\(\int \frac {1}{(a+b x)^2 \log (e (\frac {a+b x}{c+d x})^n)} \, dx\) [246]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 71 \[ \int \frac {1}{(a+b x)^2 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \operatorname {ExpIntegralEi}\left (-\frac {\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (a+b x)} \] Output:

(e*((b*x+a)/(d*x+c))^n)^(1/n)*(d*x+c)*Ei(-ln(e*((b*x+a)/(d*x+c))^n)/n)/(-a 
*d+b*c)/n/(b*x+a)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x)^2 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \operatorname {ExpIntegralEi}\left (-\frac {\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (a+b x)} \] Input:

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

Output:

((e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x)*ExpIntegralEi[-(Log[e*((a + 
b*x)/(c + d*x))^n]/n)])/((b*c - a*d)*n*(a + b*x))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, 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.

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

\(\Big \downarrow \) 2949

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

\(\Big \downarrow \) 2747

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

\(\Big \downarrow \) 2609

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

Input:

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

Output:

((e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x)*ExpIntegralEi[-(Log[e*((a + 
b*x)/(c + d*x))^n]/n)])/((b*c - a*d)*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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(a+b x)^2 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {e^{\left (\frac {1}{n}\right )} \operatorname {log\_integral}\left (\frac {d x + c}{{\left (b x + a\right )} e^{\left (\frac {1}{n}\right )}}\right )}{{\left (b c - a d\right )} n} \] Input:

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

Output:

e^(1/n)*log_integral((d*x + c)/((b*x + a)*e^(1/n)))/((b*c - a*d)*n)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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