\(\int \frac {1}{(c g+d g x)^3 (A+B \log (e (\frac {a+b x}{c+d x})^n))} \, dx\) [51]

Optimal result

Integrand size = 35, antiderivative size = 199 \[ \int \frac {1}{(c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {b e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B (b c-a d)^2 g^3 n (c+d x)}-\frac {d e^{-\frac {2 A}{B n}} (a+b x)^2 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B (b c-a d)^2 g^3 n (c+d x)^2} \] Output:

b*(b*x+a)*Ei((A+B*ln(e*((b*x+a)/(d*x+c))^n))/B/n)/B/(-a*d+b*c)^2/exp(A/B/n 
)/g^3/n/((e*((b*x+a)/(d*x+c))^n)^(1/n))/(d*x+c)-d*(b*x+a)^2*Ei(2*(A+B*ln(e 
*((b*x+a)/(d*x+c))^n))/B/n)/B/(-a*d+b*c)^2/exp(2*A/B/n)/g^3/n/((e*((b*x+a) 
/(d*x+c))^n)^(2/n))/(d*x+c)^2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{-\frac {2 A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-2/n} \left (b e^{\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )-d (a+b x) \operatorname {ExpIntegralEi}\left (\frac {2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )\right )}{B (b c-a d)^2 g^3 n (c+d x)^2} \] Input:

Integrate[1/((c*g + d*g*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]
 

Output:

((a + b*x)*(b*E^(A/(B*n))*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x)*Exp 
IntegralEi[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(B*n)] - d*(a + b*x)*Exp 
IntegralEi[(2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(B*n)]))/(B*(b*c - a 
*d)^2*E^((2*A)/(B*n))*g^3*n*(e*((a + b*x)/(c + d*x))^n)^(2/n)*(c + d*x)^2)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2951, 2767, 2009}

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.

\(\displaystyle \int \frac {1}{(c g+d g x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )} \, dx\)

\(\Big \downarrow \) 2951

\(\displaystyle \frac {\int \frac {b-\frac {d (a+b x)}{c+d x}}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}d\frac {a+b x}{c+d x}}{g^3 (b c-a d)^2}\)

\(\Big \downarrow \) 2767

\(\displaystyle \frac {\int \left (\frac {b}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}-\frac {d (a+b x)}{(c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}\right )d\frac {a+b x}{c+d x}}{g^3 (b c-a d)^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b (a+b x) e^{-\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B n (c+d x)}-\frac {d (a+b x)^2 e^{-\frac {2 A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B n (c+d x)^2}}{g^3 (b c-a d)^2}\)

Input:

Int[1/((c*g + d*g*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]
 

Output:

((b*(a + b*x)*ExpIntegralEi[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(B*n)]) 
/(B*E^(A/(B*n))*n*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x)) - (d*(a + 
b*x)^2*ExpIntegralEi[(2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(B*n)])/(B 
*E^((2*A)/(B*n))*n*(e*((a + b*x)/(c + d*x))^n)^(2/n)*(c + d*x)^2))/((b*c - 
 a*d)^2*g^3)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2767

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^( 
q_.), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (d + e*x 
^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x] 
&& IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))
 

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]

\[\int \frac {1}{\left (d g x +c g \right )^{3} \left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}d x\]

Input:

int(1/(d*g*x+c*g)^3/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)
 

Output:

int(1/(d*g*x+c*g)^3/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {{\left (b e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )} \operatorname {log\_integral}\left (\frac {{\left (b x + a\right )} e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )}}{d x + c}\right ) - d \operatorname {log\_integral}\left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} e^{\left (\frac {2 \, {\left (B \log \left (e\right ) + A\right )}}{B n}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )\right )} e^{\left (-\frac {2 \, {\left (B \log \left (e\right ) + A\right )}}{B n}\right )}}{{\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} g^{3} n} \] Input:

integrate(1/(d*g*x+c*g)^3/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="f 
ricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{(c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {\int \frac {1}{A c^{3} + 3 A c^{2} d x + 3 A c d^{2} x^{2} + A d^{3} x^{3} + B c^{3} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + 3 B c^{2} d x \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + 3 B c d^{2} x^{2} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + B d^{3} x^{3} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}\, dx}{g^{3}} \] Input:

integrate(1/(d*g*x+c*g)**3/(A+B*ln(e*((b*x+a)/(d*x+c))**n)),x)
 

Output:

Integral(1/(A*c**3 + 3*A*c**2*d*x + 3*A*c*d**2*x**2 + A*d**3*x**3 + B*c**3 
*log(e*(a/(c + d*x) + b*x/(c + d*x))**n) + 3*B*c**2*d*x*log(e*(a/(c + d*x) 
 + b*x/(c + d*x))**n) + 3*B*c*d**2*x**2*log(e*(a/(c + d*x) + b*x/(c + d*x) 
)**n) + B*d**3*x**3*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)), x)/g**3
 

Maxima [F]

\[ \int \frac {1}{(c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int { \frac {1}{{\left (d g x + c g\right )}^{3} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}} \,d x } \] Input:

integrate(1/(d*g*x+c*g)^3/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="m 
axima")
 

Output:

integrate(1/((d*g*x + c*g)^3*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)
 

Giac [F]

\[ \int \frac {1}{(c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int { \frac {1}{{\left (d g x + c g\right )}^{3} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}} \,d x } \] Input:

integrate(1/(d*g*x+c*g)^3/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="g 
iac")
 

Output:

integrate(1/((d*g*x + c*g)^3*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int \frac {1}{{\left (c\,g+d\,g\,x\right )}^3\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )} \,d x \] Input:

int(1/((c*g + d*g*x)^3*(A + B*log(e*((a + b*x)/(c + d*x))^n))),x)
 

Output:

int(1/((c*g + d*g*x)^3*(A + B*log(e*((a + b*x)/(c + d*x))^n))), x)
 

Reduce [F]

\[ \int \frac {1}{(c g+d g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {\int \frac {1}{\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) b \,c^{3}+3 \,\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) b \,c^{2} d x +3 \,\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) b c \,d^{2} x^{2}+\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) b \,d^{3} x^{3}+a \,c^{3}+3 a \,c^{2} d x +3 a c \,d^{2} x^{2}+a \,d^{3} x^{3}}d x}{g^{3}} \] Input:

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