Integrand size = 29, antiderivative size = 323 \[ \int (f x)^m \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^p \, dx=\frac {d (f x)^{1+m} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1+m}{n},-p,-p,\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (1+m)}+\frac {e x^{1+n} (f x)^m \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1+m+n}{n},-p,-p,\frac {1+m+2 n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{1+m+n} \]
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Time = 0.22 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1574, 1399, 524, 20} \[ \int (f x)^m \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^p \, dx=\frac {d (f x)^{m+1} \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {m+1}{n},-p,-p,\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1)}+\frac {e x^{n+1} (f x)^m \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {m+n+1}{n},-p,-p,\frac {m+2 n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{m+n+1} \]
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Rule 20
Rule 524
Rule 1399
Rule 1574
Rubi steps \begin{align*} \text {integral}& = \int \left (d (f x)^m \left (a+b x^n+c x^{2 n}\right )^p+e x^n (f x)^m \left (a+b x^n+c x^{2 n}\right )^p\right ) \, dx \\ & = d \int (f x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx+e \int x^n (f x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx \\ & = \left (e x^{-m} (f x)^m\right ) \int x^{m+n} \left (a+b x^n+c x^{2 n}\right )^p \, dx+\left (d \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p\right ) \int (f x)^m \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^p \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^p \, dx \\ & = \frac {d (f x)^{1+m} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac {1+m}{n};-p,-p;\frac {1+m+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (1+m)}+\left (e x^{-m} (f x)^m \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p\right ) \int x^{m+n} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^p \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^p \, dx \\ & = \frac {d (f x)^{1+m} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac {1+m}{n};-p,-p;\frac {1+m+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (1+m)}+\frac {e x^{1+n} (f x)^m \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac {1+m+n}{n};-p,-p;\frac {1+m+2 n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{1+m+n} \\ \end{align*}
Time = 0.83 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.85 \[ \int (f x)^m \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^p \, dx=\frac {x (f x)^m \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+x^n \left (b+c x^n\right )\right )^p \left (d (1+m+n) \operatorname {AppellF1}\left (\frac {1+m}{n},-p,-p,\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+e (1+m) x^n \operatorname {AppellF1}\left (\frac {1+m+n}{n},-p,-p,\frac {1+m+2 n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )}{(1+m) (1+m+n)} \]
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\[\int \left (f x \right )^{m} \left (d +e \,x^{n}\right ) \left (a +b \,x^{n}+c \,x^{2 n}\right )^{p}d x\]
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\[ \int (f x)^m \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (e x^{n} + d\right )} {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (f x\right )^{m} \,d x } \]
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Timed out. \[ \int (f x)^m \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^p \, dx=\text {Timed out} \]
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\[ \int (f x)^m \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (e x^{n} + d\right )} {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (f x\right )^{m} \,d x } \]
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\[ \int (f x)^m \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (e x^{n} + d\right )} {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (f x\right )^{m} \,d x } \]
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Timed out. \[ \int (f x)^m \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int {\left (f\,x\right )}^m\,\left (d+e\,x^n\right )\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^p \,d x \]
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