Integrand size = 49, antiderivative size = 128 \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {e^{\frac {A (1+m)}{B n}} (a+b x) (g (a+b x))^{-2-m} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1+m}{n}} (i (c+d x))^{2+m} \operatorname {ExpIntegralEi}\left (-\frac {(1+m) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B (b c-a d) i^2 n (c+d x)} \]
exp(A*(1+m)/B/n)*(b*x+a)*(g*(b*x+a))^(-2-m)*(e*((b*x+a)/(d*x+c))^n)^((1+m) /n)*(i*(d*x+c))^(2+m)*Ei(-(1+m)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/B/n)/B/(-a *d+b*c)/i^2/n/(d*x+c)
\[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx \]
Integrate[((a*g + b*g*x)^(-2 - m)*(c*i + d*i*x)^m)/(A + B*Log[e*((a + b*x) /(c + d*x))^n]), x]
Time = 0.51 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2963, 2747, 2609}
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.
\(\displaystyle \int \frac {(a g+b g x)^{-m-2} (c i+d i x)^m}{B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A} \, dx\) |
\(\Big \downarrow \) 2963 |
\(\displaystyle \frac {(g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (\frac {a+b x}{c+d x}\right )^{m+2} \int \frac {\left (\frac {a+b x}{c+d x}\right )^{-m-2}}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}d\frac {a+b x}{c+d x}}{i^2 (b c-a d)}\) |
\(\Big \downarrow \) 2747 |
\(\displaystyle \frac {(a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {m+1}{n}} \int \frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-\frac {m+1}{n}}}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}d\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{i^2 n (c+d x) (b c-a d)}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {(a+b x) e^{\frac {A (m+1)}{B n}} (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {m+1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {(m+1) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B i^2 n (c+d x) (b c-a d)}\) |
(E^((A*(1 + m))/(B*n))*(a + b*x)*(g*(a + b*x))^(-2 - m)*(e*((a + b*x)/(c + d*x))^n)^((1 + m)/n)*(i*(c + d*x))^(2 + m)*ExpIntegralEi[-(((1 + m)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(B*n))])/(B*(b*c - a*d)*i^2*n*(c + d*x) )
3.3.21.3.1 Defintions of rubi rules used
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol ] :> Simp[(d*x)^(m + 1)/(d*n*(c*x^n)^((m + 1)/n)) Subst[Int[E^(((m + 1)/n )*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}, x]
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]
\[\int \frac {\left (b g x +a g \right )^{-2-m} \left (d i x +c i \right )^{m}}{A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}d x\]
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.73 \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {{\rm Ei}\left (-\frac {{\left (B m + B\right )} n \log \left (\frac {b x + a}{d x + c}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right ) e^{\left (\frac {B m n \log \left (\frac {i}{g}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right )}}{{\left (B b c - B a d\right )} g^{2} n} \]
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) ),x, algorithm="fricas")
Ei(-((B*m + B)*n*log((b*x + a)/(d*x + c)) + A*m + (B*m + B)*log(e) + A)/(B *n))*e^((B*m*n*log(i/g) + A*m + (B*m + B)*log(e) + A)/(B*n))/((B*b*c - B*a *d)*g^2*n)
Exception generated. \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int { \frac {{\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m}}{B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A} \,d x } \]
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) ),x, algorithm="maxima")
\[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int { \frac {{\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m}}{B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A} \,d x } \]
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) ),x, algorithm="giac")
Timed out. \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int \frac {{\left (c\,i+d\,i\,x\right )}^m}{{\left (a\,g+b\,g\,x\right )}^{m+2}\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )} \,d x \]