Integrand size = 33, antiderivative size = 587 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^5} \, dx=\frac {2 B^2 d^3 n^2 (c+d x)}{(b c-a d)^4 (a+b x)}-\frac {3 b B^2 d^2 n^2 (c+d x)^2}{4 (b c-a d)^4 (a+b x)^2}+\frac {2 b^2 B^2 d n^2 (c+d x)^3}{9 (b c-a d)^4 (a+b x)^3}-\frac {b^3 B^2 n^2 (c+d x)^4}{32 (b c-a d)^4 (a+b x)^4}+\frac {2 B d^3 n (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{(b c-a d)^4 (a+b x)}-\frac {3 b B d^2 n (c+d x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{2 (b c-a d)^4 (a+b x)^2}+\frac {2 b^2 B d n (c+d x)^3 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{3 (b c-a d)^4 (a+b x)^3}-\frac {b^3 B n (c+d x)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{8 (b c-a d)^4 (a+b x)^4}+\frac {d^3 (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(b c-a d)^4 (a+b x)}-\frac {3 b d^2 (c+d x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{2 (b c-a d)^4 (a+b x)^2}+\frac {b^2 d (c+d x)^3 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(b c-a d)^4 (a+b x)^3}-\frac {b^3 (c+d x)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{4 (b c-a d)^4 (a+b x)^4} \]
2*B^2*d^3*n^2*(d*x+c)/(-a*d+b*c)^4/(b*x+a)-3/4*b*B^2*d^2*n^2*(d*x+c)^2/(-a *d+b*c)^4/(b*x+a)^2+2/9*b^2*B^2*d*n^2*(d*x+c)^3/(-a*d+b*c)^4/(b*x+a)^3-1/3 2*b^3*B^2*n^2*(d*x+c)^4/(-a*d+b*c)^4/(b*x+a)^4+2*B*d^3*n*(d*x+c)*(A+B*ln(e *(b*x+a)^n/((d*x+c)^n)))/(-a*d+b*c)^4/(b*x+a)-3/2*b*B*d^2*n*(d*x+c)^2*(A+B *ln(e*(b*x+a)^n/((d*x+c)^n)))/(-a*d+b*c)^4/(b*x+a)^2+2/3*b^2*B*d*n*(d*x+c) ^3*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))/(-a*d+b*c)^4/(b*x+a)^3-1/8*b^3*B*n*(d *x+c)^4*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))/(-a*d+b*c)^4/(b*x+a)^4+d^3*(d*x+ c)*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2/(-a*d+b*c)^4/(b*x+a)-3/2*b*d^2*(d*x +c)^2*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2/(-a*d+b*c)^4/(b*x+a)^2+b^2*d*(d* x+c)^3*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2/(-a*d+b*c)^4/(b*x+a)^3-1/4*b^3* (d*x+c)^4*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2/(-a*d+b*c)^4/(b*x+a)^4
Time = 0.60 (sec) , antiderivative size = 1011, normalized size of antiderivative = 1.72 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^5} \, dx=-\frac {72 b B^2 n^2 \left (-4 a^3 d^3 (c+d x)+6 a^2 b d^2 \left (c^2-d^2 x^2\right )-4 a b^2 d \left (c^3+d^3 x^3\right )+b^3 \left (c^4-d^4 x^4\right )\right ) \log ^2(a+b x)+72 b B^2 n^2 \left (-4 a^3 d^3 (c+d x)+6 a^2 b d^2 \left (c^2-d^2 x^2\right )-4 a b^2 d \left (c^3+d^3 x^3\right )+b^3 \left (c^4-d^4 x^4\right )\right ) \log ^2(c+d x)-4 B d (b c-a d)^3 n (a+b x) \left (12 A+7 B n+12 B \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )\right )+6 B d^2 (b c-a d)^2 n (a+b x)^2 \left (12 A+13 B n+12 B \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )\right )-12 B d^3 (b c-a d) n (a+b x)^3 \left (12 A+25 B n+12 B \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )\right )-12 B d^4 n (a+b x)^4 \log (a+b x) \left (12 A+25 B n+12 B \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )\right )+12 B d^4 n (a+b x)^4 \log (c+d x) \left (12 A+25 B n+12 B \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )\right )+9 (b c-a d)^4 \left (8 A^2+4 A B n+B^2 n^2+16 A B \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )+4 B^2 n \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )+8 B^2 \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2\right )-12 B (b c-a d) n \log (a+b x) \left (4 B d (b c-a d)^2 n (a+b x)+6 B d^2 (-b c+a d) n (a+b x)^2+12 B d^3 n (a+b x)^3-3 (b c-a d)^3 \left (4 A+B n+4 B \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )\right )\right )+12 B n \log (c+d x) \left (4 B d (b c-a d)^3 n (a+b x)-6 B d^2 (b c-a d)^2 n (a+b x)^2+12 B d^3 (b c-a d) n (a+b x)^3-12 B (b c-a d)^4 n \log (a+b x)+12 B d^4 n (a+b x)^4 \log (a+b x)-3 (b c-a d)^4 \left (4 A+B n+4 B \left (-n \log (a+b x)+n \log (c+d x)+\log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )\right )\right )}{288 b (b c-a d)^4 (a+b x)^4} \]
-1/288*(72*b*B^2*n^2*(-4*a^3*d^3*(c + d*x) + 6*a^2*b*d^2*(c^2 - d^2*x^2) - 4*a*b^2*d*(c^3 + d^3*x^3) + b^3*(c^4 - d^4*x^4))*Log[a + b*x]^2 + 72*b*B^ 2*n^2*(-4*a^3*d^3*(c + d*x) + 6*a^2*b*d^2*(c^2 - d^2*x^2) - 4*a*b^2*d*(c^3 + d^3*x^3) + b^3*(c^4 - d^4*x^4))*Log[c + d*x]^2 - 4*B*d*(b*c - a*d)^3*n* (a + b*x)*(12*A + 7*B*n + 12*B*(-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[( e*(a + b*x)^n)/(c + d*x)^n])) + 6*B*d^2*(b*c - a*d)^2*n*(a + b*x)^2*(12*A + 13*B*n + 12*B*(-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[(e*(a + b*x)^n)/ (c + d*x)^n])) - 12*B*d^3*(b*c - a*d)*n*(a + b*x)^3*(12*A + 25*B*n + 12*B* (-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[(e*(a + b*x)^n)/(c + d*x)^n])) - 12*B*d^4*n*(a + b*x)^4*Log[a + b*x]*(12*A + 25*B*n + 12*B*(-(n*Log[a + b* x]) + n*Log[c + d*x] + Log[(e*(a + b*x)^n)/(c + d*x)^n])) + 12*B*d^4*n*(a + b*x)^4*Log[c + d*x]*(12*A + 25*B*n + 12*B*(-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[(e*(a + b*x)^n)/(c + d*x)^n])) + 9*(b*c - a*d)^4*(8*A^2 + 4*A* B*n + B^2*n^2 + 16*A*B*(-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[(e*(a + b *x)^n)/(c + d*x)^n]) + 4*B^2*n*(-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[( e*(a + b*x)^n)/(c + d*x)^n]) + 8*B^2*(-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[(e*(a + b*x)^n)/(c + d*x)^n])^2) - 12*B*(b*c - a*d)*n*Log[a + b*x]*(4 *B*d*(b*c - a*d)^2*n*(a + b*x) + 6*B*d^2*(-(b*c) + a*d)*n*(a + b*x)^2 + 12 *B*d^3*n*(a + b*x)^3 - 3*(b*c - a*d)^3*(4*A + B*n + 4*B*(-(n*Log[a + b*x]) + n*Log[c + d*x] + Log[(e*(a + b*x)^n)/(c + d*x)^n]))) + 12*B*n*Log[c ...
Time = 0.66 (sec) , antiderivative size = 470, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2973, 2949, 2795, 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.
| \(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{(a+b x)^5} \, dx\) |
| \(\Big \downarrow \) 2973 |
| \(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{(a+b x)^5}dx\) |
| \(\Big \downarrow \) 2949 |
| \(\displaystyle \frac {\int \frac {(c+d x)^5 \left (b-\frac {d (a+b x)}{c+d x}\right )^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a+b x)^5}d\frac {a+b x}{c+d x}}{(b c-a d)^4}\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \frac {\int \left (\frac {b^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 (c+d x)^5}{(a+b x)^5}-\frac {3 b^2 d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 (c+d x)^4}{(a+b x)^4}+\frac {3 b d^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 (c+d x)^3}{(a+b x)^3}-\frac {d^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 (c+d x)^2}{(a+b x)^2}\right )d\frac {a+b x}{c+d x}}{(b c-a d)^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b^3 (c+d x)^4 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{4 (a+b x)^4}-\frac {b^3 B n (c+d x)^4 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{8 (a+b x)^4}+\frac {b^2 d (c+d x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{(a+b x)^3}+\frac {2 b^2 B d n (c+d x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{3 (a+b x)^3}+\frac {d^3 (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{a+b x}+\frac {2 B d^3 n (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{a+b x}-\frac {3 b d^2 (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{2 (a+b x)^2}-\frac {3 b B d^2 n (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 (a+b x)^2}-\frac {b^3 B^2 n^2 (c+d x)^4}{32 (a+b x)^4}+\frac {2 b^2 B^2 d n^2 (c+d x)^3}{9 (a+b x)^3}+\frac {2 B^2 d^3 n^2 (c+d x)}{a+b x}-\frac {3 b B^2 d^2 n^2 (c+d x)^2}{4 (a+b x)^2}}{(b c-a d)^4}\) |
((2*B^2*d^3*n^2*(c + d*x))/(a + b*x) - (3*b*B^2*d^2*n^2*(c + d*x)^2)/(4*(a + b*x)^2) + (2*b^2*B^2*d*n^2*(c + d*x)^3)/(9*(a + b*x)^3) - (b^3*B^2*n^2* (c + d*x)^4)/(32*(a + b*x)^4) + (2*B*d^3*n*(c + d*x)*(A + B*Log[e*((a + b* x)/(c + d*x))^n]))/(a + b*x) - (3*b*B*d^2*n*(c + d*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*(a + b*x)^2) + (2*b^2*B*d*n*(c + d*x)^3*(A + B*Lo g[e*((a + b*x)/(c + d*x))^n]))/(3*(a + b*x)^3) - (b^3*B*n*(c + d*x)^4*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(8*(a + b*x)^4) + (d^3*(c + d*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(a + b*x) - (3*b*d^2*(c + d*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(2*(a + b*x)^2) + (b^2*d*(c + d*x)^3 *(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(a + b*x)^3 - (b^3*(c + d*x)^4* (A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(4*(a + b*x)^4))/(b*c - a*d)^4
3.2.63.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[ c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b , c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0 ] && IntegerQ[m] && IntegerQ[r]))
Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*c - a*d)^(m + 1)*(g/b)^m Subst[Int[x^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && Ne Q[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[b*f - a*g, 0] && (GtQ[p, 0] || Lt Q[m, -1])
Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] :> Subst[Int[w*(A + B*Log[e*(u/v)^n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; Fr eeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] && !Intege rQ[n]
Leaf count of result is larger than twice the leaf count of optimal. \(4945\) vs. \(2(571)=1142\).
Time = 142.59 (sec) , antiderivative size = 4946, normalized size of antiderivative = 8.43
| method | result | size |
| parallelrisch | \(\text {Expression too large to display}\) | \(4946\) |
| risch | \(\text {Expression too large to display}\) | \(33370\) |
1/288*(144*B^2*x*ln(e*(b*x+a)^n/((d*x+c)^n))*a^5*b^4*c^5*n-1008*B^2*x*a^8* b*c^2*d^3*n^2+624*B^2*x*a^7*b^2*c^3*d^2*n^2-228*B^2*x*a^6*b^3*c^4*d*n^2+28 8*A^2*x*a^5*b^4*c^5-36*B^2*ln(b*x+a)*x^4*a^2*b^7*c^5*n^2+36*B^2*ln(d*x+c)* x^4*a^2*b^7*c^5*n^2-144*B^2*ln(b*x+a)*x^3*a^3*b^6*c^5*n^2+144*B^2*ln(d*x+c )*x^3*a^3*b^6*c^5*n^2-216*B^2*ln(b*x+a)*x^2*a^4*b^5*c^5*n^2+216*B^2*ln(d*x +c)*x^2*a^4*b^5*c^5*n^2-144*B^2*ln(b*x+a)*x*a^5*b^4*c^5*n^2+3456*A*B*ln(b* x+a)*x^2*a^5*b^4*c^4*d*n-3456*A*B*ln(d*x+c)*x^2*a^7*b^2*c^2*d^3*n+5184*A*B *ln(d*x+c)*x^2*a^6*b^3*c^3*d^2*n-3456*A*B*ln(d*x+c)*x^2*a^5*b^4*c^4*d*n-57 6*A*B*ln(d*x+c)*x^4*a^3*b^6*c^4*d*n+576*A*B*ln(b*x+a)*x^4*a^5*b^4*c^2*d^3* n-864*A*B*ln(b*x+a)*x^4*a^4*b^5*c^3*d^2*n+576*A*B*ln(b*x+a)*x^4*a^3*b^6*c^ 4*d*n-576*A*B*ln(d*x+c)*x^4*a^5*b^4*c^2*d^3*n+864*A*B*ln(d*x+c)*x^4*a^4*b^ 5*c^3*d^2*n+3456*A*B*ln(d*x+c)*x*a^7*b^2*c^3*d^2*n-2304*A*B*ln(d*x+c)*x*a^ 6*b^3*c^4*d*n+2304*A*B*ln(b*x+a)*x*a^8*b*c^2*d^3*n-3456*A*B*ln(b*x+a)*x*a^ 7*b^2*c^3*d^2*n+2304*A*B*ln(b*x+a)*x^3*a^6*b^3*c^2*d^3*n-3456*A*B*ln(b*x+a )*x^3*a^5*b^4*c^3*d^2*n+2304*A*B*ln(b*x+a)*x^3*a^4*b^5*c^4*d*n-2304*A*B*ln (d*x+c)*x^3*a^6*b^3*c^2*d^3*n+3456*A*B*ln(d*x+c)*x^3*a^5*b^4*c^3*d^2*n-230 4*A*B*ln(d*x+c)*x^3*a^4*b^5*c^4*d*n+3456*A*B*ln(b*x+a)*x^2*a^7*b^2*c^2*d^3 *n-5184*A*B*ln(b*x+a)*x^2*a^6*b^3*c^3*d^2*n+2304*A*B*ln(b*x+a)*x*a^6*b^3*c ^4*d*n-2304*A*B*ln(d*x+c)*x*a^8*b*c^2*d^3*n-2304*B^2*ln(d*x+c)*x*a^8*b*c^2 *d^3*n^2+1728*B^2*ln(d*x+c)*x*a^7*b^2*c^3*d^2*n^2-768*B^2*ln(d*x+c)*x*a...
Leaf count of result is larger than twice the leaf count of optimal. 2458 vs. \(2 (571) = 1142\).
Time = 0.37 (sec) , antiderivative size = 2458, normalized size of antiderivative = 4.19 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^5} \, dx=\text {Too large to display} \]
-1/288*(72*A^2*b^4*c^4 - 288*A^2*a*b^3*c^3*d + 432*A^2*a^2*b^2*c^2*d^2 - 2 88*A^2*a^3*b*c*d^3 + 72*A^2*a^4*d^4 - 12*(25*(B^2*b^4*c*d^3 - B^2*a*b^3*d^ 4)*n^2 + 12*(A*B*b^4*c*d^3 - A*B*a*b^3*d^4)*n)*x^3 + (9*B^2*b^4*c^4 - 64*B ^2*a*b^3*c^3*d + 216*B^2*a^2*b^2*c^2*d^2 - 576*B^2*a^3*b*c*d^3 + 415*B^2*a ^4*d^4)*n^2 + 6*((13*B^2*b^4*c^2*d^2 - 176*B^2*a*b^3*c*d^3 + 163*B^2*a^2*b ^2*d^4)*n^2 + 12*(A*B*b^4*c^2*d^2 - 8*A*B*a*b^3*c*d^3 + 7*A*B*a^2*b^2*d^4) *n)*x^2 - 72*(B^2*b^4*d^4*n^2*x^4 + 4*B^2*a*b^3*d^4*n^2*x^3 + 6*B^2*a^2*b^ 2*d^4*n^2*x^2 + 4*B^2*a^3*b*d^4*n^2*x - (B^2*b^4*c^4 - 4*B^2*a*b^3*c^3*d + 6*B^2*a^2*b^2*c^2*d^2 - 4*B^2*a^3*b*c*d^3)*n^2)*log(b*x + a)^2 - 72*(B^2* b^4*d^4*n^2*x^4 + 4*B^2*a*b^3*d^4*n^2*x^3 + 6*B^2*a^2*b^2*d^4*n^2*x^2 + 4* B^2*a^3*b*d^4*n^2*x - (B^2*b^4*c^4 - 4*B^2*a*b^3*c^3*d + 6*B^2*a^2*b^2*c^2 *d^2 - 4*B^2*a^3*b*c*d^3)*n^2)*log(d*x + c)^2 + 72*(B^2*b^4*c^4 - 4*B^2*a* b^3*c^3*d + 6*B^2*a^2*b^2*c^2*d^2 - 4*B^2*a^3*b*c*d^3 + B^2*a^4*d^4)*log(e )^2 + 12*(3*A*B*b^4*c^4 - 16*A*B*a*b^3*c^3*d + 36*A*B*a^2*b^2*c^2*d^2 - 48 *A*B*a^3*b*c*d^3 + 25*A*B*a^4*d^4)*n - 4*((7*B^2*b^4*c^3*d - 60*B^2*a*b^3* c^2*d^2 + 324*B^2*a^2*b^2*c*d^3 - 271*B^2*a^3*b*d^4)*n^2 + 12*(A*B*b^4*c^3 *d - 6*A*B*a*b^3*c^2*d^2 + 18*A*B*a^2*b^2*c*d^3 - 13*A*B*a^3*b*d^4)*n)*x - 12*((25*B^2*b^4*d^4*n^2 + 12*A*B*b^4*d^4*n)*x^4 + 4*(12*A*B*a*b^3*d^4*n + (3*B^2*b^4*c*d^3 + 22*B^2*a*b^3*d^4)*n^2)*x^3 - (3*B^2*b^4*c^4 - 16*B^2*a *b^3*c^3*d + 36*B^2*a^2*b^2*c^2*d^2 - 48*B^2*a^3*b*c*d^3)*n^2 + 6*(12*A...
Timed out. \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^5} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2238 vs. \(2 (571) = 1142\).
Time = 0.34 (sec) , antiderivative size = 2238, normalized size of antiderivative = 3.81 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^5} \, dx=\text {Too large to display} \]
1/288*B^2*(12*(12*d^4*e*n*log(b*x + a)/(b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^ 3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4) - 12*d^4*e*n*log(d*x + c)/(b^5*c^ 4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4) + (12 *b^3*d^3*e*n*x^3 - 3*b^3*c^3*e*n + 13*a*b^2*c^2*d*e*n - 23*a^2*b*c*d^2*e*n + 25*a^3*d^3*e*n - 6*(b^3*c*d^2*e*n - 7*a*b^2*d^3*e*n)*x^2 + 4*(b^3*c^2*d *e*n - 5*a*b^2*c*d^2*e*n + 13*a^2*b*d^3*e*n)*x)/(a^4*b^4*c^3 - 3*a^5*b^3*c ^2*d + 3*a^6*b^2*c*d^2 - a^7*b*d^3 + (b^8*c^3 - 3*a*b^7*c^2*d + 3*a^2*b^6* c*d^2 - a^3*b^5*d^3)*x^4 + 4*(a*b^7*c^3 - 3*a^2*b^6*c^2*d + 3*a^3*b^5*c*d^ 2 - a^4*b^4*d^3)*x^3 + 6*(a^2*b^6*c^3 - 3*a^3*b^5*c^2*d + 3*a^4*b^4*c*d^2 - a^5*b^3*d^3)*x^2 + 4*(a^3*b^5*c^3 - 3*a^4*b^4*c^2*d + 3*a^5*b^3*c*d^2 - a^6*b^2*d^3)*x))*log((b*x + a)^n*e/(d*x + c)^n)/e - (9*b^4*c^4*e^2*n^2 - 6 4*a*b^3*c^3*d*e^2*n^2 + 216*a^2*b^2*c^2*d^2*e^2*n^2 - 576*a^3*b*c*d^3*e^2* n^2 + 415*a^4*d^4*e^2*n^2 - 300*(b^4*c*d^3*e^2*n^2 - a*b^3*d^4*e^2*n^2)*x^ 3 + 6*(13*b^4*c^2*d^2*e^2*n^2 - 176*a*b^3*c*d^3*e^2*n^2 + 163*a^2*b^2*d^4* e^2*n^2)*x^2 + 72*(b^4*d^4*e^2*n^2*x^4 + 4*a*b^3*d^4*e^2*n^2*x^3 + 6*a^2*b ^2*d^4*e^2*n^2*x^2 + 4*a^3*b*d^4*e^2*n^2*x + a^4*d^4*e^2*n^2)*log(b*x + a) ^2 + 72*(b^4*d^4*e^2*n^2*x^4 + 4*a*b^3*d^4*e^2*n^2*x^3 + 6*a^2*b^2*d^4*e^2 *n^2*x^2 + 4*a^3*b*d^4*e^2*n^2*x + a^4*d^4*e^2*n^2)*log(d*x + c)^2 - 4*(7* b^4*c^3*d*e^2*n^2 - 60*a*b^3*c^2*d^2*e^2*n^2 + 324*a^2*b^2*c*d^3*e^2*n^2 - 271*a^3*b*d^4*e^2*n^2)*x - 300*(b^4*d^4*e^2*n^2*x^4 + 4*a*b^3*d^4*e^2*...
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^5} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{{\left (b x + a\right )}^{5}} \,d x } \]
Time = 6.74 (sec) , antiderivative size = 1579, normalized size of antiderivative = 2.69 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^5} \, dx=\text {Too large to display} \]
(B*d^4*n*atan((B*d^4*n*(12*A + 25*B*n)*((b^5*c^4 - a^4*b*d^4 + 2*a^3*b^2*c *d^3 - 2*a*b^4*c^3*d)/(b^4*c^3 - a^3*b*d^3 + 3*a^2*b^2*c*d^2 - 3*a*b^3*c^2 *d) + 2*b*d*x)*(b^4*c^3 - a^3*b*d^3 + 3*a^2*b^2*c*d^2 - 3*a*b^3*c^2*d)*1i) /(b*(25*B^2*d^4*n^2 + 12*A*B*d^4*n)*(a*d - b*c)^4))*(12*A + 25*B*n)*1i)/(1 2*b*(a*d - b*c)^4) - log((e*(a + b*x)^n)/(c + d*x)^n)^2*(B^2/(4*b*(a^4 + b ^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x)) - (B^2*d^4)/(4*b*(a^4*d ^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3))) - ((72 *A^2*a^3*d^3 - 72*A^2*b^3*c^3 + 415*B^2*a^3*d^3*n^2 - 9*B^2*b^3*c^3*n^2 + 216*A^2*a*b^2*c^2*d - 216*A^2*a^2*b*c*d^2 + 300*A*B*a^3*d^3*n - 36*A*B*b^3 *c^3*n + 55*B^2*a*b^2*c^2*d*n^2 - 161*B^2*a^2*b*c*d^2*n^2 + 156*A*B*a*b^2* c^2*d*n - 276*A*B*a^2*b*c*d^2*n)/(12*(a*d - b*c)) + (x^2*(163*B^2*a*b^2*d^ 3*n^2 - 13*B^2*b^3*c*d^2*n^2 + 84*A*B*a*b^2*d^3*n - 12*A*B*b^3*c*d^2*n))/( 2*(a*d - b*c)) + (x*(271*B^2*a^2*b*d^3*n^2 + 7*B^2*b^3*c^2*d*n^2 - 53*B^2* a*b^2*c*d^2*n^2 + 156*A*B*a^2*b*d^3*n + 12*A*B*b^3*c^2*d*n - 60*A*B*a*b^2* c*d^2*n))/(3*(a*d - b*c)) + (d*x^3*(25*B^2*b^3*d^2*n^2 + 12*A*B*b^3*d^2*n) )/(a*d - b*c))/(x*(96*a^3*b^4*c^2 + 96*a^5*b^2*d^2 - 192*a^4*b^3*c*d) + x^ 3*(96*a*b^6*c^2 + 96*a^3*b^4*d^2 - 192*a^2*b^5*c*d) + x^4*(24*b^7*c^2 + 24 *a^2*b^5*d^2 - 48*a*b^6*c*d) + x^2*(144*a^2*b^5*c^2 + 144*a^4*b^3*d^2 - 28 8*a^3*b^4*c*d) + 24*a^6*b*d^2 + 24*a^4*b^3*c^2 - 48*a^5*b^2*c*d) - log((e* (a + b*x)^n)/(c + d*x)^n)*((A*B)/(2*(a^4*b + b^5*x^4 + 4*a^3*b^2*x + 4*...