\(\int \frac {1}{(a g+b g x)^2 (A+B \log (e (\frac {a+b x}{c+d x})^n))} \, dx\) [22]

Optimal result

Integrand size = 35, antiderivative size = 94 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B (b c-a d) g^2 n (a+b x)} \] Output:

exp(A/B/n)*(e*((b*x+a)/(d*x+c))^n)^(1/n)*(d*x+c)*Ei(-(A+B*ln(e*((b*x+a)/(d 
*x+c))^n))/B/n)/B/(-a*d+b*c)/g^2/n/(b*x+a)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B (b c-a d) g^2 n (a+b x)} \] Input:

Integrate[1/((a*g + b*g*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]
 

Output:

(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))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 94, 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 = {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.

\(\displaystyle \int \frac {1}{(a g+b g x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )} \, dx\)

\(\Big \downarrow \) 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

\(\displaystyle \frac {e^{\frac {A}{B n}} (c+d x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B g^2 n (a+b x) (b c-a d)}\)

Input:

Int[1/((a*g + b*g*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]
 

Output:

(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))
 

Defintions of rubi rules used

rule 2609

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]
 

rule 2747

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]
 

rule 2949

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])
 

Maple [F]

\[\int \frac {1}{\left (b g x +a g \right )^{2} \left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}d x\]

Input:

int(1/(b*g*x+a*g)^2/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)
 

Output:

int(1/(b*g*x+a*g)^2/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )} \operatorname {log\_integral}\left (\frac {{\left (d x + c\right )} e^{\left (-\frac {B \log \left (e\right ) + A}{B n}\right )}}{b x + a}\right )}{{\left (B b c - B a d\right )} g^{2} n} \] Input:

integrate(1/(b*g*x+a*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((d*x + c)*e^(-(B*log(e) + A)/(B*n))/ 
(b*x + a))/((B*b*c - B*a*d)*g^2*n)
 

Sympy [F]

\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {\int \frac {1}{A a^{2} + 2 A a b x + A b^{2} x^{2} + B a^{2} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + 2 B a b x \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + B b^{2} x^{2} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}\, dx}{g^{2}} \] Input:

integrate(1/(b*g*x+a*g)**2/(A+B*ln(e*((b*x+a)/(d*x+c))**n)),x)
 

Output:

Integral(1/(A*a**2 + 2*A*a*b*x + A*b**2*x**2 + B*a**2*log(e*(a/(c + d*x) + 
 b*x/(c + d*x))**n) + 2*B*a*b*x*log(e*(a/(c + d*x) + b*x/(c + d*x))**n) + 
B*b**2*x**2*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)), x)/g**2
 

Maxima [F]

\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{2} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}} \,d x } \] Input:

integrate(1/(b*g*x+a*g)^2/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="m 
axima")
 

Output:

integrate(1/((b*g*x + a*g)^2*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)
 

Giac [F]

\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{2} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}} \,d x } \] Input:

integrate(1/(b*g*x+a*g)^2/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="g 
iac")
 

Output:

integrate(1/((b*g*x + a*g)^2*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int \frac {1}{{\left (a\,g+b\,g\,x\right )}^2\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )} \,d x \] Input:

int(1/((a*g + b*g*x)^2*(A + B*log(e*((a + b*x)/(c + d*x))^n))),x)
 

Output:

int(1/((a*g + b*g*x)^2*(A + B*log(e*((a + b*x)/(c + d*x))^n))), x)
 

Reduce [F]

\[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx =\text {Too large to display} \] Input:

int(1/(b*g*x+a*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**2*b*c + log(((a + b*x)**n 
*e)/(c + d*x)**n)*a**2*b*d*x + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**2 
*c*x + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**2*d*x**2 + log(((a + b*x) 
**n*e)/(c + d*x)**n)*b**3*c*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)*b**3 
*d*x**3 + a**3*c + a**3*d*x + 2*a**2*b*c*x + 2*a**2*b*d*x**2 + a*b**2*c*x* 
*2 + a*b**2*d*x**3),x)*a**2*b*d**2*n + 2*int(1/(log(((a + b*x)**n*e)/(c + 
d*x)**n)*a**2*b*c + log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b*d*x + 2*log( 
((a + b*x)**n*e)/(c + d*x)**n)*a*b**2*c*x + 2*log(((a + b*x)**n*e)/(c + d* 
x)**n)*a*b**2*d*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)*b**3*c*x**2 + lo 
g(((a + b*x)**n*e)/(c + d*x)**n)*b**3*d*x**3 + a**3*c + a**3*d*x + 2*a**2* 
b*c*x + 2*a**2*b*d*x**2 + a*b**2*c*x**2 + a*b**2*d*x**3),x)*a*b**2*c*d*n - 
 int(1/(log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b*c + log(((a + b*x)**n*e) 
/(c + d*x)**n)*a**2*b*d*x + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**2*c* 
x + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**2*d*x**2 + log(((a + b*x)**n 
*e)/(c + d*x)**n)*b**3*c*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)*b**3*d* 
x**3 + a**3*c + a**3*d*x + 2*a**2*b*c*x + 2*a**2*b*d*x**2 + a*b**2*c*x**2 
+ a*b**2*d*x**3),x)*b**3*c**2*n - log(log(((a + b*x)**n*e)/(c + d*x)**n)*b 
 + a)*d)/(b**2*g**2*n*(a*d - b*c))