Integrand size = 49, antiderivative size = 190 \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \, dx=-\frac {e^{\frac {A (1+m)}{B n}} (a+b x) (g (a+b x))^{-2-m} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1+m}{n}} (i (c+d x))^{2+m} \Gamma \left (1+p,\frac {(1+m) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right ) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \left (\frac {(1+m) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )^{-p}}{(b c-a d) i^2 (1+m) (c+d x)} \]
-exp(A*(1+m)/B/n)*(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)*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)/i^2/(1+m)/(d*x+c)/(((1+m)*( A+B*ln(e*((b*x+a)/(d*x+c))^n))/B/n)^p)
\[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \, dx=\int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \, dx \]
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])^p, x]
Time = 0.55 (sec) , antiderivative size = 190, 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.
\(\displaystyle \int (a g+b g x)^{-m-2} (c i+d i x)^m \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^p \, dx\) |
\(\Big \downarrow \) 2963 |
\(\displaystyle \frac {(g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (\frac {a+b x}{c+d x}\right )^{m+2} \int \left (\frac {a+b x}{c+d x}\right )^{-m-2} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^pd\frac {a+b x}{c+d x}}{i^2 (b c-a d)}\) |
\(\Big \downarrow \) 2747 |
\(\displaystyle \frac {(a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {m+1}{n}} \int \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-\frac {m+1}{n}} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^pd\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{i^2 n (c+d x) (b c-a d)}\) |
\(\Big \downarrow \) 2612 |
\(\displaystyle -\frac {(a+b x) e^{\frac {A (m+1)}{B n}} (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {m+1}{n}} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^p \left (\frac {(m+1) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{B n}\right )^{-p} \Gamma \left (p+1,\frac {(m+1) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{i^2 (m+1) (c+d x) (b c-a d)}\) |
-((E^((A*(1 + m))/(B*n))*(a + b*x)*(g*(a + b*x))^(-2 - m)*(e*((a + b*x)/(c + d*x))^n)^((1 + m)/n)*(i*(c + d*x))^(2 + 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)*i^2*(1 + m)*(c + d*x)*(((1 + m)*(A + B*Log[e*((a + b *x)/(c + d*x))^n]))/(B*n))^p))
3.3.11.3.1 Defintions of rubi rules used
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]
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]
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]
\[\int \left (b g x +a g \right )^{-2-m} \left (d i x +c i \right )^{m} {\left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}^{p}d x\]
\[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \, dx=\int { {\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{p} \,d x } \]
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) )^p,x, algorithm="fricas")
Timed out. \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \, dx=\text {Timed out} \]
\[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \, dx=\int { {\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{p} \,d x } \]
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) )^p,x, algorithm="maxima")
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)^p, x)
\[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \, dx=\int { {\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{p} \,d x } \]
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) )^p,x, algorithm="giac")
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)^p, x)
Timed out. \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p \, dx=\int \frac {{\left (c\,i+d\,i\,x\right )}^m\,{\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^p}{{\left (a\,g+b\,g\,x\right )}^{m+2}} \,d x \]