\(\int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{(A+B \log (e (\frac {a+b x}{c+d x})^n))^2} \, dx\) [222]

Optimal result

Integrand size = 49, antiderivative size = 214 \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-\frac {e^{\frac {A (1+m)}{B n}} (1+m) (a+b x) (g (a+b x))^{-2-m} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1+m}{n}} (i (c+d x))^{2+m} \operatorname {ExpIntegralEi}\left (-\frac {(1+m) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B^2 (b c-a d) i^2 n^2 (c+d x)}-\frac {(a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m}}{B (b c-a d) i^2 n (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {2963, 2743, 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.

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

\(\Big \downarrow \) 2963

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

\(\Big \downarrow \) 2743

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

\(\Big \downarrow \) 2747

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

\(\Big \downarrow \) 2609

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

Input:

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

Output:

((g*(a + b*x))^(-2 - m)*((a + b*x)/(c + d*x))^(2 + m)*(i*(c + d*x))^(2 + m 
)*(-((E^((A*(1 + m))/(B*n))*(1 + m)*(e*((a + b*x)/(c + d*x))^n)^((1 + m)/n 
)*((a + b*x)/(c + d*x))^(-1 - m)*ExpIntegralEi[-(((1 + m)*(A + B*Log[e*((a 
 + b*x)/(c + d*x))^n]))/(B*n))])/(B^2*n^2)) - ((a + b*x)/(c + d*x))^(-1 - 
m)/(B*n*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))))/((b*c - a*d)*i^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 2743

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

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.33 \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-\frac {{\left (B b d g^{2} n x^{2} + B a c g^{2} n + {\left (B b c + B a d\right )} g^{2} n x\right )} {\left (b g x + a g\right )}^{-m - 2} e^{\left (m \log \left (b g x + a g\right ) - m \log \left (\frac {b x + a}{d x + c}\right ) + m \log \left (\frac {i}{g}\right )\right )} + {\left ({\left (B m + B\right )} n \log \left (\frac {b x + a}{d x + c}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A\right )} {\rm Ei}\left (-\frac {{\left (B m + B\right )} n \log \left (\frac {b x + a}{d x + c}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right ) e^{\left (\frac {B m n \log \left (\frac {i}{g}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right )}}{{\left (B^{3} b c - B^{3} a d\right )} g^{2} n^{3} \log \left (\frac {b x + a}{d x + c}\right ) + {\left (B^{3} b c - B^{3} a d\right )} g^{2} n^{2} \log \left (e\right ) + {\left (A B^{2} b c - A B^{2} a d\right )} g^{2} n^{2}} \] Input:

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=\frac {\int \frac {\left (d i x +c i \right )^{m}}{\left (b g x +a g \right )^{m} \mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right )^{2} a^{2} b^{2}+2 \left (b g x +a g \right )^{m} \mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right )^{2} a \,b^{3} x +\left (b g x +a g \right )^{m} \mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right )^{2} b^{4} x^{2}+2 \left (b g x +a g \right )^{m} \mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) a^{3} b +4 \left (b g x +a g \right )^{m} \mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) a^{2} b^{2} x +2 \left (b g x +a g \right )^{m} \mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) a \,b^{3} x^{2}+\left (b g x +a g \right )^{m} a^{4}+2 \left (b g x +a g \right )^{m} a^{3} b x +\left (b g x +a g \right )^{m} a^{2} b^{2} x^{2}}d x}{g^{2}} \] Input:

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

Output:

int((c*i + d*i*x)**m/((a*g + b*g*x)**m*log(((a + b*x)**n*e)/(c + d*x)**n)* 
*2*a**2*b**2 + 2*(a*g + b*g*x)**m*log(((a + b*x)**n*e)/(c + d*x)**n)**2*a* 
b**3*x + (a*g + b*g*x)**m*log(((a + b*x)**n*e)/(c + d*x)**n)**2*b**4*x**2 
+ 2*(a*g + b*g*x)**m*log(((a + b*x)**n*e)/(c + d*x)**n)*a**3*b + 4*(a*g + 
b*g*x)**m*log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b**2*x + 2*(a*g + b*g*x) 
**m*log(((a + b*x)**n*e)/(c + d*x)**n)*a*b**3*x**2 + (a*g + b*g*x)**m*a**4 
 + 2*(a*g + b*g*x)**m*a**3*b*x + (a*g + b*g*x)**m*a**2*b**2*x**2),x)/g**2