Integrand size = 40, antiderivative size = 88 \[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (e (a+b x)^n (c+d x)^{-n}\right )^{-\frac {1+m}{n}} \operatorname {ExpIntegralEi}\left (\frac {(1+m) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{n}\right )}{(b c-a d) n} \]
(b*x+a)^(1+m)*(d*x+c)^(-1-m)*Ei((1+m)*ln(e*(b*x+a)^n/((d*x+c)^n))/n)/(-a*d +b*c)/n/((e*(b*x+a)^n/((d*x+c)^n))^((1+m)/n))
\[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx \]
Time = 0.54 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2973, 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.
\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{-m-2}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx\) |
\(\Big \downarrow \) 2973 |
\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{-m-2}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )}dx\) |
\(\Big \downarrow \) 2963 |
\(\displaystyle \frac {(a+b x)^m (c+d x)^{-m} \left (\frac {a+b x}{c+d x}\right )^{-m} \int \frac {\left (\frac {a+b x}{c+d x}\right )^m}{\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}d\frac {a+b x}{c+d x}}{b c-a d}\) |
\(\Big \downarrow \) 2747 |
\(\displaystyle \frac {(a+b x)^{m+1} (c+d x)^{-m-1} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-\frac {m+1}{n}} \int \frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {m+1}{n}} (c+d x)}{a+b x}d\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n (b c-a d)}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {(a+b x)^{m+1} (c+d x)^{-m-1} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-\frac {m+1}{n}} \operatorname {ExpIntegralEi}\left (\frac {(m+1) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{n (b c-a d)}\) |
((a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*ExpIntegralEi[((1 + m)*Log[e*((a + b *x)/(c + d*x))^n])/n])/((b*c - a*d)*n*(e*((a + b*x)/(c + d*x))^n)^((1 + m) /n))
3.3.40.3.1 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[((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]
\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-2-m}}{\ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )}d x\]
\[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2}}{\log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2}}{\log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )} \,d x } \]
\[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2}}{\log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\,{\left (c+d\,x\right )}^{m+2}} \,d x \]