Integrand size = 15, antiderivative size = 240 \[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {8}{135} x \left (-2+3 x^2\right )^{3/4}+\frac {2}{27} x^3 \left (-2+3 x^2\right )^{3/4}+\frac {32 x \sqrt [4]{-2+3 x^2}}{135 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}-\frac {32 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x}+\frac {16 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{135 \sqrt {3} x} \]
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Time = 0.09 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {327, 236, 311, 226, 1210} \[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {16 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{135 \sqrt {3} x}-\frac {32 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x}+\frac {8}{135} \left (3 x^2-2\right )^{3/4} x+\frac {32 \sqrt [4]{3 x^2-2} x}{135 \left (\sqrt {3 x^2-2}+\sqrt {2}\right )}+\frac {2}{27} \left (3 x^2-2\right )^{3/4} x^3 \]
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Rule 226
Rule 236
Rule 311
Rule 327
Rule 1210
Rubi steps \begin{align*} \text {integral}& = \frac {2}{27} x^3 \left (-2+3 x^2\right )^{3/4}+\frac {4}{9} \int \frac {x^2}{\sqrt [4]{-2+3 x^2}} \, dx \\ & = \frac {8}{135} x \left (-2+3 x^2\right )^{3/4}+\frac {2}{27} x^3 \left (-2+3 x^2\right )^{3/4}+\frac {16}{135} \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx \\ & = \frac {8}{135} x \left (-2+3 x^2\right )^{3/4}+\frac {2}{27} x^3 \left (-2+3 x^2\right )^{3/4}+\frac {\left (16 \sqrt {\frac {2}{3}} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{135 x} \\ & = \frac {8}{135} x \left (-2+3 x^2\right )^{3/4}+\frac {2}{27} x^3 \left (-2+3 x^2\right )^{3/4}+\frac {\left (32 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{135 \sqrt {3} x}-\frac {\left (32 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {x^2}{\sqrt {2}}}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{135 \sqrt {3} x} \\ & = \frac {8}{135} x \left (-2+3 x^2\right )^{3/4}+\frac {2}{27} x^3 \left (-2+3 x^2\right )^{3/4}+\frac {32 x \sqrt [4]{-2+3 x^2}}{135 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}-\frac {32 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x}+\frac {16 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.26 \[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2 x \left (-8+2 x^2+15 x^4+4\ 2^{3/4} \sqrt [4]{2-3 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {3 x^2}{2}\right )\right )}{135 \sqrt [4]{-2+3 x^2}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.18
method | result | size |
meijerg | \(\frac {2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x^{5} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {5}{2};\frac {7}{2};\frac {3 x^{2}}{2}\right )}{10 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\) | \(42\) |
risch | \(\frac {2 x \left (5 x^{2}+4\right ) \left (3 x^{2}-2\right )^{\frac {3}{4}}}{135}+\frac {8 \,2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {3 x^{2}}{2}\right )}{135 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\) | \(60\) |
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\[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {x^{4}}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}}} \,d x } \]
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Result contains complex when optimal does not.
Time = 0.47 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.12 \[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2^{\frac {3}{4}} x^{5} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{10} \]
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\[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {x^{4}}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {x^{4}}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\int \frac {x^4}{{\left (3\,x^2-2\right )}^{1/4}} \,d x \]
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