Integrand size = 33, antiderivative size = 129 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^2} \, dx=-\frac {2 B^2 n^2 (c+d x)}{(b c-a d) (a+b x)}-\frac {2 B n (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{(b c-a d) (a+b x)}-\frac {(c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(b c-a d) (a+b x)} \]
-2*B^2*n^2*(d*x+c)/(-a*d+b*c)/(b*x+a)-2*B*n*(d*x+c)*(A+B*ln(e*(b*x+a)^n/(( d*x+c)^n)))/(-a*d+b*c)/(b*x+a)-(d*x+c)*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2 /(-a*d+b*c)/(b*x+a)
Time = 0.27 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.83 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^2} \, dx=\frac {B^2 d n^2 (a+b x) \log ^2(a+b x)+B^2 d n^2 (a+b x) \log ^2(c+d x)+2 B d n (a+b x) \log (c+d x) \left (A+B n+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )-2 B d n (a+b x) \log (a+b x) \left (A+B n+B n \log (c+d x)+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )-(b c-a d) \left (A^2+2 A B n+2 B^2 n^2+2 B (A+B n) \log \left (e (a+b x)^n (c+d x)^{-n}\right )+B^2 \log ^2\left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{b (b c-a d) (a+b x)} \]
(B^2*d*n^2*(a + b*x)*Log[a + b*x]^2 + B^2*d*n^2*(a + b*x)*Log[c + d*x]^2 + 2*B*d*n*(a + b*x)*Log[c + d*x]*(A + B*n + B*Log[(e*(a + b*x)^n)/(c + d*x) ^n]) - 2*B*d*n*(a + b*x)*Log[a + b*x]*(A + B*n + B*n*Log[c + d*x] + B*Log[ (e*(a + b*x)^n)/(c + d*x)^n]) - (b*c - a*d)*(A^2 + 2*A*B*n + 2*B^2*n^2 + 2 *B*(A + B*n)*Log[(e*(a + b*x)^n)/(c + d*x)^n] + B^2*Log[(e*(a + b*x)^n)/(c + d*x)^n]^2))/(b*(b*c - a*d)*(a + b*x))
Time = 0.38 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2973, 2949, 2742, 2741}
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.
| \(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{(a+b x)^2} \, dx\) |
| \(\Big \downarrow \) 2973 |
| \(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{(a+b x)^2}dx\) |
| \(\Big \downarrow \) 2949 |
| \(\displaystyle \frac {\int \frac {(c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a+b x)^2}d\frac {a+b x}{c+d x}}{b c-a d}\) |
| \(\Big \downarrow \) 2742 |
| \(\displaystyle \frac {2 B n \int \frac {(c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a+b x)^2}d\frac {a+b x}{c+d x}-\frac {(c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{a+b x}}{b c-a d}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {2 B n \left (-\frac {(c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{a+b x}-\frac {B n (c+d x)}{a+b x}\right )-\frac {(c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{a+b x}}{b c-a d}\) |
(-(((c + d*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(a + b*x)) + 2*B*n *(-((B*n*(c + d*x))/(a + b*x)) - ((c + d*x)*(A + B*Log[e*((a + b*x)/(c + d *x))^n]))/(a + b*x)))/(b*c - a*d)
3.2.60.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbo l] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])^p/(d*(m + 1))), x] - Simp[b*n* (p/(m + 1)) Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b , c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]
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. \(298\) vs. \(2(129)=258\).
Time = 7.26 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.32
| method | result | size |
| parallelrisch | \(-\frac {-A^{2} b^{3} c d n +2 B^{2} a \,b^{2} d^{2} n^{3}-2 B^{2} b^{3} c d \,n^{3}+A^{2} a \,b^{2} d^{2} n +2 A B a \,b^{2} d^{2} n^{2}-2 A B \,b^{3} c d \,n^{2}-B^{2} x \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} b^{3} d^{2} n -2 B^{2} x \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) b^{3} d^{2} n^{2}-B^{2} \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} b^{3} c d n -2 B^{2} \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) b^{3} c d \,n^{2}-2 A B x \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) b^{3} d^{2} n -2 A B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) b^{3} c d n}{\left (b x +a \right ) b^{3} d n \left (a d -c b \right )}\) | \(299\) |
| risch | \(\text {Expression too large to display}\) | \(10098\) |
-(-A^2*b^3*c*d*n+2*B^2*a*b^2*d^2*n^3-2*B^2*b^3*c*d*n^3+A^2*a*b^2*d^2*n+2*A *B*a*b^2*d^2*n^2-2*A*B*b^3*c*d*n^2-B^2*x*ln(e*(b*x+a)^n/((d*x+c)^n))^2*b^3 *d^2*n-2*B^2*x*ln(e*(b*x+a)^n/((d*x+c)^n))*b^3*d^2*n^2-B^2*ln(e*(b*x+a)^n/ ((d*x+c)^n))^2*b^3*c*d*n-2*B^2*ln(e*(b*x+a)^n/((d*x+c)^n))*b^3*c*d*n^2-2*A *B*x*ln(e*(b*x+a)^n/((d*x+c)^n))*b^3*d^2*n-2*A*B*ln(e*(b*x+a)^n/((d*x+c)^n ))*b^3*c*d*n)/(b*x+a)/b^3/d/n/(a*d-b*c)
Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (129) = 258\).
Time = 0.28 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.63 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^2} \, dx=-\frac {A^{2} b c - A^{2} a d + 2 \, {\left (B^{2} b c - B^{2} a d\right )} n^{2} + {\left (B^{2} b d n^{2} x + B^{2} b c n^{2}\right )} \log \left (b x + a\right )^{2} + {\left (B^{2} b d n^{2} x + B^{2} b c n^{2}\right )} \log \left (d x + c\right )^{2} + {\left (B^{2} b c - B^{2} a d\right )} \log \left (e\right )^{2} + 2 \, {\left (A B b c - A B a d\right )} n + 2 \, {\left (B^{2} b c n^{2} + A B b c n + {\left (B^{2} b d n^{2} + A B b d n\right )} x + {\left (B^{2} b d n x + B^{2} b c n\right )} \log \left (e\right )\right )} \log \left (b x + a\right ) - 2 \, {\left (B^{2} b c n^{2} + A B b c n + {\left (B^{2} b d n^{2} + A B b d n\right )} x + {\left (B^{2} b d n^{2} x + B^{2} b c n^{2}\right )} \log \left (b x + a\right ) + {\left (B^{2} b d n x + B^{2} b c n\right )} \log \left (e\right )\right )} \log \left (d x + c\right ) + 2 \, {\left (A B b c - A B a d + {\left (B^{2} b c - B^{2} a d\right )} n\right )} \log \left (e\right )}{a b^{2} c - a^{2} b d + {\left (b^{3} c - a b^{2} d\right )} x} \]
-(A^2*b*c - A^2*a*d + 2*(B^2*b*c - B^2*a*d)*n^2 + (B^2*b*d*n^2*x + B^2*b*c *n^2)*log(b*x + a)^2 + (B^2*b*d*n^2*x + B^2*b*c*n^2)*log(d*x + c)^2 + (B^2 *b*c - B^2*a*d)*log(e)^2 + 2*(A*B*b*c - A*B*a*d)*n + 2*(B^2*b*c*n^2 + A*B* b*c*n + (B^2*b*d*n^2 + A*B*b*d*n)*x + (B^2*b*d*n*x + B^2*b*c*n)*log(e))*lo g(b*x + a) - 2*(B^2*b*c*n^2 + A*B*b*c*n + (B^2*b*d*n^2 + A*B*b*d*n)*x + (B ^2*b*d*n^2*x + B^2*b*c*n^2)*log(b*x + a) + (B^2*b*d*n*x + B^2*b*c*n)*log(e ))*log(d*x + c) + 2*(A*B*b*c - A*B*a*d + (B^2*b*c - B^2*a*d)*n)*log(e))/(a *b^2*c - a^2*b*d + (b^3*c - a*b^2*d)*x)
Timed out. \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^2} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (129) = 258\).
Time = 0.22 (sec) , antiderivative size = 449, normalized size of antiderivative = 3.48 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^2} \, dx=-B^{2} {\left (\frac {2 \, {\left (\frac {d e n \log \left (b x + a\right )}{b^{2} c - a b d} - \frac {d e n \log \left (d x + c\right )}{b^{2} c - a b d} + \frac {e n}{b^{2} x + a b}\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}{e} + \frac {2 \, b c e^{2} n^{2} - 2 \, a d e^{2} n^{2} - {\left (b d e^{2} n^{2} x + a d e^{2} n^{2}\right )} \log \left (b x + a\right )^{2} - {\left (b d e^{2} n^{2} x + a d e^{2} n^{2}\right )} \log \left (d x + c\right )^{2} + 2 \, {\left (b d e^{2} n^{2} x + a d e^{2} n^{2}\right )} \log \left (b x + a\right ) - 2 \, {\left (b d e^{2} n^{2} x + a d e^{2} n^{2} - {\left (b d e^{2} n^{2} x + a d e^{2} n^{2}\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{{\left (a b^{2} c - a^{2} b d + {\left (b^{3} c - a b^{2} d\right )} x\right )} e^{2}}\right )} - \frac {B^{2} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )^{2}}{b^{2} x + a b} - \frac {2 \, {\left (\frac {d e n \log \left (b x + a\right )}{b^{2} c - a b d} - \frac {d e n \log \left (d x + c\right )}{b^{2} c - a b d} + \frac {e n}{b^{2} x + a b}\right )} A B}{e} - \frac {2 \, A B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}{b^{2} x + a b} - \frac {A^{2}}{b^{2} x + a b} \]
-B^2*(2*(d*e*n*log(b*x + a)/(b^2*c - a*b*d) - d*e*n*log(d*x + c)/(b^2*c - a*b*d) + e*n/(b^2*x + a*b))*log((b*x + a)^n*e/(d*x + c)^n)/e + (2*b*c*e^2* n^2 - 2*a*d*e^2*n^2 - (b*d*e^2*n^2*x + a*d*e^2*n^2)*log(b*x + a)^2 - (b*d* e^2*n^2*x + a*d*e^2*n^2)*log(d*x + c)^2 + 2*(b*d*e^2*n^2*x + a*d*e^2*n^2)* log(b*x + a) - 2*(b*d*e^2*n^2*x + a*d*e^2*n^2 - (b*d*e^2*n^2*x + a*d*e^2*n ^2)*log(b*x + a))*log(d*x + c))/((a*b^2*c - a^2*b*d + (b^3*c - a*b^2*d)*x) *e^2)) - B^2*log((b*x + a)^n*e/(d*x + c)^n)^2/(b^2*x + a*b) - 2*(d*e*n*log (b*x + a)/(b^2*c - a*b*d) - d*e*n*log(d*x + c)/(b^2*c - a*b*d) + e*n/(b^2* x + a*b))*A*B/e - 2*A*B*log((b*x + a)^n*e/(d*x + c)^n)/(b^2*x + a*b) - A^2 /(b^2*x + a*b)
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^2} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{{\left (b x + a\right )}^{2}} \,d x } \]
Time = 1.98 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.55 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{(a+b x)^2} \, dx=-\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\,\left (\frac {2\,A\,B}{x\,b^2+a\,b}+\frac {2\,B^2\,n}{x\,b^2+a\,b}\right )-{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^2\,\left (\frac {B^2}{b\,\left (a+b\,x\right )}-\frac {B^2\,d}{b\,\left (a\,d-b\,c\right )}\right )-\frac {A^2+2\,A\,B\,n+2\,B^2\,n^2}{x\,b^2+a\,b}-\frac {B\,d\,n\,\mathrm {atan}\left (\frac {\left (\frac {c\,b^2+a\,d\,b}{b}+2\,b\,d\,x\right )\,1{}\mathrm {i}}{a\,d-b\,c}\right )\,\left (A+B\,n\right )\,4{}\mathrm {i}}{b\,\left (a\,d-b\,c\right )} \]
- log((e*(a + b*x)^n)/(c + d*x)^n)*((2*A*B)/(a*b + b^2*x) + (2*B^2*n)/(a*b + b^2*x)) - log((e*(a + b*x)^n)/(c + d*x)^n)^2*(B^2/(b*(a + b*x)) - (B^2* d)/(b*(a*d - b*c))) - (A^2 + 2*B^2*n^2 + 2*A*B*n)/(a*b + b^2*x) - (B*d*n*a tan((((b^2*c + a*b*d)/b + 2*b*d*x)*1i)/(a*d - b*c))*(A + B*n)*4i)/(b*(a*d - b*c))