Integrand size = 20, antiderivative size = 234 \[ \int (f+g x)^3 \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e f^3 n p x^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{n},2+\frac {1}{n},-\frac {e x^n}{d}\right )}{d (1+n)}-\frac {3 e f^2 g n p x^{2+n} \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{n},2 \left (1+\frac {1}{n}\right ),-\frac {e x^n}{d}\right )}{2 d (2+n)}-\frac {e f g^2 n p x^{3+n} \operatorname {Hypergeometric2F1}\left (1,\frac {3+n}{n},2+\frac {3}{n},-\frac {e x^n}{d}\right )}{d (3+n)}-\frac {e g^3 n p x^{4+n} \operatorname {Hypergeometric2F1}\left (1,\frac {4+n}{n},2 \left (1+\frac {2}{n}\right ),-\frac {e x^n}{d}\right )}{4 d (4+n)}-\frac {f^4 p \log \left (d+e x^n\right )}{4 g}+\frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g} \]
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Time = 0.15 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2513, 1858, 266, 371} \[ \int (f+g x)^3 \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}-\frac {f^4 p \log \left (d+e x^n\right )}{4 g}-\frac {e f^3 n p x^{n+1} \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{n},2+\frac {1}{n},-\frac {e x^n}{d}\right )}{d (n+1)}-\frac {3 e f^2 g n p x^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{n},2 \left (1+\frac {1}{n}\right ),-\frac {e x^n}{d}\right )}{2 d (n+2)}-\frac {e f g^2 n p x^{n+3} \operatorname {Hypergeometric2F1}\left (1,\frac {n+3}{n},2+\frac {3}{n},-\frac {e x^n}{d}\right )}{d (n+3)}-\frac {e g^3 n p x^{n+4} \operatorname {Hypergeometric2F1}\left (1,\frac {n+4}{n},2 \left (1+\frac {2}{n}\right ),-\frac {e x^n}{d}\right )}{4 d (n+4)} \]
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Rule 266
Rule 371
Rule 1858
Rule 2513
Rubi steps \begin{align*} \text {integral}& = \frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}-\frac {(e n p) \int \frac {x^{-1+n} (f+g x)^4}{d+e x^n} \, dx}{4 g} \\ & = \frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}-\frac {(e n p) \int \left (\frac {f^4 x^{-1+n}}{d+e x^n}+\frac {4 f^3 g x^n}{d+e x^n}+\frac {6 f^2 g^2 x^{1+n}}{d+e x^n}+\frac {4 f g^3 x^{2+n}}{d+e x^n}+\frac {g^4 x^{3+n}}{d+e x^n}\right ) \, dx}{4 g} \\ & = \frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}-\left (e f^3 n p\right ) \int \frac {x^n}{d+e x^n} \, dx-\frac {\left (e f^4 n p\right ) \int \frac {x^{-1+n}}{d+e x^n} \, dx}{4 g}-\frac {1}{2} \left (3 e f^2 g n p\right ) \int \frac {x^{1+n}}{d+e x^n} \, dx-\left (e f g^2 n p\right ) \int \frac {x^{2+n}}{d+e x^n} \, dx-\frac {1}{4} \left (e g^3 n p\right ) \int \frac {x^{3+n}}{d+e x^n} \, dx \\ & = -\frac {e f^3 n p x^{1+n} \, _2F_1\left (1,1+\frac {1}{n};2+\frac {1}{n};-\frac {e x^n}{d}\right )}{d (1+n)}-\frac {3 e f^2 g n p x^{2+n} \, _2F_1\left (1,\frac {2+n}{n};2 \left (1+\frac {1}{n}\right );-\frac {e x^n}{d}\right )}{2 d (2+n)}-\frac {e f g^2 n p x^{3+n} \, _2F_1\left (1,\frac {3+n}{n};2+\frac {3}{n};-\frac {e x^n}{d}\right )}{d (3+n)}-\frac {e g^3 n p x^{4+n} \, _2F_1\left (1,\frac {4+n}{n};2 \left (1+\frac {2}{n}\right );-\frac {e x^n}{d}\right )}{4 d (4+n)}-\frac {f^4 p \log \left (d+e x^n\right )}{4 g}+\frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.96 \[ \int (f+g x)^3 \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {-e n p \left (\frac {4 f^3 g x^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{n},2+\frac {1}{n},-\frac {e x^n}{d}\right )}{d (1+n)}+\frac {6 f^2 g^2 x^{2+n} \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{n},2 \left (1+\frac {1}{n}\right ),-\frac {e x^n}{d}\right )}{d (2+n)}+\frac {4 f g^3 x^{3+n} \operatorname {Hypergeometric2F1}\left (1,\frac {3+n}{n},2+\frac {3}{n},-\frac {e x^n}{d}\right )}{d (3+n)}+\frac {g^4 x^{4+n} \operatorname {Hypergeometric2F1}\left (1,\frac {4+n}{n},2+\frac {4}{n},-\frac {e x^n}{d}\right )}{d (4+n)}+\frac {f^4 \log \left (d+e x^n\right )}{e n}\right )+(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g} \]
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\[\int \left (g x +f \right )^{3} \ln \left (c \left (d +e \,x^{n}\right )^{p}\right )d x\]
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\[ \int (f+g x)^3 \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int { {\left (g x + f\right )}^{3} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ) \,d x } \]
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Result contains complex when optimal does not.
Time = 12.91 (sec) , antiderivative size = 515, normalized size of antiderivative = 2.20 \[ \int (f+g x)^3 \log \left (c \left (d+e x^n\right )^p\right ) \, dx=- \frac {d^{-2 - \frac {4}{n}} d^{1 + \frac {4}{n}} e g^{3} p x^{n + 4} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {4}{n}\right ) \Gamma \left (1 + \frac {4}{n}\right )}{4 \Gamma \left (2 + \frac {4}{n}\right )} - \frac {d^{-2 - \frac {4}{n}} d^{1 + \frac {4}{n}} e g^{3} p x^{n + 4} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {4}{n}\right ) \Gamma \left (1 + \frac {4}{n}\right )}{n \Gamma \left (2 + \frac {4}{n}\right )} - \frac {d^{-2 - \frac {3}{n}} d^{1 + \frac {3}{n}} e f g^{2} p x^{n + 3} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {3}{n}\right ) \Gamma \left (1 + \frac {3}{n}\right )}{\Gamma \left (2 + \frac {3}{n}\right )} - \frac {3 d^{-2 - \frac {3}{n}} d^{1 + \frac {3}{n}} e f g^{2} p x^{n + 3} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {3}{n}\right ) \Gamma \left (1 + \frac {3}{n}\right )}{n \Gamma \left (2 + \frac {3}{n}\right )} - \frac {3 d^{-2 - \frac {2}{n}} d^{1 + \frac {2}{n}} e f^{2} g p x^{n + 2} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {2}{n}\right ) \Gamma \left (1 + \frac {2}{n}\right )}{2 \Gamma \left (2 + \frac {2}{n}\right )} - \frac {3 d^{-2 - \frac {2}{n}} d^{1 + \frac {2}{n}} e f^{2} g p x^{n + 2} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {2}{n}\right ) \Gamma \left (1 + \frac {2}{n}\right )}{n \Gamma \left (2 + \frac {2}{n}\right )} + f^{3} x \log {\left (c \left (d + e x^{n}\right )^{p} \right )} + \frac {3 f^{2} g x^{2} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{2} + f g^{2} x^{3} \log {\left (c \left (d + e x^{n}\right )^{p} \right )} + \frac {g^{3} x^{4} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{4} + \frac {d^{- \frac {1}{n}} d^{1 + \frac {1}{n}} e e^{\frac {1}{n}} e^{-1 - \frac {1}{n}} f^{3} p x \Phi \left (\frac {d x^{- n} e^{i \pi }}{e}, 1, \frac {e^{i \pi }}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{d n \Gamma \left (1 + \frac {1}{n}\right )} \]
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\[ \int (f+g x)^3 \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int { {\left (g x + f\right )}^{3} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ) \,d x } \]
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\[ \int (f+g x)^3 \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int { {\left (g x + f\right )}^{3} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ) \,d x } \]
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Timed out. \[ \int (f+g x)^3 \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f+g\,x\right )}^3 \,d x \]
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