Integrand size = 18, antiderivative size = 81 \[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {2 e p (f x)^{3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},-\frac {e x^2}{d}\right )}{d f^3 (1+m) (3+m)}+\frac {(f x)^{1+m} \log \left (c \left (d+e x^2\right )^p\right )}{f (1+m)} \]
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Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2505, 16, 371} \[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {(f x)^{m+1} \log \left (c \left (d+e x^2\right )^p\right )}{f (m+1)}-\frac {2 e p (f x)^{m+3} \operatorname {Hypergeometric2F1}\left (1,\frac {m+3}{2},\frac {m+5}{2},-\frac {e x^2}{d}\right )}{d f^3 (m+1) (m+3)} \]
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Rule 16
Rule 371
Rule 2505
Rubi steps \begin{align*} \text {integral}& = \frac {(f x)^{1+m} \log \left (c \left (d+e x^2\right )^p\right )}{f (1+m)}-\frac {(2 e p) \int \frac {x (f x)^{1+m}}{d+e x^2} \, dx}{f (1+m)} \\ & = \frac {(f x)^{1+m} \log \left (c \left (d+e x^2\right )^p\right )}{f (1+m)}-\frac {(2 e p) \int \frac {(f x)^{2+m}}{d+e x^2} \, dx}{f^2 (1+m)} \\ & = -\frac {2 e p (f x)^{3+m} \, _2F_1\left (1,\frac {3+m}{2};\frac {5+m}{2};-\frac {e x^2}{d}\right )}{d f^3 (1+m) (3+m)}+\frac {(f x)^{1+m} \log \left (c \left (d+e x^2\right )^p\right )}{f (1+m)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {x (f x)^m \left (-2 e p x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},-\frac {e x^2}{d}\right )+d (3+m) \log \left (c \left (d+e x^2\right )^p\right )\right )}{d (1+m) (3+m)} \]
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\[\int \left (f x \right )^{m} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )d x\]
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\[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \,d x } \]
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Time = 28.95 (sec) , antiderivative size = 377, normalized size of antiderivative = 4.65 \[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=- 2 e p \left (\begin {cases} \frac {0^{m} \sqrt {- \frac {d}{e^{3}}} \log {\left (- e \sqrt {- \frac {d}{e^{3}}} + x \right )}}{2} - \frac {0^{m} \sqrt {- \frac {d}{e^{3}}} \log {\left (e \sqrt {- \frac {d}{e^{3}}} + x \right )}}{2} + \frac {0^{m} x}{e} & \text {for}\: \left (f = 0 \wedge m \neq -1\right ) \vee f = 0 \\\frac {f^{m + 1} m x^{m + 3} \Phi \left (\frac {e x^{2} e^{i \pi }}{d}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 d f m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 4 d f \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 f^{m + 1} x^{m + 3} \Phi \left (\frac {e x^{2} e^{i \pi }}{d}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 d f m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 4 d f \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} & \text {for}\: m > -\infty \wedge m < \infty \wedge m \neq -1 \\- \frac {\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}}{2 e f} + \frac {\log {\left (f x \right )} \log {\left (d + e x^{2} \right )}}{2 e f} & \text {otherwise} \end {cases}\right ) + \left (\begin {cases} 0^{m} x & \text {for}\: f = 0 \\\frac {\begin {cases} \frac {\left (f x\right )^{m + 1}}{m + 1} & \text {for}\: m \neq -1 \\\log {\left (f x \right )} & \text {otherwise} \end {cases}}{f} & \text {otherwise} \end {cases}\right ) \log {\left (c \left (d + e x^{2}\right )^{p} \right )} \]
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\[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \,d x } \]
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\[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \,d x } \]
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Timed out. \[ \int (f x)^m \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\int \ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,{\left (f\,x\right )}^m \,d x \]
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