Integrand size = 35, antiderivative size = 136 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=-\frac {2 B^2 n^2 (c+d x)}{(b c-a d) g^2 (a+b x)}-\frac {2 B n (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) g^2 (a+b x)}-\frac {(c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b c-a d) g^2 (a+b x)} \]
-2*B^2*n^2*(d*x+c)/(-a*d+b*c)/g^2/(b*x+a)-2*B*n*(d*x+c)*(A+B*ln(e*((b*x+a) /(d*x+c))^n))/(-a*d+b*c)/g^2/(b*x+a)-(d*x+c)*(A+B*ln(e*((b*x+a)/(d*x+c))^n ))^2/(-a*d+b*c)/g^2/(b*x+a)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.33 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.43 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=-\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2+\frac {B n \left (2 (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+2 d (a+b x) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-2 d (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)+2 B n (b c-a d+d (a+b x) \log (a+b x)-d (a+b x) \log (c+d x))-B d n (a+b x) \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )+B d n (a+b x) \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{b c-a d}}{b g^2 (a+b x)} \]
-(((A + B*Log[e*((a + b*x)/(c + d*x))^n])^2 + (B*n*(2*(b*c - a*d)*(A + B*L og[e*((a + b*x)/(c + d*x))^n]) + 2*d*(a + b*x)*Log[a + b*x]*(A + B*Log[e*( (a + b*x)/(c + d*x))^n]) - 2*d*(a + b*x)*(A + B*Log[e*((a + b*x)/(c + d*x) )^n])*Log[c + d*x] + 2*B*n*(b*c - a*d + d*(a + b*x)*Log[a + b*x] - d*(a + b*x)*Log[c + d*x]) - B*d*n*(a + b*x)*(Log[a + b*x]*(Log[a + b*x] - 2*Log[( b*(c + d*x))/(b*c - a*d)]) - 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)]) + B*d*n*(a + b*x)*((2*Log[(d*(a + b*x))/(-(b*c) + a*d)] - Log[c + d*x])*Log [c + d*x] + 2*PolyLog[2, (b*(c + d*x))/(b*c - a*d)])))/(b*c - a*d))/(b*g^2 *(a + b*x)))
Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {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.
| \(\displaystyle \int \frac {\left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{(a g+b g x)^2} \, dx\) |
| \(\Big \downarrow \) 2949 |
| \(\displaystyle \frac {\int \frac {(c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a+b x)^2}d\frac {a+b x}{c+d x}}{g^2 (b c-a d)}\) |
| \(\Big \downarrow \) 2742 |
| \(\displaystyle \frac {2 B n \int \frac {(c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a+b x)^2}d\frac {a+b x}{c+d x}-\frac {(c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{a+b x}}{g^2 (b c-a d)}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {2 B n \left (-\frac {(c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{a+b x}-\frac {B n (c+d x)}{a+b x}\right )-\frac {(c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{a+b x}}{g^2 (b c-a d)}\) |
(-(((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)*g^2)
3.1.15.3.1 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])
Leaf count of result is larger than twice the leaf count of optimal. \(295\) vs. \(2(136)=272\).
Time = 3.14 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.18
| method | result | size |
| parallelrisch | \(-\frac {-A^{2} b^{3} c d n +2 B^{2} a \,b^{2} d^{2} n^{3}-2 B^{2} b^{3} c d \,n^{3}+A^{2} a \,b^{2} d^{2} n +2 A B a \,b^{2} d^{2} n^{2}-2 A B \,b^{3} c d \,n^{2}-B^{2} x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} b^{3} d^{2} n -2 B^{2} x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} d^{2} n^{2}-B^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} b^{3} c d n -2 B^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} c d \,n^{2}-2 A B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} d^{2} n -2 A B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} c d n}{g^{2} \left (b x +a \right ) b^{3} d n \left (a d -c b \right )}\) | \(296\) |
-(-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)/(d*x+c))^n)^2*b^3*d ^2*n-2*B^2*x*ln(e*((b*x+a)/(d*x+c))^n)*b^3*d^2*n^2-B^2*ln(e*((b*x+a)/(d*x+ c))^n)^2*b^3*c*d*n-2*B^2*ln(e*((b*x+a)/(d*x+c))^n)*b^3*c*d*n^2-2*A*B*x*ln( e*((b*x+a)/(d*x+c))^n)*b^3*d^2*n-2*A*B*ln(e*((b*x+a)/(d*x+c))^n)*b^3*c*d*n )/g^2/(b*x+a)/b^3/d/n/(a*d-b*c)
Time = 0.27 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.90 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=-\frac {A^{2} b c - A^{2} a d + 2 \, {\left (B^{2} b c - B^{2} a d\right )} n^{2} + {\left (B^{2} b c - B^{2} a d\right )} \log \left (e\right )^{2} + {\left (B^{2} b d n^{2} x + B^{2} b c n^{2}\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2} + 2 \, {\left (A B b c - A B a d\right )} n + 2 \, {\left (A B b c - A B a d + {\left (B^{2} b c - B^{2} a d\right )} n + {\left (B^{2} b d n x + B^{2} b c n\right )} \log \left (\frac {b x + a}{d x + c}\right )\right )} \log \left (e\right ) + 2 \, {\left (B^{2} b c n^{2} + A B b c n + {\left (B^{2} b d n^{2} + A B b d n\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{{\left (b^{3} c - a b^{2} d\right )} g^{2} x + {\left (a b^{2} c - a^{2} b d\right )} g^{2}} \]
-(A^2*b*c - A^2*a*d + 2*(B^2*b*c - B^2*a*d)*n^2 + (B^2*b*c - B^2*a*d)*log( e)^2 + (B^2*b*d*n^2*x + B^2*b*c*n^2)*log((b*x + a)/(d*x + c))^2 + 2*(A*B*b *c - A*B*a*d)*n + 2*(A*B*b*c - A*B*a*d + (B^2*b*c - B^2*a*d)*n + (B^2*b*d* n*x + B^2*b*c*n)*log((b*x + a)/(d*x + c)))*log(e) + 2*(B^2*b*c*n^2 + A*B*b *c*n + (B^2*b*d*n^2 + A*B*b*d*n)*x)*log((b*x + a)/(d*x + c)))/((b^3*c - a* b^2*d)*g^2*x + (a*b^2*c - a^2*b*d)*g^2)
\[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=\frac {\int \frac {A^{2}}{a^{2} + 2 a b x + b^{2} x^{2}}\, dx + \int \frac {B^{2} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}^{2}}{a^{2} + 2 a b x + b^{2} x^{2}}\, dx + \int \frac {2 A B \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{a^{2} + 2 a b x + b^{2} x^{2}}\, dx}{g^{2}} \]
(Integral(A**2/(a**2 + 2*a*b*x + b**2*x**2), x) + Integral(B**2*log(e*(a/( c + d*x) + b*x/(c + d*x))**n)**2/(a**2 + 2*a*b*x + b**2*x**2), x) + Integr al(2*A*B*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/(a**2 + 2*a*b*x + b**2*x* *2), x))/g**2
Leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (136) = 272\).
Time = 0.22 (sec) , antiderivative size = 430, normalized size of antiderivative = 3.16 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=-2 \, A B n {\left (\frac {1}{b^{2} g^{2} x + a b g^{2}} + \frac {d \log \left (b x + a\right )}{{\left (b^{2} c - a b d\right )} g^{2}} - \frac {d \log \left (d x + c\right )}{{\left (b^{2} c - a b d\right )} g^{2}}\right )} - {\left (2 \, n {\left (\frac {1}{b^{2} g^{2} x + a b g^{2}} + \frac {d \log \left (b x + a\right )}{{\left (b^{2} c - a b d\right )} g^{2}} - \frac {d \log \left (d x + c\right )}{{\left (b^{2} c - a b d\right )} g^{2}}\right )} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) - \frac {{\left ({\left (b d x + a d\right )} \log \left (b x + a\right )^{2} + {\left (b d x + a d\right )} \log \left (d x + c\right )^{2} - 2 \, b c + 2 \, a d - 2 \, {\left (b d x + a d\right )} \log \left (b x + a\right ) + 2 \, {\left (b d x + a d - {\left (b d x + a d\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )\right )} n^{2}}{a b^{2} c g^{2} - a^{2} b d g^{2} + {\left (b^{3} c g^{2} - a b^{2} d g^{2}\right )} x}\right )} B^{2} - \frac {B^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )^{2}}{b^{2} g^{2} x + a b g^{2}} - \frac {2 \, A B \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{b^{2} g^{2} x + a b g^{2}} - \frac {A^{2}}{b^{2} g^{2} x + a b g^{2}} \]
-2*A*B*n*(1/(b^2*g^2*x + a*b*g^2) + d*log(b*x + a)/((b^2*c - a*b*d)*g^2) - d*log(d*x + c)/((b^2*c - a*b*d)*g^2)) - (2*n*(1/(b^2*g^2*x + a*b*g^2) + d *log(b*x + a)/((b^2*c - a*b*d)*g^2) - d*log(d*x + c)/((b^2*c - a*b*d)*g^2) )*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) - ((b*d*x + a*d)*log(b*x + a)^2 + (b*d*x + a*d)*log(d*x + c)^2 - 2*b*c + 2*a*d - 2*(b*d*x + a*d)*log(b*x + a) + 2*(b*d*x + a*d - (b*d*x + a*d)*log(b*x + a))*log(d*x + c))*n^2/(a*b^2 *c*g^2 - a^2*b*d*g^2 + (b^3*c*g^2 - a*b^2*d*g^2)*x))*B^2 - B^2*log(e*(b*x/ (d*x + c) + a/(d*x + c))^n)^2/(b^2*g^2*x + a*b*g^2) - 2*A*B*log(e*(b*x/(d* x + c) + a/(d*x + c))^n)/(b^2*g^2*x + a*b*g^2) - A^2/(b^2*g^2*x + a*b*g^2)
Time = 0.99 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.28 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=-{\left (\frac {{\left (d x + c\right )} B^{2} n^{2} \log \left (\frac {b x + a}{d x + c}\right )^{2}}{{\left (b x + a\right )} g^{2}} + \frac {2 \, {\left (B^{2} n^{2} + B^{2} n \log \left (e\right ) + A B n\right )} {\left (d x + c\right )} \log \left (\frac {b x + a}{d x + c}\right )}{{\left (b x + a\right )} g^{2}} + \frac {{\left (2 \, B^{2} n^{2} + 2 \, B^{2} n \log \left (e\right ) + B^{2} \log \left (e\right )^{2} + 2 \, A B n + 2 \, A B \log \left (e\right ) + A^{2}\right )} {\left (d x + c\right )}}{{\left (b x + a\right )} g^{2}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]
-((d*x + c)*B^2*n^2*log((b*x + a)/(d*x + c))^2/((b*x + a)*g^2) + 2*(B^2*n^ 2 + B^2*n*log(e) + A*B*n)*(d*x + c)*log((b*x + a)/(d*x + c))/((b*x + a)*g^ 2) + (2*B^2*n^2 + 2*B^2*n*log(e) + B^2*log(e)^2 + 2*A*B*n + 2*A*B*log(e) + A^2)*(d*x + c)/((b*x + a)*g^2))*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)
Time = 2.61 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.75 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=-\frac {A^2+2\,A\,B\,n+2\,B^2\,n^2}{x\,b^2\,g^2+a\,b\,g^2}-{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2\,\left (\frac {B^2}{b\,\left (a\,g^2+b\,g^2\,x\right )}-\frac {B^2\,d}{b\,g^2\,\left (a\,d-b\,c\right )}\right )-\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {2\,B^2\,n}{x\,b^2\,g^2+a\,b\,g^2}+\frac {2\,A\,B}{x\,b^2\,g^2+a\,b\,g^2}\right )-\frac {B\,d\,n\,\mathrm {atan}\left (\frac {\left (2\,b\,d\,x+\frac {c\,b^2\,g^2+a\,d\,b\,g^2}{b\,g^2}\right )\,1{}\mathrm {i}}{a\,d-b\,c}\right )\,\left (A+B\,n\right )\,4{}\mathrm {i}}{b\,g^2\,\left (a\,d-b\,c\right )} \]
- (A^2 + 2*B^2*n^2 + 2*A*B*n)/(b^2*g^2*x + a*b*g^2) - log(e*((a + b*x)/(c + d*x))^n)^2*(B^2/(b*(a*g^2 + b*g^2*x)) - (B^2*d)/(b*g^2*(a*d - b*c))) - l og(e*((a + b*x)/(c + d*x))^n)*((2*B^2*n)/(b^2*g^2*x + a*b*g^2) + (2*A*B)/( b^2*g^2*x + a*b*g^2)) - (B*d*n*atan(((2*b*d*x + (b^2*c*g^2 + a*b*d*g^2)/(b *g^2))*1i)/(a*d - b*c))*(A + B*n)*4i)/(b*g^2*(a*d - b*c))