Integrand size = 36, antiderivative size = 96 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {e^{\frac {A}{B n}} (c+d x) \left (e (a+b x)^n (c+d x)^{-n}\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{B n}\right )}{B (b c-a d) g^2 n (a+b x)} \]
exp(A/B/n)*(d*x+c)*(e*(b*x+a)^n/((d*x+c)^n))^(1/n)*Ei((-A-B*ln(e*(b*x+a)^n /((d*x+c)^n)))/B/n)/B/(-a*d+b*c)/g^2/n/(b*x+a)
\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx \]
Time = 0.41 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2973, 2949, 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.
\(\displaystyle \int \frac {1}{(a g+b g x)^2 \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )} \, dx\) |
\(\Big \downarrow \) 2973 |
\(\displaystyle \int \frac {1}{(a g+b g x)^2 \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}dx\) |
\(\Big \downarrow \) 2949 |
\(\displaystyle \frac {\int \frac {(c+d x)^2}{(a+b x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}d\frac {a+b x}{c+d x}}{g^2 (b c-a d)}\) |
\(\Big \downarrow \) 2747 |
\(\displaystyle \frac {(c+d x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} \int \frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n}}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}d\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{g^2 n (a+b x) (b c-a d)}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {e^{\frac {A}{B n}} (c+d x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B g^2 n (a+b x) (b c-a d)}\) |
(E^(A/(B*n))*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x)*ExpIntegralEi[-( (A + B*Log[e*((a + b*x)/(c + d*x))^n])/(B*n))])/(B*(b*c - a*d)*g^2*n*(a + b*x))
3.3.29.3.1 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_.), 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]
\[\int \frac {1}{\left (b g x +a g \right )^{2} \left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}d x\]
Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )} \operatorname {log\_integral}\left (\frac {{\left (d x + c\right )} e^{\left (-\frac {B \log \left (e\right ) + A}{B n}\right )}}{b x + a}\right )}{{\left (B b c - B a d\right )} g^{2} n} \]
e^((B*log(e) + A)/(B*n))*log_integral((d*x + c)*e^(-(B*log(e) + A)/(B*n))/ (b*x + a))/((B*b*c - B*a*d)*g^2*n)
Timed out. \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{2} {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}} \,d x } \]
\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{2} {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\int \frac {1}{{\left (a\,g+b\,g\,x\right )}^2\,\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )} \,d x \]