\(\int \frac {(a g+b g x)^m (c i+d i x)^{-2-m}}{A+B \log (e (\frac {a+b x}{c+d x})^n)} \, dx\) [215]

Optimal result

Integrand size = 49, antiderivative size = 125 \[ \int \frac {(a g+b g x)^m (c i+d i x)^{-2-m}}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {e^{-\frac {A (1+m)}{B n}} (a+b x) (g (a+b x))^m \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-\frac {1+m}{n}} (i (c+d x))^{-m} \operatorname {ExpIntegralEi}\left (\frac {(1+m) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B (b c-a d) i^2 n (c+d x)} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2963, 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.

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

\(\Big \downarrow \) 2963

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

\(\Big \downarrow \) 2747

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

\(\Big \downarrow \) 2609

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

Input:

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

Output:

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

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 2963

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol 
] :> Simp[d^2*((g*((a + b*x)/b))^m/(i^2*(b*c - a*d)*(i*((c + d*x)/d))^m*((a 
 + b*x)/(c + d*x))^m))   Subst[Int[x^m*(A + B*Log[e*x^n])^p, x], x, (a + b* 
x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, A, B, m, n, p, q}, x 
] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && EqQ[m + 
 q + 2, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.78 \[ \int \frac {(a g+b g x)^m (c i+d i x)^{-2-m}}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {{\rm Ei}\left (\frac {{\left (B m + B\right )} n \log \left (\frac {b x + a}{d x + c}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right ) e^{\left (-\frac {{\left (B m + 2 \, B\right )} n \log \left (\frac {i}{g}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right )}}{{\left (B b c - B a d\right )} g^{2} n} \] Input:

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {(a g+b g x)^m (c i+d i x)^{-2-m}}{A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

 - int((a*g + b*g*x)**m/((c*i + d*i*x)**m*log(((a + b*x)**n*e)/(c + d*x)** 
n)*b*c**2 + 2*(c*i + d*i*x)**m*log(((a + b*x)**n*e)/(c + d*x)**n)*b*c*d*x 
+ (c*i + d*i*x)**m*log(((a + b*x)**n*e)/(c + d*x)**n)*b*d**2*x**2 + (c*i + 
 d*i*x)**m*a*c**2 + 2*(c*i + d*i*x)**m*a*c*d*x + (c*i + d*i*x)**m*a*d**2*x 
**2),x)