Integrand size = 35, antiderivative size = 96 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B (b c-a d) g^2 n (c+d x)} \] Output:
(b*x+a)*Ei((A+B*ln(e*((b*x+a)/(d*x+c))^n))/B/n)/B/(-a*d+b*c)/exp(A/B/n)/g^ 2/n/((e*((b*x+a)/(d*x+c))^n)^(1/n))/(d*x+c)
Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B (b c-a d) g^2 n (c+d x)} \] Input:
Integrate[1/((c*g + d*g*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]
Output:
((a + b*x)*ExpIntegralEi[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(B*n)])/(B *(b*c - a*d)*E^(A/(B*n))*g^2*n*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x ))
Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2951, 2737, 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.
\(\displaystyle \int \frac {1}{(c g+d g x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )} \, dx\) |
\(\Big \downarrow \) 2951 |
\(\displaystyle \frac {\int \frac {1}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}d\frac {a+b x}{c+d x}}{g^2 (b c-a d)}\) |
\(\Big \downarrow \) 2737 |
\(\displaystyle \frac {(a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \int \frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}}}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}d\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{g^2 n (c+d x) (b c-a d)}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {(a+b x) e^{-\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B g^2 n (c+d x) (b c-a d)}\) |
Input:
Int[1/((c*g + d*g*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]
Output:
((a + b*x)*ExpIntegralEi[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(B*n)])/(B *(b*c - a*d)*E^(A/(B*n))*g^2*n*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x ))
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_), x_Symbol] :> Simp[x/(n*(c*x ^n)^(1/n)) Subst[Int[E^(x/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ [{a, b, c, 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/d)^m Subst[Int[(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] && NeQ[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[d*f - c*g, 0] && (GtQ[p, 0] || LtQ[m, - 1])
\[\int \frac {1}{\left (d g x +c g \right )^{2} \left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}d x\]
Input:
int(1/(d*g*x+c*g)^2/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)
Output:
int(1/(d*g*x+c*g)^2/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)
Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{\left (-\frac {B \log \left (e\right ) + A}{B n}\right )} \operatorname {log\_integral}\left (\frac {{\left (b x + a\right )} e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )}}{d x + c}\right )}{{\left (B b c - B a d\right )} g^{2} n} \] Input:
integrate(1/(d*g*x+c*g)^2/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="f ricas")
Output:
e^(-(B*log(e) + A)/(B*n))*log_integral((b*x + a)*e^((B*log(e) + A)/(B*n))/ (d*x + c))/((B*b*c - B*a*d)*g^2*n)
\[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {\int \frac {1}{A c^{2} + 2 A c d x + A d^{2} x^{2} + B c^{2} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + 2 B c d x \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + B d^{2} x^{2} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}\, dx}{g^{2}} \] Input:
integrate(1/(d*g*x+c*g)**2/(A+B*ln(e*((b*x+a)/(d*x+c))**n)),x)
Output:
Integral(1/(A*c**2 + 2*A*c*d*x + A*d**2*x**2 + B*c**2*log(e*(a/(c + d*x) + b*x/(c + d*x))**n) + 2*B*c*d*x*log(e*(a/(c + d*x) + b*x/(c + d*x))**n) + B*d**2*x**2*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)), x)/g**2
\[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int { \frac {1}{{\left (d g x + c g\right )}^{2} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}} \,d x } \] Input:
integrate(1/(d*g*x+c*g)^2/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="m axima")
Output:
integrate(1/((d*g*x + c*g)^2*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)
\[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int { \frac {1}{{\left (d g x + c g\right )}^{2} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}} \,d x } \] Input:
integrate(1/(d*g*x+c*g)^2/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="g iac")
Output:
integrate(1/((d*g*x + c*g)^2*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)
Timed out. \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int \frac {1}{{\left (c\,g+d\,g\,x\right )}^2\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )} \,d x \] Input:
int(1/((c*g + d*g*x)^2*(A + B*log(e*((a + b*x)/(c + d*x))^n))),x)
Output:
int(1/((c*g + d*g*x)^2*(A + B*log(e*((a + b*x)/(c + d*x))^n))), x)
\[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx =\text {Too large to display} \] Input:
int(1/(d*g*x+c*g)^2/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x)
Output:
(int(1/(log(((a + b*x)**n*e)/(c + d*x)**n)*a*b*c**2 + 2*log(((a + b*x)**n* e)/(c + d*x)**n)*a*b*c*d*x + log(((a + b*x)**n*e)/(c + d*x)**n)*a*b*d**2*x **2 + log(((a + b*x)**n*e)/(c + d*x)**n)*b**2*c**2*x + 2*log(((a + b*x)**n *e)/(c + d*x)**n)*b**2*c*d*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)*b**2* d**2*x**3 + a**2*c**2 + 2*a**2*c*d*x + a**2*d**2*x**2 + a*b*c**2*x + 2*a*b *c*d*x**2 + a*b*d**2*x**3),x)*a**2*d**2*n - 2*int(1/(log(((a + b*x)**n*e)/ (c + d*x)**n)*a*b*c**2 + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b*c*d*x + log(((a + b*x)**n*e)/(c + d*x)**n)*a*b*d**2*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)*b**2*c**2*x + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*b**2*c*d*x* *2 + log(((a + b*x)**n*e)/(c + d*x)**n)*b**2*d**2*x**3 + a**2*c**2 + 2*a** 2*c*d*x + a**2*d**2*x**2 + a*b*c**2*x + 2*a*b*c*d*x**2 + a*b*d**2*x**3),x) *a*b*c*d*n + int(1/(log(((a + b*x)**n*e)/(c + d*x)**n)*a*b*c**2 + 2*log((( a + b*x)**n*e)/(c + d*x)**n)*a*b*c*d*x + log(((a + b*x)**n*e)/(c + d*x)**n )*a*b*d**2*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)*b**2*c**2*x + 2*log(( (a + b*x)**n*e)/(c + d*x)**n)*b**2*c*d*x**2 + log(((a + b*x)**n*e)/(c + d* x)**n)*b**2*d**2*x**3 + a**2*c**2 + 2*a**2*c*d*x + a**2*d**2*x**2 + a*b*c* *2*x + 2*a*b*c*d*x**2 + a*b*d**2*x**3),x)*b**2*c**2*n - log(log(((a + b*x) **n*e)/(c + d*x)**n)*b + a))/(d*g**2*n*(a*d - b*c))