Integrand size = 20, antiderivative size = 135 \[ \int (d+e x)^m \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {a p (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {a (d+e x)}{a d-b e}\right )}{e (a d-b e) (1+m) (2+m)}-\frac {p (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,1+\frac {e x}{d}\right )}{d e \left (2+3 m+m^2\right )}+\frac {(d+e x)^{1+m} \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (1+m)} \]
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Time = 0.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2513, 528, 88, 67, 70} \[ \int (d+e x)^m \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {(d+e x)^{m+1} \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (m+1)}+\frac {a p (d+e x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {a (d+e x)}{a d-b e}\right )}{e (m+1) (m+2) (a d-b e)}-\frac {p (d+e x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {e x}{d}+1\right )}{d e \left (m^2+3 m+2\right )} \]
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Rule 67
Rule 70
Rule 88
Rule 528
Rule 2513
Rubi steps \begin{align*} \text {integral}& = \frac {(d+e x)^{1+m} \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (1+m)}+\frac {(b p) \int \frac {(d+e x)^{1+m}}{\left (a+\frac {b}{x}\right ) x^2} \, dx}{e (1+m)} \\ & = \frac {(d+e x)^{1+m} \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (1+m)}+\frac {(b p) \int \frac {(d+e x)^{1+m}}{x (b+a x)} \, dx}{e (1+m)} \\ & = \frac {(d+e x)^{1+m} \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (1+m)}+\frac {p \int \frac {(d+e x)^{1+m}}{x} \, dx}{e (1+m)}-\frac {(a p) \int \frac {(d+e x)^{1+m}}{b+a x} \, dx}{e (1+m)} \\ & = \frac {a p (d+e x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac {a (d+e x)}{a d-b e}\right )}{e (a d-b e) (1+m) (2+m)}-\frac {p (d+e x)^{2+m} \, _2F_1\left (1,2+m;3+m;1+\frac {e x}{d}\right )}{d e \left (2+3 m+m^2\right )}+\frac {(d+e x)^{1+m} \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (1+m)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.91 \[ \int (d+e x)^m \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {(d+e x)^{1+m} \left (-a d p (d+e x) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {a (d+e x)}{a d-b e}\right )+(a d-b e) \left (p (d+e x) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,1+\frac {e x}{d}\right )-d (2+m) \log \left (c \left (a+\frac {b}{x}\right )^p\right )\right )\right )}{d e (-a d+b e) (1+m) (2+m)} \]
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\[\int \left (e x +d \right )^{m} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )d x\]
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\[ \int (d+e x)^m \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\int { {\left (e x + d\right )}^{m} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \,d x } \]
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\[ \int (d+e x)^m \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\int \left (d + e x\right )^{m} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}\, dx \]
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\[ \int (d+e x)^m \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\int { {\left (e x + d\right )}^{m} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \,d x } \]
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\[ \int (d+e x)^m \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\int { {\left (e x + d\right )}^{m} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \,d x } \]
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Timed out. \[ \int (d+e x)^m \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\int \ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )\,{\left (d+e\,x\right )}^m \,d x \]
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