Integrand size = 26, antiderivative size = 221 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=-\frac {3 b e m n x^2}{16 f}+\frac {1}{16} b m n x^4+\frac {e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac {1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {b e^2 m n \log \left (e+f x^2\right )}{16 f^2}+\frac {b e^2 m n \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 f^2}-\frac {1}{16} b n x^4 \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {b e^2 m n \operatorname {PolyLog}\left (2,1+\frac {f x^2}{e}\right )}{8 f^2} \]
-3/16*b*e*m*n*x^2/f+1/16*b*m*n*x^4+1/4*e*m*x^2*(a+b*ln(c*x^n))/f-1/8*m*x^4 *(a+b*ln(c*x^n))+1/16*b*e^2*m*n*ln(f*x^2+e)/f^2+1/8*b*e^2*m*n*ln(-f*x^2/e) *ln(f*x^2+e)/f^2-1/4*e^2*m*(a+b*ln(c*x^n))*ln(f*x^2+e)/f^2-1/16*b*n*x^4*ln (d*(f*x^2+e)^m)+1/4*x^4*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)+1/8*b*e^2*m*n*po lylog(2,1+f*x^2/e)/f^2
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.47 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=-\frac {-4 a e f m x^2+3 b e f m n x^2+2 a f^2 m x^4-b f^2 m n x^4-4 b e f m x^2 \log \left (c x^n\right )+2 b f^2 m x^4 \log \left (c x^n\right )+4 b e^2 m n \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+4 b e^2 m n \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+4 a e^2 m \log \left (e+f x^2\right )-b e^2 m n \log \left (e+f x^2\right )-4 b e^2 m n \log (x) \log \left (e+f x^2\right )+4 b e^2 m \log \left (c x^n\right ) \log \left (e+f x^2\right )-4 a f^2 x^4 \log \left (d \left (e+f x^2\right )^m\right )+b f^2 n x^4 \log \left (d \left (e+f x^2\right )^m\right )-4 b f^2 x^4 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+4 b e^2 m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+4 b e^2 m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{16 f^2} \]
-1/16*(-4*a*e*f*m*x^2 + 3*b*e*f*m*n*x^2 + 2*a*f^2*m*x^4 - b*f^2*m*n*x^4 - 4*b*e*f*m*x^2*Log[c*x^n] + 2*b*f^2*m*x^4*Log[c*x^n] + 4*b*e^2*m*n*Log[x]*L og[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 4*b*e^2*m*n*Log[x]*Log[1 + (I*Sqrt[f]*x)/S qrt[e]] + 4*a*e^2*m*Log[e + f*x^2] - b*e^2*m*n*Log[e + f*x^2] - 4*b*e^2*m* n*Log[x]*Log[e + f*x^2] + 4*b*e^2*m*Log[c*x^n]*Log[e + f*x^2] - 4*a*f^2*x^ 4*Log[d*(e + f*x^2)^m] + b*f^2*n*x^4*Log[d*(e + f*x^2)^m] - 4*b*f^2*x^4*Lo g[c*x^n]*Log[d*(e + f*x^2)^m] + 4*b*e^2*m*n*PolyLog[2, ((-I)*Sqrt[f]*x)/Sq rt[e]] + 4*b*e^2*m*n*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/f^2
Time = 0.46 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2823, 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.
| \(\displaystyle \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (-\frac {m x^3}{8}+\frac {1}{4} \log \left (d \left (f x^2+e\right )^m\right ) x^3+\frac {e m x}{4 f}-\frac {e^2 m \log \left (f x^2+e\right )}{4 f^2 x}\right )dx+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac {e^2 m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{4 f^2}+\frac {e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac {1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac {e^2 m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{4 f^2}+\frac {e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac {1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {1}{16} x^4 \log \left (d \left (e+f x^2\right )^m\right )-\frac {e^2 m \operatorname {PolyLog}\left (2,\frac {f x^2}{e}+1\right )}{8 f^2}-\frac {e^2 m \log \left (e+f x^2\right )}{16 f^2}-\frac {e^2 m \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 f^2}+\frac {3 e m x^2}{16 f}-\frac {m x^4}{16}\right )\) |
(e*m*x^2*(a + b*Log[c*x^n]))/(4*f) - (m*x^4*(a + b*Log[c*x^n]))/8 - (e^2*m *(a + b*Log[c*x^n])*Log[e + f*x^2])/(4*f^2) + (x^4*(a + b*Log[c*x^n])*Log[ d*(e + f*x^2)^m])/4 - b*n*((3*e*m*x^2)/(16*f) - (m*x^4)/16 - (e^2*m*Log[e + f*x^2])/(16*f^2) - (e^2*m*Log[-((f*x^2)/e)]*Log[e + f*x^2])/(8*f^2) + (x ^4*Log[d*(e + f*x^2)^m])/16 - (e^2*m*PolyLog[2, 1 + (f*x^2)/e])/(8*f^2))
3.1.90.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* (e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[1/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 180.77 (sec) , antiderivative size = 1031, normalized size of antiderivative = 4.67
(1/4*b*x^4*ln(x^n)+1/16*x^4*(-2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n) +2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-2 *I*b*Pi*csgn(I*c*x^n)^3+4*b*ln(c)-b*n+4*a))*ln((f*x^2+e)^m)-1/8*I*m/f*e*x^ 2*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/8*x^4*a*m+(1/4*I*Pi*csgn(I*(f *x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2-1/4*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*( f*x^2+e)^m)*csgn(I*d)-1/4*I*Pi*csgn(I*d*(f*x^2+e)^m)^3+1/4*I*Pi*csgn(I*d*( f*x^2+e)^m)^2*csgn(I*d)+1/2*ln(d))*(1/4*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csg n(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^ n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)+2*a)*x^4+1/2*b*x^4*ln(x^n)-1/8*b*n*x ^4)+1/4*m/f*e*x^2*b*ln(c)+1/8*I*m/f*e*x^2*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1 /8*I*m/f*e*x^2*b*Pi*csgn(I*c*x^n)^3+1/4*m/f*b*ln(x^n)*e*x^2-1/4*m/f^2*b*n* e^2*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/4*m/f^2*b*n*e^2*ln(x)*ln( (f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/8*I*m/f^2*e^2*ln(f*x^2+e)*b*Pi*csgn(I*c )*csgn(I*c*x^n)^2-1/8*I*m/f^2*e^2*ln(f*x^2+e)*b*Pi*csgn(I*x^n)*csgn(I*c*x^ n)^2+1/8*I*m/f^2*e^2*ln(f*x^2+e)*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)- 1/8*m*x^4*b*ln(c)-1/8*m*b*ln(x^n)*x^4+1/8*m/f^2*b*n*e^2-1/4*m/f^2*e^2*ln(f *x^2+e)*a+1/16*I*m*x^4*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/8*I*m/f^ 2*e^2*ln(f*x^2+e)*b*Pi*csgn(I*c*x^n)^3+1/16*b*m*n*x^4-3/16*b*e*m*n*x^2/f+1 /8*I*m/f*e*x^2*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/4*m/f^2*b*n*e^2*ln(x)*ln (f*x^2+e)-1/16*I*m*x^4*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/16*I*m*x^4*b*Pi...
\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Timed out} \]
\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
1/16*(4*b*x^4*log(x^n) - (b*(n - 4*log(c)) - 4*a)*x^4)*log((f*x^2 + e)^m) + integrate(-1/8*((4*(f*m - 2*f*log(d))*a - (f*m*n - 4*(f*m - 2*f*log(d))* log(c))*b)*x^5 - 8*(b*e*log(c)*log(d) + a*e*log(d))*x^3 + 4*((f*m - 2*f*lo g(d))*b*x^5 - 2*b*e*x^3*log(d))*log(x^n))/(f*x^2 + e), x)
\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int x^3\,\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]