Integrand size = 49, antiderivative size = 214 \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-\frac {e^{\frac {A (1+m)}{B n}} (1+m) (a+b x) (g (a+b x))^{-2-m} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1+m}{n}} (i (c+d x))^{2+m} \operatorname {ExpIntegralEi}\left (-\frac {(1+m) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B^2 (b c-a d) i^2 n^2 (c+d x)}-\frac {(a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m}}{B (b c-a d) i^2 n (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \]
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Time = 0.22 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {2563, 2343, 2347, 2209} \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-\frac {(m+1) (a+b x) e^{\frac {A (m+1)}{B n}} (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {m+1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {(m+1) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B^2 i^2 n^2 (c+d x) (b c-a d)}-\frac {(a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2}}{B i^2 n (c+d x) (b c-a d) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )} \]
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Rule 2209
Rule 2343
Rule 2347
Rule 2563
Rubi steps \begin{align*} \text {integral}& = \frac {\left ((g (a+b x))^{-2-m} \left (\frac {a+b x}{c+d x}\right )^{2+m} (i (c+d x))^{2+m}\right ) \text {Subst}\left (\int \frac {x^{-2-m}}{\left (A+B \log \left (e x^n\right )\right )^2} \, dx,x,\frac {a+b x}{c+d x}\right )}{(b c-a d) i^2} \\ & = -\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 )}-\frac {\left ((1+m) (g (a+b x))^{-2-m} \left (\frac {a+b x}{c+d x}\right )^{2+m} (i (c+d x))^{2+m}\right ) \text {Subst}\left (\int \frac {x^{-2-m}}{A+B \log \left (e x^n\right )} \, dx,x,\frac {a+b x}{c+d x}\right )}{B (b c-a d) i^2 n} \\ & = -\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 )}-\frac {\left ((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}\right ) \text {Subst}\left (\int \frac {e^{\frac {(-1-m) x}{n}}}{A+B x} \, dx,x,\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B (b c-a d) i^2 n^2 (c+d x)} \\ & = -\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} \text {Ei}\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 )} \\ \end{align*}
\[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=\int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx \]
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\[\int \frac {\left (b g x +a g \right )^{-2-m} \left (d i x +c i \right )^{m}}{{\left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}^{2}}d x\]
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Time = 0.32 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.33 \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-\frac {{\left (B b d g^{2} n x^{2} + B a c g^{2} n + {\left (B b c + B a d\right )} g^{2} n x\right )} {\left (b g x + a g\right )}^{-m - 2} e^{\left (m \log \left (b g x + a g\right ) - m \log \left (\frac {b x + a}{d x + c}\right ) + m \log \left (\frac {i}{g}\right )\right )} + {\left ({\left (B m + B\right )} n \log \left (\frac {b x + a}{d x + c}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A\right )} {\rm Ei}\left (-\frac {{\left (B m + B\right )} n \log \left (\frac {b x + a}{d x + c}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right ) e^{\left (\frac {B m n \log \left (\frac {i}{g}\right ) + A m + {\left (B m + B\right )} \log \left (e\right ) + A}{B n}\right )}}{{\left (B^{3} b c - B^{3} a d\right )} g^{2} n^{3} \log \left (\frac {b x + a}{d x + c}\right ) + {\left (B^{3} b c - B^{3} a d\right )} g^{2} n^{2} \log \left (e\right ) + {\left (A B^{2} b c - A B^{2} a d\right )} g^{2} n^{2}} \]
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Exception generated. \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=\int { \frac {{\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m}}{{\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2}} \,d x } \]
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\[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=\int { \frac {{\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m}}{{\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a g+b g x)^{-2-m} (c i+d i x)^m}{\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=\int \frac {{\left (c\,i+d\,i\,x\right )}^m}{{\left (a\,g+b\,g\,x\right )}^{m+2}\,{\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^2} \,d x \]
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