Integrand size = 25, antiderivative size = 88 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}} \]
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Time = 0.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {4, 226} \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}} \]
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Rule 4
Rule 226
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {a+c x^4}} \, dx \\ & = \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}} \\ \end{align*}
Result contains complex when optimal does not.
Time = 10.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {i \sqrt {1+\frac {c x^4}{a}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right ),-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} \sqrt {a+c x^4}} \]
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Result contains complex when optimal does not.
Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {\sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) | \(70\) |
elliptic | \(\frac {\sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) | \(70\) |
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Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.33 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {a} \left (-\frac {c}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {c}{a}\right )^{\frac {1}{4}}\right )\,|\,-1)}{c} \]
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Result contains complex when optimal does not.
Time = 0.39 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} \]
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\[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + a}} \,d x } \]
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\[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + a}} \,d x } \]
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Time = 12.90 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {x\,\sqrt {\frac {c\,x^4}{a}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {5}{4};\ -\frac {c\,x^4}{a}\right )}{\sqrt {c\,x^4+a}} \]
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