Integrand size = 29, antiderivative size = 196 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f (1+m)}+\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f (1+m)} \]
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Time = 0.17 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1574, 371} \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\frac {(f x)^{m+1} \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(f x)^{m+1} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (\sqrt {b^2-4 a c}+b\right )} \]
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Rule 371
Rule 1574
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n}+\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n}\right ) \, dx \\ & = \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {(f x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx+\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {(f x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx \\ & = \frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f (1+m)}+\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f (1+m)} \\ \end{align*}
Time = 0.65 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.81 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\frac {x (f x)^m \left (\left (b d+\sqrt {b^2-4 a c} d-2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+\left (-b d+\sqrt {b^2-4 a c} d+2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )\right )}{2 a \sqrt {b^2-4 a c} (1+m)} \]
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\[\int \frac {\left (f x \right )^{m} \left (d +e \,x^{n}\right )}{a +b \,x^{n}+c \,x^{2 n}}d x\]
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\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
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\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int \frac {\left (f x\right )^{m} \left (d + e x^{n}\right )}{a + b x^{n} + c x^{2 n}}\, dx \]
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\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
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\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
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Timed out. \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int \frac {{\left (f\,x\right )}^m\,\left (d+e\,x^n\right )}{a+b\,x^n+c\,x^{2\,n}} \,d x \]
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