Optimal. Leaf size=49 \[ 2 \tan ^{-1}\left (\frac {x}{\sqrt {x^3-x^2+2}}\right )-2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^3-x^2+2}}\right ) \]
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Rubi [F] time = 51.17, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\left (-4+x^3\right ) \sqrt {2-x^2+x^3}}{\left (2+x^3\right ) \left (2+x^2+x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {\left (-4+x^3\right ) \sqrt {2-x^2+x^3}}{\left (2+x^3\right ) \left (2+x^2+x^3\right )} \, dx &=\int \left (\frac {3 x \sqrt {2-x^2+x^3}}{2+x^3}+\frac {(-2-3 x) \sqrt {2-x^2+x^3}}{2+x^2+x^3}\right ) \, dx\\ &=3 \int \frac {x \sqrt {2-x^2+x^3}}{2+x^3} \, dx+\int \frac {(-2-3 x) \sqrt {2-x^2+x^3}}{2+x^2+x^3} \, dx\\ &=3 \int \left (-\frac {\sqrt {2-x^2+x^3}}{3 \sqrt [3]{2} \left (\sqrt [3]{2}+x\right )}-\frac {(-1)^{2/3} \sqrt {2-x^2+x^3}}{3 \sqrt [3]{2} \left (\sqrt [3]{2}-\sqrt [3]{-1} x\right )}+\frac {\sqrt [3]{-\frac {1}{2}} \sqrt {2-x^2+x^3}}{3 \left (\sqrt [3]{2}+(-1)^{2/3} x\right )}\right ) \, dx+\int \left (-\frac {2 \sqrt {2-x^2+x^3}}{2+x^2+x^3}-\frac {3 x \sqrt {2-x^2+x^3}}{2+x^2+x^3}\right ) \, dx\\ &=-\left (2 \int \frac {\sqrt {2-x^2+x^3}}{2+x^2+x^3} \, dx\right )-3 \int \frac {x \sqrt {2-x^2+x^3}}{2+x^2+x^3} \, dx+\sqrt [3]{-\frac {1}{2}} \int \frac {\sqrt {2-x^2+x^3}}{\sqrt [3]{2}+(-1)^{2/3} x} \, dx-\frac {\int \frac {\sqrt {2-x^2+x^3}}{\sqrt [3]{2}+x} \, dx}{\sqrt [3]{2}}-\frac {(-1)^{2/3} \int \frac {\sqrt {2-x^2+x^3}}{\sqrt [3]{2}-\sqrt [3]{-1} x} \, dx}{\sqrt [3]{2}}\\ \end {align*}
rest of steps removed due to Latex formating problem.
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Mathematica [C] time = 6.40, size = 5117, normalized size = 104.43 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
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IntegrateAlgebraic [A] time = 0.35, size = 49, normalized size = 1.00 \begin {gather*} 2 \tan ^{-1}\left (\frac {x}{\sqrt {2-x^2+x^3}}\right )-2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {2-x^2+x^3}}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.65, size = 88, normalized size = 1.80 \begin {gather*} \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {x^{3} - x^{2} + 2} {\left (x^{3} - 3 \, x^{2} + 2\right )}}{4 \, {\left (x^{4} - x^{3} + 2 \, x\right )}}\right ) - \arctan \left (\frac {\sqrt {x^{3} - x^{2} + 2} {\left (x^{3} - 2 \, x^{2} + 2\right )}}{2 \, {\left (x^{4} - x^{3} + 2 \, x\right )}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{3} - x^{2} + 2} {\left (x^{3} - 4\right )}}{{\left (x^{3} + x^{2} + 2\right )} {\left (x^{3} + 2\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.64, size = 128, normalized size = 2.61
method | result | size |
trager | \(\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+2\right ) x^{2}-4 x \sqrt {x^{3}-x^{2}+2}+2 \RootOf \left (\textit {\_Z}^{2}+2\right )}{x^{3}+x^{2}+2}\right )+\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 x \sqrt {x^{3}-x^{2}+2}+2 \RootOf \left (\textit {\_Z}^{2}+1\right )}{x^{3}+2}\right )\) | \(128\) |
default | \(\frac {\left (4-2 i\right ) \sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}\, \sqrt {\left (-\frac {2}{5}+\frac {i}{5}\right ) \left (x -1-i\right )}\, \sqrt {\left (-\frac {2}{5}-\frac {i}{5}\right ) \left (x -1+i\right )}\, \EllipticF \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )}{\sqrt {x^{3}-x^{2}+2}}+\left (-4+2 i\right ) \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+2\right )}{\sum }\frac {\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha +1\right ) \sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}\, \sqrt {\left (-\frac {2}{5}+\frac {i}{5}\right ) \left (x -1-i\right )}\, \sqrt {\left (-\frac {2}{5}-\frac {i}{5}\right ) \left (x -1+i\right )}\, \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, i \underline {\hspace {1.25 ex}}\alpha ^{2}-2 \underline {\hspace {1.25 ex}}\alpha ^{2}-i \underline {\hspace {1.25 ex}}\alpha +2 \underline {\hspace {1.25 ex}}\alpha +i-2, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )}{\sqrt {x^{3}-x^{2}+2}}\right )+\left (4-2 i\right ) \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+\textit {\_Z}^{2}+2\right )}{\sum }\frac {\underline {\hspace {1.25 ex}}\alpha ^{3} \sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}\, \sqrt {\left (-\frac {2}{5}+\frac {i}{5}\right ) \left (x -1-i\right )}\, \sqrt {\left (-\frac {2}{5}-\frac {i}{5}\right ) \left (x -1+i\right )}\, \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, \frac {1}{2} i \underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha ^{2}, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )}{\sqrt {x^{3}-x^{2}+2}}\right )\) | \(278\) |
elliptic | \(\frac {4 \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \EllipticF \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )}{\sqrt {x^{3}-x^{2}+2}}-\frac {2 i \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \EllipticF \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )}{\sqrt {x^{3}-x^{2}+2}}+\frac {4 \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha ^{3} \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, i \underline {\hspace {1.25 ex}}\alpha ^{2}-2 \underline {\hspace {1.25 ex}}\alpha ^{2}-i \underline {\hspace {1.25 ex}}\alpha +2 \underline {\hspace {1.25 ex}}\alpha +i-2, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )\right )}{\sqrt {x^{3}-x^{2}+2}}-\frac {2 i \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha ^{3} \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, i \underline {\hspace {1.25 ex}}\alpha ^{2}-2 \underline {\hspace {1.25 ex}}\alpha ^{2}-i \underline {\hspace {1.25 ex}}\alpha +2 \underline {\hspace {1.25 ex}}\alpha +i-2, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )\right )}{\sqrt {x^{3}-x^{2}+2}}-\frac {4 \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha ^{2} \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, i \underline {\hspace {1.25 ex}}\alpha ^{2}-2 \underline {\hspace {1.25 ex}}\alpha ^{2}-i \underline {\hspace {1.25 ex}}\alpha +2 \underline {\hspace {1.25 ex}}\alpha +i-2, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )\right )}{\sqrt {x^{3}-x^{2}+2}}+\frac {2 i \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha ^{2} \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, i \underline {\hspace {1.25 ex}}\alpha ^{2}-2 \underline {\hspace {1.25 ex}}\alpha ^{2}-i \underline {\hspace {1.25 ex}}\alpha +2 \underline {\hspace {1.25 ex}}\alpha +i-2, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )\right )}{\sqrt {x^{3}-x^{2}+2}}+\frac {4 \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, i \underline {\hspace {1.25 ex}}\alpha ^{2}-2 \underline {\hspace {1.25 ex}}\alpha ^{2}-i \underline {\hspace {1.25 ex}}\alpha +2 \underline {\hspace {1.25 ex}}\alpha +i-2, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )\right )}{\sqrt {x^{3}-x^{2}+2}}-\frac {2 i \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, i \underline {\hspace {1.25 ex}}\alpha ^{2}-2 \underline {\hspace {1.25 ex}}\alpha ^{2}-i \underline {\hspace {1.25 ex}}\alpha +2 \underline {\hspace {1.25 ex}}\alpha +i-2, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )\right )}{\sqrt {x^{3}-x^{2}+2}}-\frac {4 \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+\textit {\_Z}^{2}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha ^{3} \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, \frac {1}{2} i \underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha ^{2}, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )\right )}{\sqrt {x^{3}-x^{2}+2}}+\frac {2 i \sqrt {\frac {2}{5}+\frac {1}{5} i+\frac {2}{5} x +\frac {1}{5} i x}\, \sqrt {-\frac {2}{5} x +\frac {1}{5} i x +\frac {3}{5}+\frac {1}{5} i}\, \sqrt {-\frac {2}{5} x -\frac {1}{5} i x +\frac {3}{5}-\frac {1}{5} i}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{3}+\textit {\_Z}^{2}+2\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha ^{3} \EllipticPi \left (\sqrt {\left (\frac {2}{5}+\frac {i}{5}\right ) \left (1+x \right )}, \frac {1}{2} i \underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha ^{2}, \frac {2 \sqrt {5}}{5}-\frac {i \sqrt {5}}{5}\right )\right )}{\sqrt {x^{3}-x^{2}+2}}\) | \(1019\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{3} - x^{2} + 2} {\left (x^{3} - 4\right )}}{{\left (x^{3} + x^{2} + 2\right )} {\left (x^{3} + 2\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.60, size = 235, normalized size = 4.80 \begin {gather*} \left (\sum _{_{\mathrm {X264}}\in \left \{-2^{1/3},2^{1/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right ),-2^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right \}\cup \mathrm {root}\left (z^3+z^2+2,z\right )}\frac {\sqrt {5}\,\sqrt {x\,\left (2-\mathrm {i}\right )+2-\mathrm {i}}\,\sqrt {3+x\,\left (-2+1{}\mathrm {i}\right )+1{}\mathrm {i}}\,\sqrt {3+x\,\left (-2-\mathrm {i}\right )-\mathrm {i}}\,\Pi \left (\frac {2+1{}\mathrm {i}}{_{\mathrm {X264}}+1};\mathrm {asin}\left (\frac {\sqrt {5}\,\sqrt {x\,\left (2-\mathrm {i}\right )+2-\mathrm {i}}}{5}\right )\middle |\frac {3}{5}+\frac {4}{5}{}\mathrm {i}\right )\,\left (2\,{_{\mathrm {X264}}}^5+6\,{_{\mathrm {X264}}}^3-2\,{_{\mathrm {X264}}}^2+12\right )\,\left (\frac {4}{25}+\frac {2}{25}{}\mathrm {i}\right )}{_{\mathrm {X264}}\,\left (_{\mathrm {X264}}+1\right )\,\sqrt {x^3-x^2+2}\,\left (6\,{_{\mathrm {X264}}}^4+5\,{_{\mathrm {X264}}}^3+12\,_{\mathrm {X264}}+4\right )}\right )+\frac {\sqrt {x\,\left (\frac {2}{5}-\frac {1}{5}{}\mathrm {i}\right )+\frac {2}{5}-\frac {1}{5}{}\mathrm {i}}\,\sqrt {\frac {3}{5}+x\,\left (-\frac {2}{5}+\frac {1}{5}{}\mathrm {i}\right )+\frac {1}{5}{}\mathrm {i}}\,\sqrt {\frac {3}{5}+x\,\left (-\frac {2}{5}-\frac {1}{5}{}\mathrm {i}\right )-\frac {1}{5}{}\mathrm {i}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {x\,\left (\frac {2}{5}-\frac {1}{5}{}\mathrm {i}\right )+\frac {2}{5}-\frac {1}{5}{}\mathrm {i}}\right )\middle |\frac {3}{5}+\frac {4}{5}{}\mathrm {i}\right )\,\left (4+2{}\mathrm {i}\right )}{\sqrt {x^3-x^2+2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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