Integrand size = 18, antiderivative size = 59 \[ \int \left (a+b x^n\right )^{-\frac {1-n}{n}} \, dx=x \left (a+b x^n\right )^{-\frac {1-n}{n}} \left (1+\frac {b x^n}{a}\right )^{-1+\frac {1}{n}} \operatorname {Hypergeometric2F1}\left (-1+\frac {1}{n},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {252, 251} \[ \int \left (a+b x^n\right )^{-\frac {1-n}{n}} \, dx=x \left (a+b x^n\right )^{-\frac {1-n}{n}} \left (\frac {b x^n}{a}+1\right )^{\frac {1}{n}-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{n}-1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right ) \]
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Rule 251
Rule 252
Rubi steps \begin{align*} \text {integral}& = \left (\left (a+b x^n\right )^{-\frac {1-n}{n}} \left (1+\frac {b x^n}{a}\right )^{\frac {1-n}{n}}\right ) \int \left (1+\frac {b x^n}{a}\right )^{-\frac {1-n}{n}} \, dx \\ & = x \left (a+b x^n\right )^{-\frac {1-n}{n}} \left (1+\frac {b x^n}{a}\right )^{-1+\frac {1}{n}} \, _2F_1\left (-1+\frac {1}{n},\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \left (a+b x^n\right )^{-\frac {1-n}{n}} \, dx=a x \left (a+b x^n\right )^{-1/n} \left (1+\frac {b x^n}{a}\right )^{\frac {1}{n}} \operatorname {Hypergeometric2F1}\left (-1+\frac {1}{n},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right ) \]
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\[\int \left (a +b \,x^{n}\right )^{-\frac {1-n}{n}}d x\]
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\[ \int \left (a+b x^n\right )^{-\frac {1-n}{n}} \, dx=\int { {\left (b x^{n} + a\right )}^{\frac {n - 1}{n}} \,d x } \]
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Result contains complex when optimal does not.
Time = 2.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.81 \[ \int \left (a+b x^n\right )^{-\frac {1-n}{n}} \, dx=\frac {a^{\frac {1}{n}} a^{1 - \frac {2}{n}} x \Gamma \left (\frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{n}, -1 + \frac {1}{n} \\ 1 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (1 + \frac {1}{n}\right )} \]
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\[ \int \left (a+b x^n\right )^{-\frac {1-n}{n}} \, dx=\int { {\left (b x^{n} + a\right )}^{\frac {n - 1}{n}} \,d x } \]
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\[ \int \left (a+b x^n\right )^{-\frac {1-n}{n}} \, dx=\int { {\left (b x^{n} + a\right )}^{\frac {n - 1}{n}} \,d x } \]
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Time = 6.00 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int \left (a+b x^n\right )^{-\frac {1-n}{n}} \, dx=\frac {a\,x\,{\left (\frac {b\,x^n}{a}+1\right )}^{1/n}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{n}-1,\frac {1}{n};\ \frac {1}{n}+1;\ -\frac {b\,x^n}{a}\right )}{{\left (a+b\,x^n\right )}^{1/n}} \]
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