Integrand size = 26, antiderivative size = 251 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=-\frac {8 b e m n x}{9 f}+\frac {4}{27} b m n x^3+\frac {2 b e^{3/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 f^{3/2}}+\frac {2 e m x \left (a+b \log \left (c x^n\right )\right )}{3 f}-\frac {2}{9} m x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^{3/2}}-\frac {1}{9} b n x^3 \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {i b e^{3/2} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}-\frac {i b e^{3/2} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}} \]
-8/9*b*e*m*n*x/f+4/27*b*m*n*x^3+2/9*b*e^(3/2)*m*n*arctan(x*f^(1/2)/e^(1/2) )/f^(3/2)+2/3*e*m*x*(a+b*ln(c*x^n))/f-2/9*m*x^3*(a+b*ln(c*x^n))-2/3*e^(3/2 )*m*arctan(x*f^(1/2)/e^(1/2))*(a+b*ln(c*x^n))/f^(3/2)-1/9*b*n*x^3*ln(d*(f* x^2+e)^m)+1/3*x^3*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)+1/3*I*b*e^(3/2)*m*n*po lylog(2,-I*x*f^(1/2)/e^(1/2))/f^(3/2)-1/3*I*b*e^(3/2)*m*n*polylog(2,I*x*f^ (1/2)/e^(1/2))/f^(3/2)
Time = 0.13 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.55 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\frac {18 a e \sqrt {f} m x-24 b e \sqrt {f} m n x-6 a f^{3/2} m x^3+4 b f^{3/2} m n x^3-18 a e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+6 b e^{3/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+18 b e^{3/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)+18 b e \sqrt {f} m x \log \left (c x^n\right )-6 b f^{3/2} m x^3 \log \left (c x^n\right )-18 b e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )-9 i b e^{3/2} m n \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+9 i b e^{3/2} m n \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+9 a f^{3/2} x^3 \log \left (d \left (e+f x^2\right )^m\right )-3 b f^{3/2} n x^3 \log \left (d \left (e+f x^2\right )^m\right )+9 b f^{3/2} x^3 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+9 i b e^{3/2} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-9 i b e^{3/2} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{27 f^{3/2}} \]
(18*a*e*Sqrt[f]*m*x - 24*b*e*Sqrt[f]*m*n*x - 6*a*f^(3/2)*m*x^3 + 4*b*f^(3/ 2)*m*n*x^3 - 18*a*e^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]] + 6*b*e^(3/2)*m*n* ArcTan[(Sqrt[f]*x)/Sqrt[e]] + 18*b*e^(3/2)*m*n*ArcTan[(Sqrt[f]*x)/Sqrt[e]] *Log[x] + 18*b*e*Sqrt[f]*m*x*Log[c*x^n] - 6*b*f^(3/2)*m*x^3*Log[c*x^n] - 1 8*b*e^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[c*x^n] - (9*I)*b*e^(3/2)*m*n *Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + (9*I)*b*e^(3/2)*m*n*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 9*a*f^(3/2)*x^3*Log[d*(e + f*x^2)^m] - 3*b*f^(3 /2)*n*x^3*Log[d*(e + f*x^2)^m] + 9*b*f^(3/2)*x^3*Log[c*x^n]*Log[d*(e + f*x ^2)^m] + (9*I)*b*e^(3/2)*m*n*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]] - (9*I)* b*e^(3/2)*m*n*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/(27*f^(3/2))
Time = 0.41 (sec) , antiderivative size = 244, 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.
\(\displaystyle \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx\) |
\(\Big \downarrow \) 2823 |
\(\displaystyle -b n \int \left (-\frac {2 m x^2}{9}+\frac {1}{3} \log \left (d \left (f x^2+e\right )^m\right ) x^2+\frac {2 e m}{3 f}-\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2} x}\right )dx-\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^{3/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {2 e m x \left (a+b \log \left (c x^n\right )\right )}{3 f}-\frac {2}{9} m x^3 \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^{3/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {2 e m x \left (a+b \log \left (c x^n\right )\right )}{3 f}-\frac {2}{9} m x^3 \left (a+b \log \left (c x^n\right )\right )-b n \left (-\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 f^{3/2}}+\frac {1}{9} x^3 \log \left (d \left (e+f x^2\right )^m\right )-\frac {i e^{3/2} m \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}+\frac {i e^{3/2} m \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}+\frac {8 e m x}{9 f}-\frac {4 m x^3}{27}\right )\) |
(2*e*m*x*(a + b*Log[c*x^n]))/(3*f) - (2*m*x^3*(a + b*Log[c*x^n]))/9 - (2*e ^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*(a + b*Log[c*x^n]))/(3*f^(3/2)) + (x^ 3*(a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/3 - b*n*((8*e*m*x)/(9*f) - (4*m *x^3)/27 - (2*e^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(9*f^(3/2)) + (x^3*Lo g[d*(e + f*x^2)^m])/9 - ((I/3)*e^(3/2)*m*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[ e]])/f^(3/2) + ((I/3)*e^(3/2)*m*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/f^(3/2) )
3.1.95.3.1 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 = 88.75 (sec) , antiderivative size = 1082, normalized size of antiderivative = 4.31
2/3*m/f*e*x*b*ln(c)+2/3*m/f*b*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*n*ln (x)-1/3*m/f*b*n*e^2*ln(x)/(-e*f)^(1/2)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2) )+1/3*m/f*b*n*e^2*ln(x)/(-e*f)^(1/2)*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1 /3*I*m/f*e*x*b*Pi*csgn(I*c*x^n)^3+1/9*I*m*x^3*b*Pi*csgn(I*c)*csgn(I*x^n)*c sgn(I*c*x^n)+1/3*I*m/f*e*x*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/3*I*m/f*e*x*b* Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/3*I*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^ (1/2))*b*Pi*csgn(I*c*x^n)^3-2/3*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2) )*a+1/9*I*m*x^3*b*Pi*csgn(I*c*x^n)^3+(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))*(1/3*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(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^3+2/3*b*x^3*ln(x^n)-2/9*b*n*x^3)+1/3*I*m/f*e^2/ (e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^ n)-1/3*I*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I *c*x^n)^2-1/3*I*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*x^ n)*csgn(I*c*x^n)^2-1/3*I*m/f*e*x*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+ (1/3*b*x^3*ln(x^n)+1/18*x^3*(-3*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n) +3*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+3*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-3 *I*b*Pi*csgn(I*c*x^n)^3+6*b*ln(c)-2*b*n+6*a))*ln((f*x^2+e)^m)-2/3*m/f*e...
\[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Timed out} \]
Exception generated. \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int x^2\,\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]