Integrand size = 50, antiderivative size = 193 \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \, dx=\frac {e^{-\frac {A (1+m)}{B n}} (a+b x) (g (a+b x))^m (i (c+d x))^{-m} \left (e (a+b x)^n (c+d x)^{-n}\right )^{-\frac {1+m}{n}} \Gamma \left (1+p,-\frac {(1+m) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{B n}\right ) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \left (-\frac {(1+m) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{B n}\right )^{-p}}{(b c-a d) i^2 (1+m) (c+d x)} \]
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Time = 0.31 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2573, 2563, 2347, 2212} \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \, dx=\frac {(a+b x) e^{-\frac {A (m+1)}{B n}} (g (a+b x))^m (i (c+d x))^{-m} \left (e (a+b x)^n (c+d x)^{-n}\right )^{-\frac {m+1}{n}} \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^p \left (-\frac {(m+1) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{B n}\right )^{-p} \Gamma \left (p+1,-\frac {(m+1) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{B n}\right )}{i^2 (m+1) (c+d x) (b c-a d)} \]
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Rule 2212
Rule 2347
Rule 2563
Rule 2573
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\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,e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \text {Subst}\left (\frac {\left ((g (a+b x))^m \left (\frac {a+b x}{c+d x}\right )^{-m} (i (c+d x))^{-m}\right ) \text {Subst}\left (\int x^m \left (A+B \log \left (e x^n\right )\right )^p \, dx,x,\frac {a+b x}{c+d x}\right )}{(b c-a d) i^2},e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \text {Subst}\left (\frac {\left ((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}\right ) \text {Subst}\left (\int e^{\frac {(1+m) x}{n}} (A+B x)^p \, dx,x,\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) i^2 n (c+d x)},e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \frac {e^{-\frac {A (1+m)}{B n}} (a+b x) (g (a+b x))^m (i (c+d x))^{-m} \left (e (a+b x)^n (c+d x)^{-n}\right )^{-\frac {1+m}{n}} \Gamma \left (1+p,-\frac {(1+m) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{B n}\right ) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \left (-\frac {(1+m) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )}{B n}\right )^{-p}}{(b c-a d) i^2 (1+m) (c+d x)} \\ \end{align*}
\[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \, dx=\int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \, dx \]
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\[\int \left (b g x +a g \right )^{m} \left (d i x +c i \right )^{-2-m} {\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{p}d x\]
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\[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \, dx=\int { {\left (b g x + a g\right )}^{m} {\left (d i x + c i\right )}^{-m - 2} {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{p} \,d x } \]
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Timed out. \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \, dx=\text {Timed out} \]
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\[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \, dx=\int { {\left (b g x + a g\right )}^{m} {\left (d i x + c i\right )}^{-m - 2} {\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{p} \,d x } \]
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Exception generated. \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \, dx=\text {Exception raised: RuntimeError} \]
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Timed out. \[ \int (a g+b g x)^m (c i+d i x)^{-2-m} \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p \, dx=\int \frac {{\left (a\,g+b\,g\,x\right )}^m\,{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}^p}{{\left (c\,i+d\,i\,x\right )}^{m+2}} \,d x \]
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