Integrand size = 26, antiderivative size = 420 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x^4} \, dx=-\frac {19 b^2 f m n^2}{108 e x^2}+\frac {26 b^2 f^2 m n^2}{27 e^2 x}+\frac {2 b^2 f^3 m n^2 \log (x)}{27 e^3}-\frac {5 b f m n \left (a+b \log \left (c x^n\right )\right )}{18 e x^2}+\frac {8 b f^2 m n \left (a+b \log \left (c x^n\right )\right )}{9 e^2 x}-\frac {2 b f^3 m n \log \left (1+\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {f m \left (a+b \log \left (c x^n\right )\right )^2}{6 e x^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )^2}{3 e^2 x}-\frac {f^3 m \log \left (1+\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3}-\frac {2 b^2 f^3 m n^2 \log (e+f x)}{27 e^3}-\frac {2 b^2 n^2 \log \left (d (e+f x)^m\right )}{27 x^3}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {2 b^2 f^3 m n^2 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{9 e^3}+\frac {2 b f^3 m n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{3 e^3}+\frac {2 b^2 f^3 m n^2 \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right )}{3 e^3} \]
-19/108*b^2*f*m*n^2/e/x^2+26/27*b^2*f^2*m*n^2/e^2/x+2/27*b^2*f^3*m*n^2*ln( x)/e^3-5/18*b*f*m*n*(a+b*ln(c*x^n))/e/x^2+8/9*b*f^2*m*n*(a+b*ln(c*x^n))/e^ 2/x-2/9*b*f^3*m*n*ln(1+e/f/x)*(a+b*ln(c*x^n))/e^3-1/6*f*m*(a+b*ln(c*x^n))^ 2/e/x^2+1/3*f^2*m*(a+b*ln(c*x^n))^2/e^2/x-1/3*f^3*m*ln(1+e/f/x)*(a+b*ln(c* x^n))^2/e^3-2/27*b^2*f^3*m*n^2*ln(f*x+e)/e^3-2/27*b^2*n^2*ln(d*(f*x+e)^m)/ x^3-2/9*b*n*(a+b*ln(c*x^n))*ln(d*(f*x+e)^m)/x^3-1/3*(a+b*ln(c*x^n))^2*ln(d *(f*x+e)^m)/x^3+2/9*b^2*f^3*m*n^2*polylog(2,-e/f/x)/e^3+2/3*b*f^3*m*n*(a+b *ln(c*x^n))*polylog(2,-e/f/x)/e^3+2/3*b^2*f^3*m*n^2*polylog(3,-e/f/x)/e^3
Leaf count is larger than twice the leaf count of optimal. \(909\) vs. \(2(420)=840\).
Time = 0.26 (sec) , antiderivative size = 909, normalized size of antiderivative = 2.16 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x^4} \, dx=-\frac {18 a^2 e^2 f m x+30 a b e^2 f m n x+19 b^2 e^2 f m n^2 x-36 a^2 e f^2 m x^2-96 a b e f^2 m n x^2-104 b^2 e f^2 m n^2 x^2-36 a^2 f^3 m x^3 \log (x)-24 a b f^3 m n x^3 \log (x)-8 b^2 f^3 m n^2 x^3 \log (x)+36 a b f^3 m n x^3 \log ^2(x)+12 b^2 f^3 m n^2 x^3 \log ^2(x)-12 b^2 f^3 m n^2 x^3 \log ^3(x)+36 a b e^2 f m x \log \left (c x^n\right )+30 b^2 e^2 f m n x \log \left (c x^n\right )-72 a b e f^2 m x^2 \log \left (c x^n\right )-96 b^2 e f^2 m n x^2 \log \left (c x^n\right )-72 a b f^3 m x^3 \log (x) \log \left (c x^n\right )-24 b^2 f^3 m n x^3 \log (x) \log \left (c x^n\right )+36 b^2 f^3 m n x^3 \log ^2(x) \log \left (c x^n\right )+18 b^2 e^2 f m x \log ^2\left (c x^n\right )-36 b^2 e f^2 m x^2 \log ^2\left (c x^n\right )-36 b^2 f^3 m x^3 \log (x) \log ^2\left (c x^n\right )+36 a^2 f^3 m x^3 \log (e+f x)+24 a b f^3 m n x^3 \log (e+f x)+8 b^2 f^3 m n^2 x^3 \log (e+f x)-72 a b f^3 m n x^3 \log (x) \log (e+f x)-24 b^2 f^3 m n^2 x^3 \log (x) \log (e+f x)+36 b^2 f^3 m n^2 x^3 \log ^2(x) \log (e+f x)+72 a b f^3 m x^3 \log \left (c x^n\right ) \log (e+f x)+24 b^2 f^3 m n x^3 \log \left (c x^n\right ) \log (e+f x)-72 b^2 f^3 m n x^3 \log (x) \log \left (c x^n\right ) \log (e+f x)+36 b^2 f^3 m x^3 \log ^2\left (c x^n\right ) \log (e+f x)+36 a^2 e^3 \log \left (d (e+f x)^m\right )+24 a b e^3 n \log \left (d (e+f x)^m\right )+8 b^2 e^3 n^2 \log \left (d (e+f x)^m\right )+72 a b e^3 \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )+24 b^2 e^3 n \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )+36 b^2 e^3 \log ^2\left (c x^n\right ) \log \left (d (e+f x)^m\right )+72 a b f^3 m n x^3 \log (x) \log \left (1+\frac {f x}{e}\right )+24 b^2 f^3 m n^2 x^3 \log (x) \log \left (1+\frac {f x}{e}\right )-36 b^2 f^3 m n^2 x^3 \log ^2(x) \log \left (1+\frac {f x}{e}\right )+72 b^2 f^3 m n x^3 \log (x) \log \left (c x^n\right ) \log \left (1+\frac {f x}{e}\right )+24 b f^3 m n x^3 \left (3 a+b n+3 b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )-72 b^2 f^3 m n^2 x^3 \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right )}{108 e^3 x^3} \]
-1/108*(18*a^2*e^2*f*m*x + 30*a*b*e^2*f*m*n*x + 19*b^2*e^2*f*m*n^2*x - 36* a^2*e*f^2*m*x^2 - 96*a*b*e*f^2*m*n*x^2 - 104*b^2*e*f^2*m*n^2*x^2 - 36*a^2* f^3*m*x^3*Log[x] - 24*a*b*f^3*m*n*x^3*Log[x] - 8*b^2*f^3*m*n^2*x^3*Log[x] + 36*a*b*f^3*m*n*x^3*Log[x]^2 + 12*b^2*f^3*m*n^2*x^3*Log[x]^2 - 12*b^2*f^3 *m*n^2*x^3*Log[x]^3 + 36*a*b*e^2*f*m*x*Log[c*x^n] + 30*b^2*e^2*f*m*n*x*Log [c*x^n] - 72*a*b*e*f^2*m*x^2*Log[c*x^n] - 96*b^2*e*f^2*m*n*x^2*Log[c*x^n] - 72*a*b*f^3*m*x^3*Log[x]*Log[c*x^n] - 24*b^2*f^3*m*n*x^3*Log[x]*Log[c*x^n ] + 36*b^2*f^3*m*n*x^3*Log[x]^2*Log[c*x^n] + 18*b^2*e^2*f*m*x*Log[c*x^n]^2 - 36*b^2*e*f^2*m*x^2*Log[c*x^n]^2 - 36*b^2*f^3*m*x^3*Log[x]*Log[c*x^n]^2 + 36*a^2*f^3*m*x^3*Log[e + f*x] + 24*a*b*f^3*m*n*x^3*Log[e + f*x] + 8*b^2* f^3*m*n^2*x^3*Log[e + f*x] - 72*a*b*f^3*m*n*x^3*Log[x]*Log[e + f*x] - 24*b ^2*f^3*m*n^2*x^3*Log[x]*Log[e + f*x] + 36*b^2*f^3*m*n^2*x^3*Log[x]^2*Log[e + f*x] + 72*a*b*f^3*m*x^3*Log[c*x^n]*Log[e + f*x] + 24*b^2*f^3*m*n*x^3*Lo g[c*x^n]*Log[e + f*x] - 72*b^2*f^3*m*n*x^3*Log[x]*Log[c*x^n]*Log[e + f*x] + 36*b^2*f^3*m*x^3*Log[c*x^n]^2*Log[e + f*x] + 36*a^2*e^3*Log[d*(e + f*x)^ m] + 24*a*b*e^3*n*Log[d*(e + f*x)^m] + 8*b^2*e^3*n^2*Log[d*(e + f*x)^m] + 72*a*b*e^3*Log[c*x^n]*Log[d*(e + f*x)^m] + 24*b^2*e^3*n*Log[c*x^n]*Log[d*( e + f*x)^m] + 36*b^2*e^3*Log[c*x^n]^2*Log[d*(e + f*x)^m] + 72*a*b*f^3*m*n* x^3*Log[x]*Log[1 + (f*x)/e] + 24*b^2*f^3*m*n^2*x^3*Log[x]*Log[1 + (f*x)/e] - 36*b^2*f^3*m*n^2*x^3*Log[x]^2*Log[1 + (f*x)/e] + 72*b^2*f^3*m*n*x^3*...
Time = 0.99 (sec) , antiderivative size = 403, normalized size of antiderivative = 0.96, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x^4} \, dx\) |
| \(\Big \downarrow \) 2825 |
| \(\displaystyle -f m \int \left (-\frac {2 b^2 n^2}{27 x^3 (e+f x)}-\frac {2 b \left (a+b \log \left (c x^n\right )\right ) n}{9 x^3 (e+f x)}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 x^3 (e+f x)}\right )dx-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{3 x^3}-\frac {2 b^2 n^2 \log \left (d (e+f x)^m\right )}{27 x^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -f m \left (-\frac {2 b f^2 n \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {f^2 \log \left (\frac {e}{f x}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3}+\frac {2 b f^2 n \log \left (\frac {e}{f x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {f \left (a+b \log \left (c x^n\right )\right )^2}{3 e^2 x}-\frac {8 b f n \left (a+b \log \left (c x^n\right )\right )}{9 e^2 x}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{6 e x^2}+\frac {5 b n \left (a+b \log \left (c x^n\right )\right )}{18 e x^2}-\frac {2 b^2 f^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{9 e^3}-\frac {2 b^2 f^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right )}{3 e^3}-\frac {2 b^2 f^2 n^2 \log (x)}{27 e^3}+\frac {2 b^2 f^2 n^2 \log (e+f x)}{27 e^3}-\frac {26 b^2 f n^2}{27 e^2 x}+\frac {19 b^2 n^2}{108 e x^2}\right )-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{3 x^3}-\frac {2 b^2 n^2 \log \left (d (e+f x)^m\right )}{27 x^3}\) |
(-2*b^2*n^2*Log[d*(e + f*x)^m])/(27*x^3) - (2*b*n*(a + b*Log[c*x^n])*Log[d *(e + f*x)^m])/(9*x^3) - ((a + b*Log[c*x^n])^2*Log[d*(e + f*x)^m])/(3*x^3) - f*m*((19*b^2*n^2)/(108*e*x^2) - (26*b^2*f*n^2)/(27*e^2*x) - (2*b^2*f^2* n^2*Log[x])/(27*e^3) + (5*b*n*(a + b*Log[c*x^n]))/(18*e*x^2) - (8*b*f*n*(a + b*Log[c*x^n]))/(9*e^2*x) + (2*b*f^2*n*Log[1 + e/(f*x)]*(a + b*Log[c*x^n ]))/(9*e^3) + (a + b*Log[c*x^n])^2/(6*e*x^2) - (f*(a + b*Log[c*x^n])^2)/(3 *e^2*x) + (f^2*Log[1 + e/(f*x)]*(a + b*Log[c*x^n])^2)/(3*e^3) + (2*b^2*f^2 *n^2*Log[e + f*x])/(27*e^3) - (2*b^2*f^2*n^2*PolyLog[2, -(e/(f*x))])/(9*e^ 3) - (2*b*f^2*n*(a + b*Log[c*x^n])*PolyLog[2, -(e/(f*x))])/(3*e^3) - (2*b^ 2*f^2*n^2*PolyLog[3, -(e/(f*x))])/(3*e^3))
3.1.84.3.1 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 = 33.46 (sec) , antiderivative size = 6242, normalized size of antiderivative = 14.86
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x^4} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{4}} \,d x } \]
-1/54*(9*(2*b^2*f^3*m*x^3*log(f*x + e) - 2*b^2*f^3*m*x^3*log(x) - 2*b^2*e* f^2*m*x^2 + b^2*e^2*f*m*x + 2*b^2*e^3*log(d))*log(x^n)^2 + 2*(9*b^2*e^3*lo g(x^n)^2 + 9*a^2*e^3 + 6*(e^3*n + 3*e^3*log(c))*a*b + (2*e^3*n^2 + 6*e^3*n *log(c) + 9*e^3*log(c)^2)*b^2 + 6*(3*a*b*e^3 + (e^3*n + 3*e^3*log(c))*b^2) *log(x^n))*log((f*x + e)^m))/(e^3*x^3) + integrate(1/27*(27*b^2*e^4*log(c) ^2*log(d) + 54*a*b*e^4*log(c)*log(d) + 27*a^2*e^4*log(d) + (9*(e^3*f*m + 3 *e^3*f*log(d))*a^2 + 6*(e^3*f*m*n + 3*(e^3*f*m + 3*e^3*f*log(d))*log(c))*a *b + (2*e^3*f*m*n^2 + 6*e^3*f*m*n*log(c) + 9*(e^3*f*m + 3*e^3*f*log(d))*lo g(c)^2)*b^2)*x - 3*(6*b^2*e*f^3*m*n*x^3 + 3*b^2*e^2*f^2*m*n*x^2 - 18*a*b*e ^4*log(d) - 6*(e^4*n*log(d) + 3*e^4*log(c)*log(d))*b^2 - (6*(e^3*f*m + 3*e ^3*f*log(d))*a*b + (5*e^3*f*m*n + 6*e^3*f*n*log(d) + 6*(e^3*f*m + 3*e^3*f* log(d))*log(c))*b^2)*x - 6*(b^2*f^4*m*n*x^4 + b^2*e*f^3*m*n*x^3)*log(f*x + e) + 6*(b^2*f^4*m*n*x^4 + b^2*e*f^3*m*n*x^3)*log(x))*log(x^n))/(e^3*f*x^5 + e^4*x^4), x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x^4} \, dx=\int \frac {\ln \left (d\,{\left (e+f\,x\right )}^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{x^4} \,d x \]