^3} \, dx=-\frac {45 b^3 f m n^3}{8 e x}-\frac {3 b^3 f^2 m n^3 \log (x)}{8 e^2}-\frac {21 b^2 f m n^2 \left (a+b \log \left (c x^n\right )\right )}{4 e x}+\frac {3 b^2 f^2 m n^2 \log \left (1+\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}-\frac {9 b f m n \left (a+b \log \left (c x^n\right )\right )^2}{4 e x}+\frac {3 b f^2 m n \log \left (1+\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{4 e^2}-\frac {f m \left (a+b \log \left (c x^n\right )\right )^3}{2 e x}+\frac {f^2 m \log \left (1+\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )^3}{2 e^2}+\frac {3 b^3 f^2 m n^3 \log (e+f x)}{8 e^2}-\frac {3 b^3 n^3 \log \left (d (e+f x)^m\right )}{8 x^2}-\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{4 x^2}-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{2 x^2}-\frac {3 b^3 f^2 m n^3 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{4 e^2}-\frac {3 b^2 f^2 m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{2 e^2}-\frac {3 b f^2 m n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{2 e^2}-\frac {3 b^3 f^2 m n^3 \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right )}{2 e^2}-\frac {3 b^2 f^2 m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right )}{e^2}-\frac {3 b^3 f^2 m n^3 \operatorname {PolyLog}\left (4,-\frac {e}{f x}\right )}{e^2} \]
-45/8*b^3*f*m*n^3/e/x-3/8*b^3*f^2*m*n^3*ln(x)/e^2-21/4*b^2*f*m*n^2*(a+b*ln (c*x^n))/e/x+3/4*b^2*f^2*m*n^2*ln(1+e/f/x)*(a+b*ln(c*x^n))/e^2-9/4*b*f*m*n *(a+b*ln(c*x^n))^2/e/x+3/4*b*f^2*m*n*ln(1+e/f/x)*(a+b*ln(c*x^n))^2/e^2-1/2 *f*m*(a+b*ln(c*x^n))^3/e/x+1/2*f^2*m*ln(1+e/f/x)*(a+b*ln(c*x^n))^3/e^2+3/8 *b^3*f^2*m*n^3*ln(f*x+e)/e^2-3/8*b^3*n^3*ln(d*(f*x+e)^m)/x^2-3/4*b^2*n^2*( a+b*ln(c*x^n))*ln(d*(f*x+e)^m)/x^2-3/4*b*n*(a+b*ln(c*x^n))^2*ln(d*(f*x+e)^ m)/x^2-1/2*(a+b*ln(c*x^n))^3*ln(d*(f*x+e)^m)/x^2-3/4*b^3*f^2*m*n^3*polylog (2,-e/f/x)/e^2-3/2*b^2*f^2*m*n^2*(a+b*ln(c*x^n))*polylog(2,-e/f/x)/e^2-3/2 *b*f^2*m*n*(a+b*ln(c*x^n))^2*polylog(2,-e/f/x)/e^2-3/2*b^3*f^2*m*n^3*polyl og(3,-e/f/x)/e^2-3*b^2*f^2*m*n^2*(a+b*ln(c*x^n))*polylog(3,-e/f/x)/e^2-3*b ^3*f^2*m*n^3*polylog(4,-e/f/x)/e^2
Leaf count is larger than twice the leaf count of optimal. \(1736\) vs. \(2(555)=1110\).
Time = 0.55 (sec) , antiderivative size = 1736, normalized size of antiderivative = 3.13 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^3} \, dx =\text {Too large to display} \]
-1/8*(4*a^3*e*f*m*x + 18*a^2*b*e*f*m*n*x + 42*a*b^2*e*f*m*n^2*x + 45*b^3*e *f*m*n^3*x + 4*a^3*f^2*m*x^2*Log[x] + 6*a^2*b*f^2*m*n*x^2*Log[x] + 6*a*b^2 *f^2*m*n^2*x^2*Log[x] + 3*b^3*f^2*m*n^3*x^2*Log[x] - 6*a^2*b*f^2*m*n*x^2*L og[x]^2 - 6*a*b^2*f^2*m*n^2*x^2*Log[x]^2 - 3*b^3*f^2*m*n^3*x^2*Log[x]^2 + 4*a*b^2*f^2*m*n^2*x^2*Log[x]^3 + 2*b^3*f^2*m*n^3*x^2*Log[x]^3 - b^3*f^2*m* n^3*x^2*Log[x]^4 + 12*a^2*b*e*f*m*x*Log[c*x^n] + 36*a*b^2*e*f*m*n*x*Log[c* x^n] + 42*b^3*e*f*m*n^2*x*Log[c*x^n] + 12*a^2*b*f^2*m*x^2*Log[x]*Log[c*x^n ] + 12*a*b^2*f^2*m*n*x^2*Log[x]*Log[c*x^n] + 6*b^3*f^2*m*n^2*x^2*Log[x]*Lo g[c*x^n] - 12*a*b^2*f^2*m*n*x^2*Log[x]^2*Log[c*x^n] - 6*b^3*f^2*m*n^2*x^2* Log[x]^2*Log[c*x^n] + 4*b^3*f^2*m*n^2*x^2*Log[x]^3*Log[c*x^n] + 12*a*b^2*e *f*m*x*Log[c*x^n]^2 + 18*b^3*e*f*m*n*x*Log[c*x^n]^2 + 12*a*b^2*f^2*m*x^2*L og[x]*Log[c*x^n]^2 + 6*b^3*f^2*m*n*x^2*Log[x]*Log[c*x^n]^2 - 6*b^3*f^2*m*n *x^2*Log[x]^2*Log[c*x^n]^2 + 4*b^3*e*f*m*x*Log[c*x^n]^3 + 4*b^3*f^2*m*x^2* Log[x]*Log[c*x^n]^3 - 4*a^3*f^2*m*x^2*Log[e + f*x] - 6*a^2*b*f^2*m*n*x^2*L og[e + f*x] - 6*a*b^2*f^2*m*n^2*x^2*Log[e + f*x] - 3*b^3*f^2*m*n^3*x^2*Log [e + f*x] + 12*a^2*b*f^2*m*n*x^2*Log[x]*Log[e + f*x] + 12*a*b^2*f^2*m*n^2* x^2*Log[x]*Log[e + f*x] + 6*b^3*f^2*m*n^3*x^2*Log[x]*Log[e + f*x] - 12*a*b ^2*f^2*m*n^2*x^2*Log[x]^2*Log[e + f*x] - 6*b^3*f^2*m*n^3*x^2*Log[x]^2*Log[ e + f*x] + 4*b^3*f^2*m*n^3*x^2*Log[x]^3*Log[e + f*x] - 12*a^2*b*f^2*m*x^2* Log[c*x^n]*Log[e + f*x] - 12*a*b^2*f^2*m*n*x^2*Log[c*x^n]*Log[e + f*x] ...
Time = 1.12 (sec) , antiderivative size = 519, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2825, 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.
\(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2825 |
\(\displaystyle -f m \int \left (-\frac {3 b^3 n^3}{8 x^2 (e+f x)}-\frac {3 b^2 \left (a+b \log \left (c x^n\right )\right ) n^2}{4 x^2 (e+f x)}-\frac {3 b \left (a+b \log \left (c x^n\right )\right )^2 n}{4 x^2 (e+f x)}-\frac {\left (a+b \log \left (c x^n\right )\right )^3}{2 x^2 (e+f x)}\right )dx-\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{4 x^2}-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{2 x^2}-\frac {3 b^3 n^3 \log \left (d (e+f x)^m\right )}{8 x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{4 x^2}-f m \left (\frac {3 b^2 f n^2 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {3 b^2 f n^2 \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {3 b^2 f n^2 \log \left (\frac {e}{f x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}+\frac {21 b^2 n^2 \left (a+b \log \left (c x^n\right )\right )}{4 e x}+\frac {3 b f n \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{2 e^2}-\frac {3 b f n \log \left (\frac {e}{f x}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{4 e^2}-\frac {f \log \left (\frac {e}{f x}+1\right ) \left (a+b \log \left (c x^n\right )\right )^3}{2 e^2}+\frac {9 b n \left (a+b \log \left (c x^n\right )\right )^2}{4 e x}+\frac {\left (a+b \log \left (c x^n\right )\right )^3}{2 e x}+\frac {3 b^3 f n^3 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{4 e^2}+\frac {3 b^3 f n^3 \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right )}{2 e^2}+\frac {3 b^3 f n^3 \operatorname {PolyLog}\left (4,-\frac {e}{f x}\right )}{e^2}+\frac {3 b^3 f n^3 \log (x)}{8 e^2}-\frac {3 b^3 f n^3 \log (e+f x)}{8 e^2}+\frac {45 b^3 n^3}{8 e x}\right )-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{2 x^2}-\frac {3 b^3 n^3 \log \left (d (e+f x)^m\right )}{8 x^2}\) |
(-3*b^3*n^3*Log[d*(e + f*x)^m])/(8*x^2) - (3*b^2*n^2*(a + b*Log[c*x^n])*Lo g[d*(e + f*x)^m])/(4*x^2) - (3*b*n*(a + b*Log[c*x^n])^2*Log[d*(e + f*x)^m] )/(4*x^2) - ((a + b*Log[c*x^n])^3*Log[d*(e + f*x)^m])/(2*x^2) - f*m*((45*b ^3*n^3)/(8*e*x) + (3*b^3*f*n^3*Log[x])/(8*e^2) + (21*b^2*n^2*(a + b*Log[c* x^n]))/(4*e*x) - (3*b^2*f*n^2*Log[1 + e/(f*x)]*(a + b*Log[c*x^n]))/(4*e^2) + (9*b*n*(a + b*Log[c*x^n])^2)/(4*e*x) - (3*b*f*n*Log[1 + e/(f*x)]*(a + b *Log[c*x^n])^2)/(4*e^2) + (a + b*Log[c*x^n])^3/(2*e*x) - (f*Log[1 + e/(f*x )]*(a + b*Log[c*x^n])^3)/(2*e^2) - (3*b^3*f*n^3*Log[e + f*x])/(8*e^2) + (3 *b^3*f*n^3*PolyLog[2, -(e/(f*x))])/(4*e^2) + (3*b^2*f*n^2*(a + b*Log[c*x^n ])*PolyLog[2, -(e/(f*x))])/(2*e^2) + (3*b*f*n*(a + b*Log[c*x^n])^2*PolyLog [2, -(e/(f*x))])/(2*e^2) + (3*b^3*f*n^3*PolyLog[3, -(e/(f*x))])/(2*e^2) + (3*b^2*f*n^2*(a + b*Log[c*x^n])*PolyLog[3, -(e/(f*x))])/e^2 + (3*b^3*f*n^3 *PolyLog[4, -(e/(f*x))])/e^2)
3.1.89.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))^(p_.)*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q* (a + b*Log[c*x^n])^p, x]}, Simp[Log[d*(e + f*x^m)^r] u, x] - Simp[f*m*r Int[x^(m - 1)/(e + f*x^m) u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m , n, q}, x] && IGtQ[p, 0] && RationalQ[m] && RationalQ[q]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 149.09 (sec) , antiderivative size = 21008, normalized size of antiderivative = 37.85
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{3}} \,d x } \]
integral((b^3*log(c*x^n)^3 + 3*a*b^2*log(c*x^n)^2 + 3*a^2*b*log(c*x^n) + a ^3)*log((f*x + e)^m*d)/x^3, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^3} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{3}} \,d x } \]
1/8*(4*(b^3*f^2*m*x^2*log(f*x + e) - b^3*f^2*m*x^2*log(x) - b^3*e*f*m*x - b^3*e^2*log(d))*log(x^n)^3 - (4*b^3*e^2*log(x^n)^3 + 4*a^3*e^2 + 6*(e^2*n + 2*e^2*log(c))*a^2*b + 6*(e^2*n^2 + 2*e^2*n*log(c) + 2*e^2*log(c)^2)*a*b^ 2 + (3*e^2*n^3 + 6*e^2*n^2*log(c) + 6*e^2*n*log(c)^2 + 4*e^2*log(c)^3)*b^3 + 6*(2*a*b^2*e^2 + (e^2*n + 2*e^2*log(c))*b^3)*log(x^n)^2 + 6*(2*a^2*b*e^ 2 + 2*(e^2*n + 2*e^2*log(c))*a*b^2 + (e^2*n^2 + 2*e^2*n*log(c) + 2*e^2*log (c)^2)*b^3)*log(x^n))*log((f*x + e)^m))/(e^2*x^2) - integrate(-1/8*(8*b^3* e^3*log(c)^3*log(d) + 24*a*b^2*e^3*log(c)^2*log(d) + 24*a^2*b*e^3*log(c)*l og(d) + 8*a^3*e^3*log(d) + 6*(2*b^3*e*f^2*m*n*x^2 + 4*a*b^2*e^3*log(d) + 2 *(e^3*n*log(d) + 2*e^3*log(c)*log(d))*b^3 + (2*(e^2*f*m + 2*e^2*f*log(d))* a*b^2 + (3*e^2*f*m*n + 2*e^2*f*n*log(d) + 2*(e^2*f*m + 2*e^2*f*log(d))*log (c))*b^3)*x - 2*(b^3*f^3*m*n*x^3 + b^3*e*f^2*m*n*x^2)*log(f*x + e) + 2*(b^ 3*f^3*m*n*x^3 + b^3*e*f^2*m*n*x^2)*log(x))*log(x^n)^2 + (4*(e^2*f*m + 2*e^ 2*f*log(d))*a^3 + 6*(e^2*f*m*n + 2*(e^2*f*m + 2*e^2*f*log(d))*log(c))*a^2* b + 6*(e^2*f*m*n^2 + 2*e^2*f*m*n*log(c) + 2*(e^2*f*m + 2*e^2*f*log(d))*log (c)^2)*a*b^2 + (3*e^2*f*m*n^3 + 6*e^2*f*m*n^2*log(c) + 6*e^2*f*m*n*log(c)^ 2 + 4*(e^2*f*m + 2*e^2*f*log(d))*log(c)^3)*b^3)*x + 6*(4*b^3*e^3*log(c)^2* log(d) + 8*a*b^2*e^3*log(c)*log(d) + 4*a^2*b*e^3*log(d) + (2*(e^2*f*m + 2* e^2*f*log(d))*a^2*b + 2*(e^2*f*m*n + 2*(e^2*f*m + 2*e^2*f*log(d))*log(c))* a*b^2 + (e^2*f*m*n^2 + 2*e^2*f*m*n*log(c) + 2*(e^2*f*m + 2*e^2*f*log(d)...
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^3} \, dx=\int \frac {\ln \left (d\,{\left (e+f\,x\right )}^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{x^3} \,d x \]