Integrand size = 21, antiderivative size = 43 \[ \int \frac {1+b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\frac {\sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}} \]
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Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1214, 1213, 435} \[ \int \frac {1+b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\frac {\sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {b^2 x^4-1}} \]
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Rule 435
Rule 1213
Rule 1214
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1-b^2 x^4} \int \frac {1+b x^2}{\sqrt {1-b^2 x^4}} \, dx}{\sqrt {-1+b^2 x^4}} \\ & = \frac {\sqrt {1-b^2 x^4} \int \frac {\sqrt {1+b x^2}}{\sqrt {1-b x^2}} \, dx}{\sqrt {-1+b^2 x^4}} \\ & = \frac {\sqrt {1-b^2 x^4} E\left (\left .\sin ^{-1}\left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.72 \[ \int \frac {1+b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\frac {\sqrt {1-b^2 x^4} \left (3 x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )+b x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right )\right )}{3 \sqrt {-1+b^2 x^4}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.93 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.05
method | result | size |
meijerg | \(\frac {b \sqrt {-\operatorname {signum}\left (b^{2} x^{4}-1\right )}\, x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};b^{2} x^{4}\right )}{3 \sqrt {\operatorname {signum}\left (b^{2} x^{4}-1\right )}}+\frac {\sqrt {-\operatorname {signum}\left (b^{2} x^{4}-1\right )}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};b^{2} x^{4}\right )}{\sqrt {\operatorname {signum}\left (b^{2} x^{4}-1\right )}}\) | \(88\) |
default | \(\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, F\left (x \sqrt {-b}, i\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}+\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (F\left (x \sqrt {-b}, i\right )-E\left (x \sqrt {-b}, i\right )\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\) | \(107\) |
elliptic | \(\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, F\left (x \sqrt {-b}, i\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}+\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (F\left (x \sqrt {-b}, i\right )-E\left (x \sqrt {-b}, i\right )\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\) | \(107\) |
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none
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.30 \[ \int \frac {1+b x^2}{\sqrt {-1+b^2 x^4}} \, dx=-\frac {\frac {{\left (b + 1\right )} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} - \frac {x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} - \sqrt {b^{2} x^{4} - 1}}{b x} \]
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Time = 0.80 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.42 \[ \int \frac {1+b x^2}{\sqrt {-1+b^2 x^4}} \, dx=- \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]
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\[ \int \frac {1+b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\int { \frac {b x^{2} + 1}{\sqrt {b^{2} x^{4} - 1}} \,d x } \]
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\[ \int \frac {1+b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\int { \frac {b x^{2} + 1}{\sqrt {b^{2} x^{4} - 1}} \,d x } \]
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Timed out. \[ \int \frac {1+b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\int \frac {b\,x^2+1}{\sqrt {b^2\,x^4-1}} \,d x \]
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