Integrand size = 21, antiderivative size = 152 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=\frac {c d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )}+\frac {e^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (c d^2+a e^2\right )}-\frac {c e x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right ) (1+n)} \]
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Time = 0.07 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1439, 251, 1432, 371} \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=-\frac {c e x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a (n+1) \left (a e^2+c d^2\right )}+\frac {c d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (a e^2+c d^2\right )}+\frac {e^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (a e^2+c d^2\right )} \]
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Rule 251
Rule 371
Rule 1432
Rule 1439
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^2}{\left (c d^2+a e^2\right ) \left (d+e x^n\right )}-\frac {c \left (-d+e x^n\right )}{\left (c d^2+a e^2\right ) \left (a+c x^{2 n}\right )}\right ) \, dx \\ & = -\frac {c \int \frac {-d+e x^n}{a+c x^{2 n}} \, dx}{c d^2+a e^2}+\frac {e^2 \int \frac {1}{d+e x^n} \, dx}{c d^2+a e^2} \\ & = \frac {e^2 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (c d^2+a e^2\right )}+\frac {(c d) \int \frac {1}{a+c x^{2 n}} \, dx}{c d^2+a e^2}-\frac {(c e) \int \frac {x^n}{a+c x^{2 n}} \, dx}{c d^2+a e^2} \\ & = \frac {c d x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )}+\frac {e^2 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (c d^2+a e^2\right )}-\frac {c e x^{1+n} \, _2F_1\left (1,\frac {1+n}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right ) (1+n)} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=\frac {x \left (c d^2 (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+e \left (a e (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )-c d x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )\right )\right )}{a d \left (c d^2+a e^2\right ) (1+n)} \]
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\[\int \frac {1}{\left (d +e \,x^{n}\right ) \left (a +c \,x^{2 n}\right )}d x\]
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\[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )} {\left (e x^{n} + d\right )}} \,d x } \]
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Exception generated. \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )} {\left (e x^{n} + d\right )}} \,d x } \]
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\[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )} {\left (e x^{n} + d\right )}} \,d x } \]
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Timed out. \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=\int \frac {1}{\left (a+c\,x^{2\,n}\right )\,\left (d+e\,x^n\right )} \,d x \]
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