Integrand size = 49, antiderivative size = 292 \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=-\frac {6 B^3 n^3 (a+b x) (g (a+b x))^m (i (c+d x))^{-m}}{(b c-a d) i^2 (1+m)^4 (c+d x)}+\frac {6 B^2 n^2 (a+b x) (g (a+b x))^m (i (c+d x))^{-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) i^2 (1+m)^3 (c+d x)}-\frac {3 B n (a+b x) (g (a+b x))^m (i (c+d x))^{-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b c-a d) i^2 (1+m)^2 (c+d x)}+\frac {(a+b x) (g (a+b x))^m (i (c+d x))^{-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3}{(b c-a d) i^2 (1+m) (c+d x)} \]
-6*B^3*n^3*(b*x+a)*(g*(b*x+a))^m/(-a*d+b*c)/i^2/(1+m)^4/(d*x+c)/((i*(d*x+c ))^m)+6*B^2*n^2*(b*x+a)*(g*(b*x+a))^m*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(-a* d+b*c)/i^2/(1+m)^3/(d*x+c)/((i*(d*x+c))^m)-3*B*n*(b*x+a)*(g*(b*x+a))^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)/((i*(d*x+c)) ^m)+(b*x+a)*(g*(b*x+a))^m*(A+B*ln(e*((b*x+a)/(d*x+c))^n))^3/(-a*d+b*c)/i^2 /(1+m)/(d*x+c)/((i*(d*x+c))^m)
Time = 4.71 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.71 \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\frac {(a+b x) (g (a+b x))^m (i (c+d x))^{-1-m} \left (A^3 (1+m)^3-3 A^2 B (1+m)^2 n+6 A B^2 (1+m) n^2-6 B^3 n^3+3 B (1+m) \left (A^2 (1+m)^2-2 A B (1+m) n+2 B^2 n^2\right ) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+3 B^2 (1+m)^2 (A+A m-B n) \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B^3 (1+m)^3 \log ^3\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) i (1+m)^4} \]
((a + b*x)*(g*(a + b*x))^m*(i*(c + d*x))^(-1 - 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*L og[e*((a + b*x)/(c + d*x))^n]^3))/((b*c - a*d)*i*(1 + m)^4)
Time = 0.58 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.81, 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.
\(\displaystyle \int (a g+b g x)^m (c i+d i x)^{-m-2} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3 \, dx\) |
\(\Big \downarrow \) 2963 |
\(\displaystyle \frac {(g (a+b x))^m (i (c+d x))^{-m} \left (\frac {a+b x}{c+d x}\right )^{-m} \int \left (\frac {a+b x}{c+d x}\right )^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3d\frac {a+b x}{c+d x}}{i^2 (b c-a d)}\) |
\(\Big \downarrow \) 2742 |
\(\displaystyle \frac {(g (a+b x))^m (i (c+d x))^{-m} \left (\frac {a+b x}{c+d x}\right )^{-m} \left (\frac {\left (\frac {a+b x}{c+d x}\right )^{m+1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3}{m+1}-\frac {3 B n \int \left (\frac {a+b x}{c+d x}\right )^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2d\frac {a+b x}{c+d x}}{m+1}\right )}{i^2 (b c-a d)}\) |
\(\Big \downarrow \) 2742 |
\(\displaystyle \frac {(g (a+b x))^m (i (c+d x))^{-m} \left (\frac {a+b x}{c+d x}\right )^{-m} \left (\frac {\left (\frac {a+b x}{c+d x}\right )^{m+1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3}{m+1}-\frac {3 B n \left (\frac {\left (\frac {a+b x}{c+d x}\right )^{m+1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{m+1}-\frac {2 B n \int \left (\frac {a+b x}{c+d x}\right )^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )d\frac {a+b x}{c+d x}}{m+1}\right )}{m+1}\right )}{i^2 (b c-a d)}\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {(g (a+b x))^m (i (c+d x))^{-m} \left (\frac {a+b x}{c+d x}\right )^{-m} \left (\frac {\left (\frac {a+b x}{c+d x}\right )^{m+1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3}{m+1}-\frac {3 B n \left (\frac {\left (\frac {a+b x}{c+d x}\right )^{m+1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{m+1}-\frac {2 B n \left (\frac {\left (\frac {a+b x}{c+d x}\right )^{m+1} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{m+1}-\frac {B n \left (\frac {a+b x}{c+d x}\right )^{m+1}}{(m+1)^2}\right )}{m+1}\right )}{m+1}\right )}{i^2 (b c-a d)}\) |
((g*(a + b*x))^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*L og[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*( (a + b*x)/(c + d*x))^m*(i*(c + d*x))^m)
3.3.12.3.1 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]
\[\int \left (b g x +a g \right )^{m} \left (d i x +c i \right )^{-2-m} {\left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}^{3}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 2680 vs. \(2 (292) = 584\).
Time = 0.42 (sec) , antiderivative size = 2680, normalized size of antiderivative = 9.18 \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\text {Too large to display} \]
integrate((b*g*x+a*g)^m*(d*i*x+c*i)^(-2-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. \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\int { {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{3} {\left (b g x + a g\right )}^{m} {\left (d i x + c i\right )}^{-m - 2} \,d x } \]
integrate((b*g*x+a*g)^m*(d*i*x+c*i)^(-2-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*(d*i*x + c*i)^(-m - 2), x)
\[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\int { {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{3} {\left (b g x + a g\right )}^{m} {\left (d i x + c i\right )}^{-m - 2} \,d x } \]
integrate((b*g*x+a*g)^m*(d*i*x+c*i)^(-2-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*(d*i*x + c*i)^(-m - 2), x)
Timed out. \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\int \frac {{\left (a\,g+b\,g\,x\right )}^m\,{\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^3}{{\left (c\,i+d\,i\,x\right )}^{m+2}} \,d x \]