Integrand size = 32, antiderivative size = 558 \[ \int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=-\frac {a g i^2 m x}{3 j^2}-\frac {b d^2 f n x}{3 e^2}+\frac {4 b d^2 g m n x}{9 e^2}+\frac {4 b g i^2 m n x}{9 j^2}+\frac {b d g i m n x}{3 e j}-\frac {5 b d g m n x^2}{36 e}-\frac {5 b g i m n x^2}{36 j}+\frac {2}{27} b g m n x^3-\frac {b d^3 g m n \log (d+e x)}{9 e^3}-\frac {b d^2 g i m n \log (d+e x)}{6 e^2 j}-\frac {b g i^2 m (d+e x) \log \left (c (d+e x)^n\right )}{3 e j^2}+\frac {g i m x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{6 j}-\frac {1}{9} g m x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b g i^3 m n \log (i+j x)}{9 j^3}-\frac {b d g i^2 m n \log (i+j x)}{6 e j^2}+\frac {g i^3 m \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (i+j x)}{e i-d j}\right )}{3 j^3}-\frac {b d^2 g n (i+j x) \log \left (h (i+j x)^m\right )}{3 e^2 j}+\frac {b d n x^2 \left (f+g \log \left (h (i+j x)^m\right )\right )}{6 e}-\frac {1}{9} b n x^3 \left (f+g \log \left (h (i+j x)^m\right )\right )+\frac {b d^3 n \log \left (-\frac {j (d+e x)}{e i-d j}\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )}{3 e^3}+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )+\frac {b g i^3 m n \operatorname {PolyLog}\left (2,-\frac {j (d+e x)}{e i-d j}\right )}{3 j^3}+\frac {b d^3 g m n \operatorname {PolyLog}\left (2,\frac {e (i+j x)}{e i-d j}\right )}{3 e^3} \]
-1/3*a*g*i^2*m*x/j^2-1/3*b*d^2*f*n*x/e^2+4/9*b*d^2*g*m*n*x/e^2+4/9*b*g*i^2 *m*n*x/j^2+1/3*b*d*g*i*m*n*x/e/j-5/36*b*d*g*m*n*x^2/e-5/36*b*g*i*m*n*x^2/j +2/27*b*g*m*n*x^3-1/9*b*d^3*g*m*n*ln(e*x+d)/e^3-1/6*b*d^2*g*i*m*n*ln(e*x+d )/e^2/j-1/3*b*g*i^2*m*(e*x+d)*ln(c*(e*x+d)^n)/e/j^2+1/6*g*i*m*x^2*(a+b*ln( c*(e*x+d)^n))/j-1/9*g*m*x^3*(a+b*ln(c*(e*x+d)^n))-1/9*b*g*i^3*m*n*ln(j*x+i )/j^3-1/6*b*d*g*i^2*m*n*ln(j*x+i)/e/j^2+1/3*g*i^3*m*(a+b*ln(c*(e*x+d)^n))* ln(e*(j*x+i)/(-d*j+e*i))/j^3-1/3*b*d^2*g*n*(j*x+i)*ln(h*(j*x+i)^m)/e^2/j+1 /6*b*d*n*x^2*(f+g*ln(h*(j*x+i)^m))/e-1/9*b*n*x^3*(f+g*ln(h*(j*x+i)^m))+1/3 *b*d^3*n*ln(-j*(e*x+d)/(-d*j+e*i))*(f+g*ln(h*(j*x+i)^m))/e^3+1/3*x^3*(a+b* ln(c*(e*x+d)^n))*(f+g*ln(h*(j*x+i)^m))+1/3*b*g*i^3*m*n*polylog(2,-j*(e*x+d )/(-d*j+e*i))/j^3+1/3*b*d^3*g*m*n*polylog(2,e*(j*x+i)/(-d*j+e*i))/e^3
Time = 0.50 (sec) , antiderivative size = 492, normalized size of antiderivative = 0.88 \[ \int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\frac {6 b n \log (d+e x) \left (-6 e^3 g i^3 m \log (i+j x)+6 g \left (e^3 i^3-d^3 j^3\right ) m \log \left (\frac {e (i+j x)}{e i-d j}\right )+d j \left (-6 e^2 g i^2 m-3 d e g i j m+2 d^2 j^2 (3 f-g m)+6 d^2 g j^2 \log \left (h (i+j x)^m\right )\right )\right )+e \left (6 g i m \left (6 a e^2 i^2-b \left (2 e^2 i^2+3 d e i j+6 d^2 j^2\right ) n\right ) \log (i+j x)+6 b e^2 \log \left (c (d+e x)^n\right ) \left (6 f j^3 x^3+g j m x \left (-6 i^2+3 i j x-2 j^2 x^2\right )+6 g i^3 m \log (i+j x)+6 g j^3 x^3 \log \left (h (i+j x)^m\right )\right )+j \left (6 a e^2 x \left (6 f j^2 x^2+g m \left (-6 i^2+3 i j x-2 j^2 x^2\right )\right )+b n \left (12 d^2 j^2 (-3 f+4 g m) x+3 d e \left (6 f j^2 x^2+g m \left (12 i^2+12 i j x-5 j^2 x^2\right )\right )+e^2 x \left (-12 f j^2 x^2+g m \left (48 i^2-15 i j x+8 j^2 x^2\right )\right )\right )-6 g j^2 x \left (-6 a e^2 x^2+b n \left (6 d^2-3 d e x+2 e^2 x^2\right )\right ) \log \left (h (i+j x)^m\right )\right )\right )+36 b g \left (e^3 i^3-d^3 j^3\right ) m n \operatorname {PolyLog}\left (2,\frac {j (d+e x)}{-e i+d j}\right )}{108 e^3 j^3} \]
(6*b*n*Log[d + e*x]*(-6*e^3*g*i^3*m*Log[i + j*x] + 6*g*(e^3*i^3 - d^3*j^3) *m*Log[(e*(i + j*x))/(e*i - d*j)] + d*j*(-6*e^2*g*i^2*m - 3*d*e*g*i*j*m + 2*d^2*j^2*(3*f - g*m) + 6*d^2*g*j^2*Log[h*(i + j*x)^m])) + e*(6*g*i*m*(6*a *e^2*i^2 - b*(2*e^2*i^2 + 3*d*e*i*j + 6*d^2*j^2)*n)*Log[i + j*x] + 6*b*e^2 *Log[c*(d + e*x)^n]*(6*f*j^3*x^3 + g*j*m*x*(-6*i^2 + 3*i*j*x - 2*j^2*x^2) + 6*g*i^3*m*Log[i + j*x] + 6*g*j^3*x^3*Log[h*(i + j*x)^m]) + j*(6*a*e^2*x* (6*f*j^2*x^2 + g*m*(-6*i^2 + 3*i*j*x - 2*j^2*x^2)) + b*n*(12*d^2*j^2*(-3*f + 4*g*m)*x + 3*d*e*(6*f*j^2*x^2 + g*m*(12*i^2 + 12*i*j*x - 5*j^2*x^2)) + e^2*x*(-12*f*j^2*x^2 + g*m*(48*i^2 - 15*i*j*x + 8*j^2*x^2))) - 6*g*j^2*x*( -6*a*e^2*x^2 + b*n*(6*d^2 - 3*d*e*x + 2*e^2*x^2))*Log[h*(i + j*x)^m])) + 3 6*b*g*(e^3*i^3 - d^3*j^3)*m*n*PolyLog[2, (j*(d + e*x))/(-(e*i) + d*j)])/(1 08*e^3*j^3)
Time = 0.96 (sec) , antiderivative size = 612, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2889, 2863, 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.
| \(\displaystyle \int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2889 |
| \(\displaystyle -\frac {1}{3} g j m \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{i+j x}dx-\frac {1}{3} b e n \int \frac {x^3 \left (f+g \log \left (h (i+j x)^m\right )\right )}{d+e x}dx+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle -\frac {1}{3} g j m \int \left (-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) i^3}{j^3 (i+j x)}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) i^2}{j^3}-\frac {x \left (a+b \log \left (c (d+e x)^n\right )\right ) i}{j^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{j}\right )dx-\frac {1}{3} b e n \int \left (-\frac {\left (f+g \log \left (h (i+j x)^m\right )\right ) d^3}{e^3 (d+e x)}+\frac {\left (f+g \log \left (h (i+j x)^m\right )\right ) d^2}{e^3}-\frac {x \left (f+g \log \left (h (i+j x)^m\right )\right ) d}{e^2}+\frac {x^2 \left (f+g \log \left (h (i+j x)^m\right )\right )}{e}\right )dx+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{3} g j m \left (-\frac {i^3 \log \left (\frac {e (i+j x)}{e i-d j}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{j^4}-\frac {i x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 j^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 j}+\frac {a i^2 x}{j^3}+\frac {b i^2 (d+e x) \log \left (c (d+e x)^n\right )}{e j^3}+\frac {b d^3 n \log (d+e x)}{3 e^3 j}+\frac {b d^2 i n \log (d+e x)}{2 e^2 j^2}-\frac {b d^2 n x}{3 e^2 j}-\frac {b i^3 n \operatorname {PolyLog}\left (2,-\frac {j (d+e x)}{e i-d j}\right )}{j^4}-\frac {b d i n x}{2 e j^2}+\frac {b d n x^2}{6 e j}-\frac {b i^2 n x}{j^3}+\frac {b i n x^2}{4 j^2}-\frac {b n x^3}{9 j}\right )+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )-\frac {1}{3} b e n \left (-\frac {d^3 \log \left (-\frac {j (d+e x)}{e i-d j}\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )}{e^4}-\frac {d^3 g m \operatorname {PolyLog}\left (2,\frac {e (i+j x)}{e i-d j}\right )}{e^4}+\frac {d^2 f x}{e^3}+\frac {d^2 g (i+j x) \log \left (h (i+j x)^m\right )}{e^3 j}-\frac {d^2 g m x}{e^3}-\frac {d x^2 \left (f+g \log \left (h (i+j x)^m\right )\right )}{2 e^2}+\frac {d g i^2 m \log (i+j x)}{2 e^2 j^2}-\frac {d g i m x}{2 e^2 j}+\frac {d g m x^2}{4 e^2}+\frac {x^3 \left (f+g \log \left (h (i+j x)^m\right )\right )}{3 e}+\frac {g i^3 m \log (i+j x)}{3 e j^3}-\frac {g i^2 m x}{3 e j^2}+\frac {g i m x^2}{6 e j}-\frac {g m x^3}{9 e}\right )\) |
(x^3*(a + b*Log[c*(d + e*x)^n])*(f + g*Log[h*(i + j*x)^m]))/3 - (g*j*m*((a *i^2*x)/j^3 - (b*i^2*n*x)/j^3 - (b*d*i*n*x)/(2*e*j^2) - (b*d^2*n*x)/(3*e^2 *j) + (b*i*n*x^2)/(4*j^2) + (b*d*n*x^2)/(6*e*j) - (b*n*x^3)/(9*j) + (b*d^2 *i*n*Log[d + e*x])/(2*e^2*j^2) + (b*d^3*n*Log[d + e*x])/(3*e^3*j) + (b*i^2 *(d + e*x)*Log[c*(d + e*x)^n])/(e*j^3) - (i*x^2*(a + b*Log[c*(d + e*x)^n]) )/(2*j^2) + (x^3*(a + b*Log[c*(d + e*x)^n]))/(3*j) - (i^3*(a + b*Log[c*(d + e*x)^n])*Log[(e*(i + j*x))/(e*i - d*j)])/j^4 - (b*i^3*n*PolyLog[2, -((j* (d + e*x))/(e*i - d*j))])/j^4))/3 - (b*e*n*((d^2*f*x)/e^3 - (d^2*g*m*x)/e^ 3 - (g*i^2*m*x)/(3*e*j^2) - (d*g*i*m*x)/(2*e^2*j) + (d*g*m*x^2)/(4*e^2) + (g*i*m*x^2)/(6*e*j) - (g*m*x^3)/(9*e) + (g*i^3*m*Log[i + j*x])/(3*e*j^3) + (d*g*i^2*m*Log[i + j*x])/(2*e^2*j^2) + (d^2*g*(i + j*x)*Log[h*(i + j*x)^m ])/(e^3*j) - (d*x^2*(f + g*Log[h*(i + j*x)^m]))/(2*e^2) + (x^3*(f + g*Log[ h*(i + j*x)^m]))/(3*e) - (d^3*Log[-((j*(d + e*x))/(e*i - d*j))]*(f + g*Log [h*(i + j*x)^m]))/e^4 - (d^3*g*m*PolyLog[2, (e*(i + j*x))/(e*i - d*j)])/e^ 4))/3
3.4.87.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) ^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c , d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log [(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*(g_.))*(x_)^(r_.), x_Symbol] :> Simp[x^( r + 1)*(a + b*Log[c*(d + e*x)^n])^p*((f + g*Log[h*(i + j*x)^m])/(r + 1)), x ] + (-Simp[g*j*(m/(r + 1)) Int[x^(r + 1)*((a + b*Log[c*(d + e*x)^n])^p/(i + j*x)), x], x] - Simp[b*e*n*(p/(r + 1)) Int[x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1)*((f + g*Log[h*(i + j*x)^m])/(d + e*x)), x], x]) /; FreeQ[ {a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0] && IntegerQ[r] && (E qQ[p, 1] || GtQ[r, 0]) && NeQ[r, -1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 234.84 (sec) , antiderivative size = 1724, normalized size of antiderivative = 3.09
-1/3/j^3*b*g*i^3*m*n*dilog(((j*x+i)*e+d*j-e*i)/(d*j-e*i))+49/108*b*d^3*g*m *n/e^3-1/9*n*b*f*x^3+(-1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e* x+d)^n)+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/4*I*b*Pi*csgn(I*(e*x+ d)^n)*csgn(I*c*(e*x+d)^n)^2-1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*b*ln(c)+1 /2*a)*(1/3*(I*g*Pi*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i)^m)^2-I*g*Pi*csgn(I*( j*x+i)^m)*csgn(I*h*(j*x+i)^m)*csgn(I*h)-I*g*Pi*csgn(I*h*(j*x+i)^m)^3+I*g*P i*csgn(I*h*(j*x+i)^m)^2*csgn(I*h)+2*g*ln(h)+2*f)*x^3+2/3*g*ln((j*x+i)^m)*x ^3-2/9*g*m*x^3+1/3*g*m/j*x^2*i-2/3*g*m/j^2*x*i^2+2/3*g*m/j^3*i^3*ln(j*x+i) )-1/3/e^3*b*d^3*n*g*m*dilog(((e*x+d)*j-d*j+e*i)/(-d*j+e*i))-1/9*n*b*g*ln(( j*x+i)^m)*x^3-1/3/j^3*b*g*i^3*m*n*ln(j*x+i)*ln(((j*x+i)*e+d*j-e*i)/(d*j-e* i))+1/9*b*d*g*i^2*m*n/e/j^2+2/9*b*d^2*g*i*m*n/e^2/j-1/9/j^3*g*i^3*m*ln((e* x+d)*j-d*j+e*i)*b*n-1/18*I*n*b*Pi*x^3*g*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i) ^m)^2-1/18*I*n*b*Pi*x^3*g*csgn(I*h*(j*x+i)^m)^2*csgn(I*h)+(1/3*x^3*b*g*ln( (j*x+i)^m)+1/18*b*(3*I*Pi*g*j^3*x^3*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i)^m)^ 2-3*I*Pi*g*j^3*x^3*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i)^m)*csgn(I*h)-3*I*Pi* g*j^3*x^3*csgn(I*h*(j*x+i)^m)^3+3*I*Pi*g*j^3*x^3*csgn(I*h*(j*x+i)^m)^2*csg n(I*h)+6*j^3*x^3*ln(h)*g-2*g*j^3*m*x^3+6*f*j^3*x^3+3*g*i*j^2*m*x^2+6*g*i^3 *m*ln(j*x+i)-6*g*i^2*j*m*x)/j^3)*ln((e*x+d)^n)+1/6/e*ln(h)*x^2*b*d*g*n-1/3 /e^2*ln(h)*x*b*d^2*g*n-1/3/e^3*b*d^3*n*g*m*ln(e*x+d)*ln(((e*x+d)*j-d*j+e*i )/(-d*j+e*i))-1/9*ln(h)*x^3*b*g*n+2/27*b*g*m*n*x^3+1/6/e*n*b*g*ln((j*x+...
\[ \int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} {\left (g \log \left ({\left (j x + i\right )}^{m} h\right ) + f\right )} x^{2} \,d x } \]
integral(b*f*x^2*log((e*x + d)^n*c) + a*f*x^2 + (b*g*x^2*log((e*x + d)^n*c ) + a*g*x^2)*log((j*x + i)^m*h), x)
Timed out. \[ \int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\text {Timed out} \]
\[ \int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} {\left (g \log \left ({\left (j x + i\right )}^{m} h\right ) + f\right )} x^{2} \,d x } \]
1/3*b*f*x^3*log((e*x + d)^n*c) + 1/3*a*g*x^3*log((j*x + i)^m*h) + 1/3*a*f* x^3 + 1/18*b*e*f*n*(6*d^3*log(e*x + d)/e^4 - (2*e^2*x^3 - 3*d*e*x^2 + 6*d^ 2*x)/e^3) + 1/18*a*g*j*m*(6*i^3*log(j*x + i)/j^4 - (2*j^2*x^3 - 3*i*j*x^2 + 6*i^2*x)/j^3) - 1/18*b*g*((6*e^3*i^3*m*n*log(e*x + d)*log(j*x + i) - (3* e^3*i*j^2*m*x^2 - 6*e^3*i^2*j*m*x + 6*e^3*i^3*m*log(j*x + i) - 2*(j^3*m - 3*j^3*log(h))*e^3*x^3)*log((e*x + d)^n) - (6*e^3*j^3*x^3*log((e*x + d)^n) + 3*d*e^2*j^3*n*x^2 - 6*d^2*e*j^3*n*x + 6*d^3*j^3*n*log(e*x + d) - 2*(e^3* j^3*n - 3*e^3*j^3*log(c))*x^3)*log((j*x + i)^m))/(e^3*j^3) + 18*integrate( 1/18*(2*(3*(j^3*m - 3*j^3*log(h))*e^4*log(c) - (2*j^3*m*n - 3*j^3*n*log(h) )*e^4)*x^4 + (d*e^3*j^3*m*n + (i*j^2*m*n + 6*i*j^2*n*log(h))*e^4 - 6*(3*e^ 4*i*j^2*log(h) - (j^3*m - 3*j^3*log(h))*d*e^3)*log(c))*x^3 - 3*(e^4*i^2*j* m*n + d^2*e^2*j^3*m*n + 6*d*e^3*i*j^2*log(c)*log(h))*x^2 - 6*(e^4*i^3*m*n + d^3*e*j^3*m*n)*x - 6*(d*e^3*i^3*m*n - d^4*j^3*m*n + (e^4*i^3*m*n - d^3*e *j^3*m*n)*x)*log(e*x + d))/(e^4*j^3*x^2 + d*e^3*i*j^2 + (e^4*i*j^2 + d*e^3 *j^3)*x), x))
\[ \int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} {\left (g \log \left ({\left (j x + i\right )}^{m} h\right ) + f\right )} x^{2} \,d x } \]
Timed out. \[ \int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\int x^2\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )\,\left (f+g\,\ln \left (h\,{\left (i+j\,x\right )}^m\right )\right ) \,d x \]