\(\int (a g+b g x)^m (c i+d i x)^{-2-m} (A+B \log (e (\frac {a+b x}{c+d x})^n))^p \, dx\) [210]

Optimal result

Integrand size = 49, antiderivative size = 189 \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \, dx=\frac {e^{-\frac {A (1+m)}{B n}} (a+b x) (g (a+b x))^m \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-\frac {1+m}{n}} (i (c+d x))^{-m} \Gamma \left (1+p,-\frac {(1+m) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right ) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \left (-\frac {(1+m) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )^{-p}}{(b c-a d) i^2 (1+m) (c+d x)} \] Output:

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

Mathematica [F]

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 189, 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, 2612}

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.

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

\(\Big \downarrow \) 2963

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

\(\Big \downarrow \) 2747

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

\(\Big \downarrow \) 2612

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

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])^p,x]
 

Output:

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

Defintions of rubi rules used

rule 2612

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] 
:> Simp[(-F^(g*(e - c*(f/d))))*((c + d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d) 
)^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m + 1, 
 ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] && 
!IntegerQ[m]
 

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]

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

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))^p,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))^p,x)
 

Fricas [F]

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

Output:

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

Sympy [F(-1)]

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

Output:

Timed out
 

Maxima [F]

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

Giac [F]

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

Output:

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