Integrand size = 22, antiderivative size = 92 \[ \int (f x)^q \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right ) \, dx=-\frac {b e m n x^{1+m} (f x)^q \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+q}{m},\frac {1+2 m+q}{m},-\frac {e x^m}{d}\right )}{d (1+q) (1+m+q)}+\frac {(f x)^{1+q} \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{f (1+q)} \]
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Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2505, 20, 371} \[ \int (f x)^q \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right ) \, dx=\frac {(f x)^{q+1} \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{f (q+1)}-\frac {b e m n x^{m+1} (f x)^q \operatorname {Hypergeometric2F1}\left (1,\frac {m+q+1}{m},\frac {2 m+q+1}{m},-\frac {e x^m}{d}\right )}{d (q+1) (m+q+1)} \]
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Rule 20
Rule 371
Rule 2505
Rubi steps \begin{align*} \text {integral}& = \frac {(f x)^{1+q} \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{f (1+q)}-\frac {(b e m n) \int \frac {x^{-1+m} (f x)^{1+q}}{d+e x^m} \, dx}{f (1+q)} \\ & = \frac {(f x)^{1+q} \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{f (1+q)}-\frac {\left (b e m n x^{-q} (f x)^q\right ) \int \frac {x^{m+q}}{d+e x^m} \, dx}{1+q} \\ & = -\frac {b e m n x^{1+m} (f x)^q \, _2F_1\left (1,\frac {1+m+q}{m};\frac {1+2 m+q}{m};-\frac {e x^m}{d}\right )}{d (1+q) (1+m+q)}+\frac {(f x)^{1+q} \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{f (1+q)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int (f x)^q \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right ) \, dx=\frac {x (f x)^q \left (-b e m n x^m \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+q}{m},\frac {1+2 m+q}{m},-\frac {e x^m}{d}\right )+d (1+m+q) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )\right )}{d (1+q) (1+m+q)} \]
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\[\int \left (f x \right )^{q} \left (a +b \ln \left (c \left (d +e \,x^{m}\right )^{n}\right )\right )d x\]
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\[ \int (f x)^q \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x^{m} + d\right )}^{n} c\right ) + a\right )} \left (f x\right )^{q} \,d x } \]
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\[ \int (f x)^q \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right ) \, dx=\int \left (f x\right )^{q} \left (a + b \log {\left (c \left (d + e x^{m}\right )^{n} \right )}\right )\, dx \]
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\[ \int (f x)^q \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x^{m} + d\right )}^{n} c\right ) + a\right )} \left (f x\right )^{q} \,d x } \]
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\[ \int (f x)^q \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x^{m} + d\right )}^{n} c\right ) + a\right )} \left (f x\right )^{q} \,d x } \]
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Timed out. \[ \int (f x)^q \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right ) \, dx=\int {\left (f\,x\right )}^q\,\left (a+b\,\ln \left (c\,{\left (d+e\,x^m\right )}^n\right )\right ) \,d x \]
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