Integrand size = 24, antiderivative size = 603 \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right ) \, dx=\frac {21 a b^2 e m n^2 x}{4 f}-\frac {45 b^3 e m n^3 x}{8 f}+\frac {3}{4} b^3 m n^3 x^2+\frac {21 b^3 e m n^2 x \log \left (c x^n\right )}{4 f}-\frac {9}{8} b^2 m n^2 x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {9 b e m n x \left (a+b \log \left (c x^n\right )\right )^2}{4 f}+\frac {3}{4} b m n x^2 \left (a+b \log \left (c x^n\right )\right )^2+\frac {e m x \left (a+b \log \left (c x^n\right )\right )^3}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )^3+\frac {3 b^3 e^2 m n^3 \log (e+f x)}{8 f^2}-\frac {3}{8} b^3 n^3 x^2 \log \left (d (e+f x)^m\right )+\frac {3}{4} b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )-\frac {3}{4} b n x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )-\frac {3 b^2 e^2 m n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {f x}{e}\right )}{4 f^2}+\frac {3 b e^2 m n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x}{e}\right )}{4 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right )^3 \log \left (1+\frac {f x}{e}\right )}{2 f^2}-\frac {3 b^3 e^2 m n^3 \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{4 f^2}+\frac {3 b^2 e^2 m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{2 f^2}-\frac {3 b e^2 m n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{2 f^2}-\frac {3 b^3 e^2 m n^3 \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right )}{2 f^2}+\frac {3 b^2 e^2 m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right )}{f^2}-\frac {3 b^3 e^2 m n^3 \operatorname {PolyLog}\left (4,-\frac {f x}{e}\right )}{f^2} \]
21/4*a*b^2*e*m*n^2*x/f-45/8*b^3*e*m*n^3*x/f+3/4*b^3*m*n^3*x^2+21/4*b^3*e*m *n^2*x*ln(c*x^n)/f-9/8*b^2*m*n^2*x^2*(a+b*ln(c*x^n))-9/4*b*e*m*n*x*(a+b*ln (c*x^n))^2/f+3/4*b*m*n*x^2*(a+b*ln(c*x^n))^2+1/2*e*m*x*(a+b*ln(c*x^n))^3/f -1/4*m*x^2*(a+b*ln(c*x^n))^3+3/8*b^3*e^2*m*n^3*ln(f*x+e)/f^2-3/8*b^3*n^3*x ^2*ln(d*(f*x+e)^m)+3/4*b^2*n^2*x^2*(a+b*ln(c*x^n))*ln(d*(f*x+e)^m)-3/4*b*n *x^2*(a+b*ln(c*x^n))^2*ln(d*(f*x+e)^m)+1/2*x^2*(a+b*ln(c*x^n))^3*ln(d*(f*x +e)^m)-3/4*b^2*e^2*m*n^2*(a+b*ln(c*x^n))*ln(1+f*x/e)/f^2+3/4*b*e^2*m*n*(a+ b*ln(c*x^n))^2*ln(1+f*x/e)/f^2-1/2*e^2*m*(a+b*ln(c*x^n))^3*ln(1+f*x/e)/f^2 -3/4*b^3*e^2*m*n^3*polylog(2,-f*x/e)/f^2+3/2*b^2*e^2*m*n^2*(a+b*ln(c*x^n)) *polylog(2,-f*x/e)/f^2-3/2*b*e^2*m*n*(a+b*ln(c*x^n))^2*polylog(2,-f*x/e)/f ^2-3/2*b^3*e^2*m*n^3*polylog(3,-f*x/e)/f^2+3*b^2*e^2*m*n^2*(a+b*ln(c*x^n)) *polylog(3,-f*x/e)/f^2-3*b^3*e^2*m*n^3*polylog(4,-f*x/e)/f^2
Leaf count is larger than twice the leaf count of optimal. \(1431\) vs. \(2(603)=1206\).
Time = 0.33 (sec) , antiderivative size = 1431, normalized size of antiderivative = 2.37 \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right ) \, dx =\text {Too large to display} \]
(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 - 2*a^3*f^2*m*x^2 + 6*a^2*b*f^2*m*n*x^2 - 9*a*b^2*f^2*m*n^2*x^2 + 6* b^3*f^2*m*n^3*x^2 + 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] - 6*a^2*b*f^2*m*x^2*Log[c*x^n] + 12* a*b^2*f^2*m*n*x^2*Log[c*x^n] - 9*b^3*f^2*m*n^2*x^2*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 - 6*a*b^2*f^2*m*x^2*Lo g[c*x^n]^2 + 6*b^3*f^2*m*n*x^2*Log[c*x^n]^2 + 4*b^3*e*f*m*x*Log[c*x^n]^3 - 2*b^3*f^2*m*x^2*Log[c*x^n]^3 - 4*a^3*e^2*m*Log[e + f*x] + 6*a^2*b*e^2*m*n *Log[e + f*x] - 6*a*b^2*e^2*m*n^2*Log[e + f*x] + 3*b^3*e^2*m*n^3*Log[e + f *x] + 12*a^2*b*e^2*m*n*Log[x]*Log[e + f*x] - 12*a*b^2*e^2*m*n^2*Log[x]*Log [e + f*x] + 6*b^3*e^2*m*n^3*Log[x]*Log[e + f*x] - 12*a*b^2*e^2*m*n^2*Log[x ]^2*Log[e + f*x] + 6*b^3*e^2*m*n^3*Log[x]^2*Log[e + f*x] + 4*b^3*e^2*m*n^3 *Log[x]^3*Log[e + f*x] - 12*a^2*b*e^2*m*Log[c*x^n]*Log[e + f*x] + 12*a*b^2 *e^2*m*n*Log[c*x^n]*Log[e + f*x] - 6*b^3*e^2*m*n^2*Log[c*x^n]*Log[e + f*x] + 24*a*b^2*e^2*m*n*Log[x]*Log[c*x^n]*Log[e + f*x] - 12*b^3*e^2*m*n^2*Log[ x]*Log[c*x^n]*Log[e + f*x] - 12*b^3*e^2*m*n^2*Log[x]^2*Log[c*x^n]*Log[e + f*x] - 12*a*b^2*e^2*m*Log[c*x^n]^2*Log[e + f*x] + 6*b^3*e^2*m*n*Log[c*x^n] ^2*Log[e + f*x] + 12*b^3*e^2*m*n*Log[x]*Log[c*x^n]^2*Log[e + f*x] - 4*b^3* e^2*m*Log[c*x^n]^3*Log[e + f*x] + 4*a^3*f^2*x^2*Log[d*(e + f*x)^m] - 6*a^2 *b*f^2*n*x^2*Log[d*(e + f*x)^m] + 6*a*b^2*f^2*n^2*x^2*Log[d*(e + f*x)^m...
Time = 1.09 (sec) , antiderivative size = 601, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.
\(\displaystyle \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right ) \, dx\) |
\(\Big \downarrow \) 2825 |
\(\displaystyle -f m \int \left (-\frac {3 b^3 x^2 n^3}{8 (e+f x)}+\frac {3 b^2 x^2 \left (a+b \log \left (c x^n\right )\right ) n^2}{4 (e+f x)}-\frac {3 b x^2 \left (a+b \log \left (c x^n\right )\right )^2 n}{4 (e+f x)}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^3}{2 (e+f x)}\right )dx+\frac {3}{4} b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )-\frac {3}{4} b n x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )-\frac {3}{8} b^3 n^3 x^2 \log \left (d (e+f x)^m\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{4} b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )-f m \left (-\frac {3 b^2 e^2 n^2 \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 f^3}-\frac {3 b^2 e^2 n^2 \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right ) \left (a+b \log \left (c x^n\right )\right )}{f^3}+\frac {3 b^2 e^2 n^2 \log \left (\frac {f x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{4 f^3}+\frac {9 b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{8 f}-\frac {21 a b^2 e n^2 x}{4 f^2}+\frac {3 b e^2 n \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{2 f^3}-\frac {3 b e^2 n \log \left (\frac {f x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{4 f^3}+\frac {e^2 \log \left (\frac {f x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^3}{2 f^3}+\frac {9 b e n x \left (a+b \log \left (c x^n\right )\right )^2}{4 f^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )^3}{2 f^2}-\frac {3 b n x^2 \left (a+b \log \left (c x^n\right )\right )^2}{4 f}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^3}{4 f}-\frac {21 b^3 e n^2 x \log \left (c x^n\right )}{4 f^2}+\frac {3 b^3 e^2 n^3 \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{4 f^3}+\frac {3 b^3 e^2 n^3 \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right )}{2 f^3}+\frac {3 b^3 e^2 n^3 \operatorname {PolyLog}\left (4,-\frac {f x}{e}\right )}{f^3}-\frac {3 b^3 e^2 n^3 \log (e+f x)}{8 f^3}+\frac {45 b^3 e n^3 x}{8 f^2}-\frac {3 b^3 n^3 x^2}{4 f}\right )-\frac {3}{4} b n x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )-\frac {3}{8} b^3 n^3 x^2 \log \left (d (e+f x)^m\right )\) |
(-3*b^3*n^3*x^2*Log[d*(e + f*x)^m])/8 + (3*b^2*n^2*x^2*(a + b*Log[c*x^n])* Log[d*(e + f*x)^m])/4 - (3*b*n*x^2*(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 - f*m*((-21*a*b^2*e* n^2*x)/(4*f^2) + (45*b^3*e*n^3*x)/(8*f^2) - (3*b^3*n^3*x^2)/(4*f) - (21*b^ 3*e*n^2*x*Log[c*x^n])/(4*f^2) + (9*b^2*n^2*x^2*(a + b*Log[c*x^n]))/(8*f) + (9*b*e*n*x*(a + b*Log[c*x^n])^2)/(4*f^2) - (3*b*n*x^2*(a + b*Log[c*x^n])^ 2)/(4*f) - (e*x*(a + b*Log[c*x^n])^3)/(2*f^2) + (x^2*(a + b*Log[c*x^n])^3) /(4*f) - (3*b^3*e^2*n^3*Log[e + f*x])/(8*f^3) + (3*b^2*e^2*n^2*(a + b*Log[ c*x^n])*Log[1 + (f*x)/e])/(4*f^3) - (3*b*e^2*n*(a + b*Log[c*x^n])^2*Log[1 + (f*x)/e])/(4*f^3) + (e^2*(a + b*Log[c*x^n])^3*Log[1 + (f*x)/e])/(2*f^3) + (3*b^3*e^2*n^3*PolyLog[2, -((f*x)/e)])/(4*f^3) - (3*b^2*e^2*n^2*(a + b*L og[c*x^n])*PolyLog[2, -((f*x)/e)])/(2*f^3) + (3*b*e^2*n*(a + b*Log[c*x^n]) ^2*PolyLog[2, -((f*x)/e)])/(2*f^3) + (3*b^3*e^2*n^3*PolyLog[3, -((f*x)/e)] )/(2*f^3) - (3*b^2*e^2*n^2*(a + b*Log[c*x^n])*PolyLog[3, -((f*x)/e)])/f^3 + (3*b^3*e^2*n^3*PolyLog[4, -((f*x)/e)])/f^3)
3.1.85.3.1 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 = 0.10 (sec) , antiderivative size = 19601, normalized size of antiderivative = 32.51
\[\text {output too large to display}\]
\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x \log \left ({\left (f x + e\right )}^{m} d\right ) \,d x } \]
integral((b^3*x*log(c*x^n)^3 + 3*a*b^2*x*log(c*x^n)^2 + 3*a^2*b*x*log(c*x^ n) + a^3*x)*log((f*x + e)^m*d), x)
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right ) \, dx=\text {Timed out} \]
\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x \log \left ({\left (f x + e\right )}^{m} d\right ) \,d x } \]
1/8*(2*(2*b^3*e*f*m*x - 2*b^3*e^2*m*log(f*x + e) - (f^2*m - 2*f^2*log(d))* b^3*x^2)*log(x^n)^3 + (4*b^3*f^2*x^2*log(x^n)^3 + 6*(2*a*b^2*f^2 - (f^2*n - 2*f^2*log(c))*b^3)*x^2*log(x^n)^2 + 6*(2*a^2*b*f^2 - 2*(f^2*n - 2*f^2*lo g(c))*a*b^2 + (f^2*n^2 - 2*f^2*n*log(c) + 2*f^2*log(c)^2)*b^3)*x^2*log(x^n ) + (4*a^3*f^2 - 6*(f^2*n - 2*f^2*log(c))*a^2*b + 6*(f^2*n^2 - 2*f^2*n*log (c) + 2*f^2*log(c)^2)*a*b^2 - (3*f^2*n^3 - 6*f^2*n^2*log(c) + 6*f^2*n*log( c)^2 - 4*f^2*log(c)^3)*b^3)*x^2)*log((f*x + e)^m))/f^2 + integrate(-1/8*(( 4*(f^3*m - 2*f^3*log(d))*a^3 - 6*(f^3*m*n - 2*(f^3*m - 2*f^3*log(d))*log(c ))*a^2*b + 6*(f^3*m*n^2 - 2*f^3*m*n*log(c) + 2*(f^3*m - 2*f^3*log(d))*log( c)^2)*a*b^2 - (3*f^3*m*n^3 - 6*f^3*m*n^2*log(c) + 6*f^3*m*n*log(c)^2 - 4*( f^3*m - 2*f^3*log(d))*log(c)^3)*b^3)*x^3 - 8*(b^3*e*f^2*log(c)^3*log(d) + 3*a*b^2*e*f^2*log(c)^2*log(d) + 3*a^2*b*e*f^2*log(c)*log(d) + a^3*e*f^2*lo g(d))*x^2 + 6*(2*b^3*e^2*f*m*n*x + 2*((f^3*m - 2*f^3*log(d))*a*b^2 - (f^3* m*n - f^3*n*log(d) - (f^3*m - 2*f^3*log(d))*log(c))*b^3)*x^3 - (4*a*b^2*e* f^2*log(d) - (e*f^2*m*n + 2*e*f^2*n*log(d) - 4*e*f^2*log(c)*log(d))*b^3)*x ^2 - 2*(b^3*e^2*f*m*n*x + b^3*e^3*m*n)*log(f*x + e))*log(x^n)^2 + 6*((2*(f ^3*m - 2*f^3*log(d))*a^2*b - 2*(f^3*m*n - 2*(f^3*m - 2*f^3*log(d))*log(c)) *a*b^2 + (f^3*m*n^2 - 2*f^3*m*n*log(c) + 2*(f^3*m - 2*f^3*log(d))*log(c)^2 )*b^3)*x^3 - 4*(b^3*e*f^2*log(c)^2*log(d) + 2*a*b^2*e*f^2*log(c)*log(d) + a^2*b*e*f^2*log(d))*x^2)*log(x^n))/(f^3*x^2 + e*f^2*x), x)
\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x \log \left ({\left (f x + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right ) \, dx=\int x\,\ln \left (d\,{\left (e+f\,x\right )}^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3 \,d x \]