\(\int \frac {1}{(c g+d g x)^2 (A+B \log (e (\frac {a+b x}{c+d x})^n))^2} \, dx\) [55]

Optimal result

Integrand size = 35, antiderivative size = 154 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=\frac {e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B^2 (b c-a d) g^2 n^2 (c+d x)}-\frac {a+b x}{B (b c-a d) g^2 n (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-\frac {e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \left (B e^{\frac {A}{B n}} n \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}}-\operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right ) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )\right )}{B^2 (b c-a d) g^2 n^2 (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \] Input:

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

Output:

-(((a + b*x)*(B*E^(A/(B*n))*n*(e*((a + b*x)/(c + d*x))^n)^n^(-1) - ExpInte 
gralEi[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(B*n)]*(A + B*Log[e*((a + b* 
x)/(c + d*x))^n])))/(B^2*(b*c - a*d)*E^(A/(B*n))*g^2*n^2*(e*((a + b*x)/(c 
+ d*x))^n)^n^(-1)*(c + d*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {2951, 2734, 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.

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

\(\Big \downarrow \) 2951

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

\(\Big \downarrow \) 2734

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

\(\Big \downarrow \) 2737

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

\(\Big \downarrow \) 2609

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

Input:

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

Output:

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

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 2734

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b 
*Log[c*x^n])^(p + 1)/(b*n*(p + 1))), x] - Simp[1/(b*n*(p + 1))   Int[(a + b 
*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] && Int 
egerQ[2*p]
 

rule 2737

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]
 

rule 2951

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])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.89 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-\frac {{\left ({\left (B b n x + B a n\right )} e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )} - {\left (A d x + A c + {\left (B d x + B c\right )} \log \left (e\right ) + {\left (B d n x + B c n\right )} \log \left (\frac {b x + a}{d x + c}\right )\right )} \operatorname {log\_integral}\left (\frac {{\left (b x + a\right )} e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )}}{d x + c}\right )\right )} e^{\left (-\frac {B \log \left (e\right ) + A}{B n}\right )}}{{\left (A B^{2} b c d - A B^{2} a d^{2}\right )} g^{2} n^{2} x + {\left (A B^{2} b c^{2} - A B^{2} a c d\right )} g^{2} n^{2} + {\left ({\left (B^{3} b c d - B^{3} a d^{2}\right )} g^{2} n^{2} x + {\left (B^{3} b c^{2} - B^{3} a c d\right )} g^{2} n^{2}\right )} \log \left (e\right ) + {\left ({\left (B^{3} b c d - B^{3} a d^{2}\right )} g^{2} n^{3} x + {\left (B^{3} b c^{2} - B^{3} a c d\right )} g^{2} n^{3}\right )} \log \left (\frac {b x + a}{d x + c}\right )} \] Input:

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

Output:

-((B*b*n*x + B*a*n)*e^((B*log(e) + A)/(B*n)) - (A*d*x + A*c + (B*d*x + B*c 
)*log(e) + (B*d*n*x + B*c*n)*log((b*x + a)/(d*x + c)))*log_integral((b*x + 
 a)*e^((B*log(e) + A)/(B*n))/(d*x + c)))*e^(-(B*log(e) + A)/(B*n))/((A*B^2 
*b*c*d - A*B^2*a*d^2)*g^2*n^2*x + (A*B^2*b*c^2 - A*B^2*a*c*d)*g^2*n^2 + (( 
B^3*b*c*d - B^3*a*d^2)*g^2*n^2*x + (B^3*b*c^2 - B^3*a*c*d)*g^2*n^2)*log(e) 
 + ((B^3*b*c*d - B^3*a*d^2)*g^2*n^3*x + (B^3*b*c^2 - B^3*a*c*d)*g^2*n^3)*l 
og((b*x + a)/(d*x + c)))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-{\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} {\left (\frac {b x + a}{{\left (B^{2} g^{2} n^{2} \log \left (\frac {b x + a}{d x + c}\right ) + B^{2} g^{2} n \log \left (e\right ) + A B g^{2} n\right )} {\left (d x + c\right )}} - \frac {{\rm Ei}\left (\frac {\log \left (e\right )}{n} + \frac {A}{B n} + \log \left (\frac {b x + a}{d x + c}\right )\right ) e^{\left (-\frac {A}{B n}\right )}}{B^{2} e^{\left (\frac {1}{n}\right )} g^{2} n^{2}}\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(int(1/(log(((a + b*x)**n*e)/(c + d*x)**n)**2*a*b**2*c**2 + 2*log(((a + b* 
x)**n*e)/(c + d*x)**n)**2*a*b**2*c*d*x + log(((a + b*x)**n*e)/(c + d*x)**n 
)**2*a*b**2*d**2*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)**2*b**3*c**2*x 
+ 2*log(((a + b*x)**n*e)/(c + d*x)**n)**2*b**3*c*d*x**2 + log(((a + b*x)** 
n*e)/(c + d*x)**n)**2*b**3*d**2*x**3 + 2*log(((a + b*x)**n*e)/(c + d*x)**n 
)*a**2*b*c**2 + 4*log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b*c*d*x + 2*log( 
((a + b*x)**n*e)/(c + d*x)**n)*a**2*b*d**2*x**2 + 2*log(((a + b*x)**n*e)/( 
c + d*x)**n)*a*b**2*c**2*x + 4*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**2*c 
*d*x**2 + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**2*d**2*x**3 + a**3*c** 
2 + 2*a**3*c*d*x + a**3*d**2*x**2 + a**2*b*c**2*x + 2*a**2*b*c*d*x**2 + a* 
*2*b*d**2*x**3),x)*log(((a + b*x)**n*e)/(c + d*x)**n)*a**4*b*d**2*n - 2*in 
t(1/(log(((a + b*x)**n*e)/(c + d*x)**n)**2*a*b**2*c**2 + 2*log(((a + b*x)* 
*n*e)/(c + d*x)**n)**2*a*b**2*c*d*x + log(((a + b*x)**n*e)/(c + d*x)**n)** 
2*a*b**2*d**2*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)**2*b**3*c**2*x + 2 
*log(((a + b*x)**n*e)/(c + d*x)**n)**2*b**3*c*d*x**2 + log(((a + b*x)**n*e 
)/(c + d*x)**n)**2*b**3*d**2*x**3 + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a 
**2*b*c**2 + 4*log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b*c*d*x + 2*log(((a 
 + b*x)**n*e)/(c + d*x)**n)*a**2*b*d**2*x**2 + 2*log(((a + b*x)**n*e)/(c + 
 d*x)**n)*a*b**2*c**2*x + 4*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**2*c*d* 
x**2 + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**2*d**2*x**3 + a**3*c**...