Integrand size = 49, antiderivative size = 309 \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=-\frac {6 B^3 n^3 (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m}}{(b c-a d) i^2 (1+m)^4 (c+d x)}-\frac {6 B^2 n^2 (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) i^2 (1+m)^3 (c+d x)}-\frac {3 B n (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b c-a d) i^2 (1+m)^2 (c+d x)}-\frac {(a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3}{(b c-a d) i^2 (1+m) (c+d x)} \]
-6*B^3*n^3*(b*x+a)*(g*(b*x+a))^(-2-m)*(i*(d*x+c))^(2+m)/(-a*d+b*c)/i^2/(1+ m)^4/(d*x+c)-6*B^2*n^2*(b*x+a)*(g*(b*x+a))^(-2-m)*(i*(d*x+c))^(2+m)*(A+B*l n(e*((b*x+a)/(d*x+c))^n))/(-a*d+b*c)/i^2/(1+m)^3/(d*x+c)-3*B*n*(b*x+a)*(g* (b*x+a))^(-2-m)*(i*(d*x+c))^(2+m)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))^2/(-a*d+ b*c)/i^2/(1+m)^2/(d*x+c)-(b*x+a)*(g*(b*x+a))^(-2-m)*(i*(d*x+c))^(2+m)*(A+B *ln(e*((b*x+a)/(d*x+c))^n))^3/(-a*d+b*c)/i^2/(1+m)/(d*x+c)
Time = 4.82 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.67 \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=-\frac {(g (a+b x))^{-1-m} (c+d x) (i (c+d x))^m \left (A^3 (1+m)^3+3 A^2 B (1+m)^2 n+6 A B^2 (1+m) n^2+6 B^3 n^3+3 B (1+m) \left (A^2 (1+m)^2+2 A B (1+m) n+2 B^2 n^2\right ) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+3 B^2 (1+m)^2 (A+A m+B n) \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B^3 (1+m)^3 \log ^3\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) g (1+m)^4} \]
-(((g*(a + b*x))^(-1 - m)*(c + d*x)*(i*(c + d*x))^m*(A^3*(1 + m)^3 + 3*A^2 *B*(1 + m)^2*n + 6*A*B^2*(1 + m)*n^2 + 6*B^3*n^3 + 3*B*(1 + m)*(A^2*(1 + m )^2 + 2*A*B*(1 + m)*n + 2*B^2*n^2)*Log[e*((a + b*x)/(c + d*x))^n] + 3*B^2* (1 + m)^2*(A + A*m + B*n)*Log[e*((a + b*x)/(c + d*x))^n]^2 + B^3*(1 + m)^3 *Log[e*((a + b*x)/(c + d*x))^n]^3))/((b*c - a*d)*g*(1 + m)^4))
Time = 0.59 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.82, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {2963, 2742, 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.
| \(\displaystyle \int (a g+b g x)^{-m-2} (c i+d i x)^m \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3 \, dx\) |
| \(\Big \downarrow \) 2963 |
| \(\displaystyle \frac {(g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (\frac {a+b x}{c+d x}\right )^{m+2} \int \left (\frac {a+b x}{c+d x}\right )^{-m-2} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3d\frac {a+b x}{c+d x}}{i^2 (b c-a d)}\) |
| \(\Big \downarrow \) 2742 |
| \(\displaystyle \frac {(g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (\frac {a+b x}{c+d x}\right )^{m+2} \left (\frac {3 B n \int \left (\frac {a+b x}{c+d x}\right )^{-m-2} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2d\frac {a+b x}{c+d x}}{m+1}-\frac {\left (\frac {a+b x}{c+d x}\right )^{-m-1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3}{m+1}\right )}{i^2 (b c-a d)}\) |
| \(\Big \downarrow \) 2742 |
| \(\displaystyle \frac {(g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (\frac {a+b x}{c+d x}\right )^{m+2} \left (\frac {3 B n \left (\frac {2 B n \int \left (\frac {a+b x}{c+d x}\right )^{-m-2} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )d\frac {a+b x}{c+d x}}{m+1}-\frac {\left (\frac {a+b x}{c+d x}\right )^{-m-1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{m+1}\right )}{m+1}-\frac {\left (\frac {a+b x}{c+d x}\right )^{-m-1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3}{m+1}\right )}{i^2 (b c-a d)}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {(g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (\frac {a+b x}{c+d x}\right )^{m+2} \left (\frac {3 B n \left (\frac {2 B n \left (-\frac {\left (\frac {a+b x}{c+d x}\right )^{-m-1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{m+1}-\frac {B n \left (\frac {a+b x}{c+d x}\right )^{-m-1}}{(m+1)^2}\right )}{m+1}-\frac {\left (\frac {a+b x}{c+d x}\right )^{-m-1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{m+1}\right )}{m+1}-\frac {\left (\frac {a+b x}{c+d x}\right )^{-m-1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3}{m+1}\right )}{i^2 (b c-a d)}\) |
((g*(a + b*x))^(-2 - m)*((a + b*x)/(c + d*x))^(2 + m)*(i*(c + d*x))^(2 + m )*(-((((a + b*x)/(c + d*x))^(-1 - m)*(A + B*Log[e*((a + b*x)/(c + d*x))^n] )^3)/(1 + m)) + (3*B*n*(-((((a + b*x)/(c + d*x))^(-1 - m)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(1 + m)) + (2*B*n*(-((B*n*((a + b*x)/(c + d*x))^ (-1 - m))/(1 + m)^2) - (((a + b*x)/(c + d*x))^(-1 - m)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(1 + m)))/(1 + m)))/(1 + m)))/((b*c - a*d)*i^2)
3.3.18.3.1 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_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol ] :> Simp[d^2*((g*((a + b*x)/b))^m/(i^2*(b*c - a*d)*(i*((c + d*x)/d))^m*((a + b*x)/(c + d*x))^m)) Subst[Int[x^m*(A + B*Log[e*x^n])^p, x], x, (a + b* x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, A, B, m, n, p, q}, x ] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && EqQ[m + q + 2, 0]
\[\int \left (b g x +a g \right )^{-2-m} \left (d i x +c i \right )^{m} {\left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}^{3}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 2669 vs. \(2 (309) = 618\).
Time = 0.44 (sec) , antiderivative size = 2669, normalized size of antiderivative = 8.64 \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\text {Too large to display} \]
integrate((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m*(A+B*log(e*((b*x+a)/(d*x+c))^n) )^3,x, algorithm="fricas")
-(A^3*a*c*m^3 + 6*B^3*a*c*n^3 + 3*A^3*a*c*m^2 + 3*A^3*a*c*m + A^3*a*c + (B ^3*a*c*m^3 + 3*B^3*a*c*m^2 + 3*B^3*a*c*m + B^3*a*c + (B^3*b*d*m^3 + 3*B^3* b*d*m^2 + 3*B^3*b*d*m + B^3*b*d)*x^2 + (B^3*b*c + B^3*a*d + (B^3*b*c + B^3 *a*d)*m^3 + 3*(B^3*b*c + B^3*a*d)*m^2 + 3*(B^3*b*c + B^3*a*d)*m)*x)*log(e) ^3 + ((B^3*b*d*m^3 + 3*B^3*b*d*m^2 + 3*B^3*b*d*m + B^3*b*d)*n^3*x^2 + (B^3 *b*c + B^3*a*d + (B^3*b*c + B^3*a*d)*m^3 + 3*(B^3*b*c + B^3*a*d)*m^2 + 3*( B^3*b*c + B^3*a*d)*m)*n^3*x + (B^3*a*c*m^3 + 3*B^3*a*c*m^2 + 3*B^3*a*c*m + B^3*a*c)*n^3)*log((b*x + a)/(d*x + c))^3 + 6*(A*B^2*a*c*m + A*B^2*a*c)*n^ 2 + (A^3*b*d*m^3 + 6*B^3*b*d*n^3 + 3*A^3*b*d*m^2 + 3*A^3*b*d*m + A^3*b*d + 6*(A*B^2*b*d*m + A*B^2*b*d)*n^2 + 3*(A^2*B*b*d*m^2 + 2*A^2*B*b*d*m + A^2* B*b*d)*n)*x^2 + 3*(A*B^2*a*c*m^3 + 3*A*B^2*a*c*m^2 + 3*A*B^2*a*c*m + A*B^2 *a*c + (A*B^2*b*d*m^3 + 3*A*B^2*b*d*m^2 + 3*A*B^2*b*d*m + A*B^2*b*d + (B^3 *b*d*m^2 + 2*B^3*b*d*m + B^3*b*d)*n)*x^2 + (B^3*a*c*m^2 + 2*B^3*a*c*m + B^ 3*a*c)*n + (A*B^2*b*c + A*B^2*a*d + (A*B^2*b*c + A*B^2*a*d)*m^3 + 3*(A*B^2 *b*c + A*B^2*a*d)*m^2 + 3*(A*B^2*b*c + A*B^2*a*d)*m + (B^3*b*c + B^3*a*d + (B^3*b*c + B^3*a*d)*m^2 + 2*(B^3*b*c + B^3*a*d)*m)*n)*x + ((B^3*b*d*m^3 + 3*B^3*b*d*m^2 + 3*B^3*b*d*m + B^3*b*d)*n*x^2 + (B^3*b*c + B^3*a*d + (B^3* b*c + B^3*a*d)*m^3 + 3*(B^3*b*c + B^3*a*d)*m^2 + 3*(B^3*b*c + B^3*a*d)*m)* n*x + (B^3*a*c*m^3 + 3*B^3*a*c*m^2 + 3*B^3*a*c*m + B^3*a*c)*n)*log((b*x + a)/(d*x + c)))*log(e)^2 + 3*((B^3*a*c*m^2 + 2*B^3*a*c*m + B^3*a*c)*n^3 ...
Exception generated. \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\int { {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{3} {\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m} \,d x } \]
integrate((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m*(A+B*log(e*((b*x+a)/(d*x+c))^n) )^3,x, algorithm="maxima")
integrate((B*log(e*((b*x + a)/(d*x + c))^n) + A)^3*(b*g*x + a*g)^(-m - 2)* (d*i*x + c*i)^m, x)
\[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\int { {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{3} {\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m} \,d x } \]
integrate((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m*(A+B*log(e*((b*x+a)/(d*x+c))^n) )^3,x, algorithm="giac")
integrate((B*log(e*((b*x + a)/(d*x + c))^n) + A)^3*(b*g*x + a*g)^(-m - 2)* (d*i*x + c*i)^m, x)
Timed out. \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\int \frac {{\left (c\,i+d\,i\,x\right )}^m\,{\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^3}{{\left (a\,g+b\,g\,x\right )}^{m+2}} \,d x \]