Optimal. Leaf size=47 \[ \tan ^{-1}\left (\frac {\sqrt {2 x^5+x}}{2 x^4+1}\right )-\tanh ^{-1}\left (\frac {\sqrt {2 x^5+x}}{2 x^4+1}\right ) \]
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Rubi [F] time = 1.80, 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 (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx &=\frac {\sqrt {x+2 x^5} \int \frac {\sqrt {x} \left (-1+6 x^4\right )}{\sqrt {1+2 x^4} \left (1-x^2+4 x^4+4 x^8\right )} \, dx}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\left (2 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (-1+6 x^8\right )}{\sqrt {1+2 x^8} \left (1-x^4+4 x^8+4 x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\left (2 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \left (\frac {-4+3 x^2}{2 \sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )}+\frac {4+3 x^2}{2 \sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\sqrt {x+2 x^5} \operatorname {Subst}\left (\int \frac {-4+3 x^2}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\sqrt {x+2 x^5} \operatorname {Subst}\left (\int \frac {4+3 x^2}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\sqrt {x+2 x^5} \operatorname {Subst}\left (\int \left (-\frac {4}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )}+\frac {3 x^2}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\sqrt {x+2 x^5} \operatorname {Subst}\left (\int \left (\frac {4}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )}+\frac {3 x^2}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ &=\frac {\left (3 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\left (3 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}-\frac {\left (4 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+2 x^8} \left (1-x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}+\frac {\left (4 \sqrt {x+2 x^5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+2 x^8} \left (1+x^2+2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x^4}}\\ \end {align*}
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Mathematica [F] time = 0.57, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1+6 x^4\right ) \sqrt {x+2 x^5}}{\left (1+2 x^4\right ) \left (1-x^2+4 x^4+4 x^8\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 0.52, size = 47, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {\sqrt {2 x^5+x}}{2 x^4+1}\right )-\tanh ^{-1}\left (\frac {\sqrt {2 x^5+x}}{2 x^4+1}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.43, size = 60, normalized size = 1.28 \begin {gather*} -\frac {1}{2} \, \arctan \left (\frac {2 \, x^{4} - x + 1}{2 \, \sqrt {2 \, x^{5} + x}}\right ) + \frac {1}{2} \, \log \left (\frac {2 \, x^{4} + x - 2 \, \sqrt {2 \, x^{5} + x} + 1}{2 \, x^{4} - x + 1}\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 {2 \, x^{5} + x} {\left (6 \, x^{4} - 1\right )}}{{\left (4 \, x^{8} + 4 \, x^{4} - x^{2} + 1\right )} {\left (2 \, x^{4} + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.98, size = 97, normalized size = 2.06 \begin {gather*} \frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {2 x^{5}+x}+\RootOf \left (\textit {\_Z}^{2}+1\right )}{2 x^{4}+x +1}\right )}{2}-\frac {\ln \left (-\frac {2 x^{4}+2 \sqrt {2 x^{5}+x}+x +1}{2 x^{4}-x +1}\right )}{2} \end {gather*}
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 {2 \, x^{5} + x} {\left (6 \, x^{4} - 1\right )}}{{\left (4 \, x^{8} + 4 \, x^{4} - x^{2} + 1\right )} {\left (2 \, x^{4} + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.34, size = 72, normalized size = 1.53 \begin {gather*} \frac {\ln \left (\frac {x}{2}-\sqrt {2\,x^5+x}+x^4+\frac {1}{2}\right )}{2}-\frac {\ln \left (2\,x^4-x+1\right )}{2}+\frac {\ln \left (x^4-\frac {x}{2}+\frac {1}{2}-\sqrt {2\,x^5+x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {\ln \left (2\,x^4+x+1\right )\,1{}\mathrm {i}}{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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