Integrand size = 25, antiderivative size = 176 \[ \int \frac {\left (f+g x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {2 f g p x^n}{n}+\frac {d g^2 p x^n}{2 e n}-\frac {g^2 p x^{2 n}}{4 n}-\frac {d^2 g^2 p \log \left (d+e x^n\right )}{2 e^2 n}+\frac {g^2 x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac {2 f g \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \]
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Time = 0.13 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2525, 45, 2463, 2436, 2332, 2441, 2352, 2442} \[ \int \frac {\left (f+g x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {2 f g \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {g^2 x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac {d^2 g^2 p \log \left (d+e x^n\right )}{2 e^2 n}+\frac {f^2 p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{n}+\frac {d g^2 p x^n}{2 e n}-\frac {2 f g p x^n}{n}-\frac {g^2 p x^{2 n}}{4 n} \]
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Rule 45
Rule 2332
Rule 2352
Rule 2436
Rule 2441
Rule 2442
Rule 2463
Rule 2525
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(f+g x)^2 \log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (2 f g \log \left (c (d+e x)^p\right )+\frac {f^2 \log \left (c (d+e x)^p\right )}{x}+g^2 x \log \left (c (d+e x)^p\right )\right ) \, dx,x,x^n\right )}{n} \\ & = \frac {f^2 \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}+\frac {(2 f g) \text {Subst}\left (\int \log \left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n}+\frac {g^2 \text {Subst}\left (\int x \log \left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n} \\ & = \frac {g^2 x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {(2 f g) \text {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^n\right )}{e n}-\frac {\left (e f^2 p\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^n\right )}{n}-\frac {\left (e g^2 p\right ) \text {Subst}\left (\int \frac {x^2}{d+e x} \, dx,x,x^n\right )}{2 n} \\ & = -\frac {2 f g p x^n}{n}+\frac {g^2 x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac {2 f g \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n}-\frac {\left (e g^2 p\right ) \text {Subst}\left (\int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx,x,x^n\right )}{2 n} \\ & = -\frac {2 f g p x^n}{n}+\frac {d g^2 p x^n}{2 e n}-\frac {g^2 p x^{2 n}}{4 n}-\frac {d^2 g^2 p \log \left (d+e x^n\right )}{2 e^2 n}+\frac {g^2 x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac {2 f g \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.70 \[ \int \frac {\left (f+g x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {-e g p x^n \left (8 e f-2 d g+e g x^n\right )-2 d^2 g^2 p \log \left (d+e x^n\right )+2 e \left (4 d f g+e g x^n \left (4 f+g x^n\right )+2 e f^2 \log \left (-\frac {e x^n}{d}\right )\right ) \log \left (c \left (d+e x^n\right )^p\right )+4 e^2 f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{4 e^2 n} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 3.21 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.78
method | result | size |
risch | \(\frac {\left (2 f^{2} \ln \left (x \right ) n +g^{2} x^{2 n}+4 f g \,x^{n}\right ) \ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{2 n}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {g^{2} x^{2 n}}{2}+2 f g \,x^{n}+f^{2} \ln \left (x^{n}\right )\right )}{n}-\frac {g^{2} p \,x^{2 n}}{4 n}+\frac {d \,g^{2} p \,x^{n}}{2 e n}-\frac {d^{2} g^{2} p \ln \left (d +e \,x^{n}\right )}{2 e^{2} n}-\frac {p \,f^{2} \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )-\frac {2 f g p \,x^{n}}{n}+\frac {2 p f g d \ln \left (d +e \,x^{n}\right )}{e n}\) | \(314\) |
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Time = 0.35 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.09 \[ \int \frac {\left (f+g x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {4 \, e^{2} f^{2} n p \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) - 4 \, e^{2} f^{2} n \log \left (c\right ) \log \left (x\right ) + 4 \, e^{2} f^{2} p {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + {\left (e^{2} g^{2} p - 2 \, e^{2} g^{2} \log \left (c\right )\right )} x^{2 \, n} - 2 \, {\left (4 \, e^{2} f g \log \left (c\right ) - {\left (4 \, e^{2} f g - d e g^{2}\right )} p\right )} x^{n} - 2 \, {\left (2 \, e^{2} f^{2} n p \log \left (x\right ) + e^{2} g^{2} p x^{2 \, n} + 4 \, e^{2} f g p x^{n} + {\left (4 \, d e f g - d^{2} g^{2}\right )} p\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \]
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\[ \int \frac {\left (f+g x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\left (f + g x^{n}\right )^{2} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{x}\, dx \]
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\[ \int \frac {\left (f+g x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (g x^{n} + f\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
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\[ \int \frac {\left (f+g x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (g x^{n} + f\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
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Timed out. \[ \int \frac {\left (f+g x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f+g\,x^n\right )}^2}{x} \,d x \]
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