Optimal. Leaf size=53 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}} \]
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Rubi [C] time = 0.65, antiderivative size = 314, normalized size of antiderivative = 5.92, number of steps used = 20, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6725, 406, 220, 409, 1217, 1707} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}+\frac {i \left (\sqrt {2}+(1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {x^4+1}}+\frac {i \left (\sqrt {2}+(-1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {x^4+1}}+\frac {\left ((-1-i)-i \sqrt {2}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {x^4+1}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \left (1+(-1)^{3/4}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {x^4+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {x^4+1}} \end {gather*}
Antiderivative was successfully verified.
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Rule 220
Rule 406
Rule 409
Rule 1217
Rule 1707
Rule 6725
Rubi steps
\begin {align*} \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+x^8} \, dx &=\int \left (-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {1+x^4}}{i-x^4}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {1+x^4}}{i+x^4}\right ) \, dx\\ &=\left (-\frac {1}{2}-\frac {i}{2}\right ) \int \frac {\sqrt {1+x^4}}{i-x^4} \, dx+\left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {\sqrt {1+x^4}}{i+x^4} \, dx\\ &=-\left (i \int \frac {1}{\left (i-x^4\right ) \sqrt {1+x^4}} \, dx\right )-i \int \frac {1}{\left (i+x^4\right ) \sqrt {1+x^4}} \, dx+\left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx+\left (\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {1}{2} \int \frac {1}{\left (1-\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1-(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (1+\sqrt [4]{-1}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{4}-\frac {i}{4}\right ) \sqrt [4]{-1} \left (1+\sqrt [4]{-1}\right )\right ) \int \frac {1+x^2}{\left (1+\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\left (\left (\frac {1}{4}-\frac {i}{4}\right ) (-1)^{3/4} \left (1-(-1)^{3/4}\right )\right ) \int \frac {1+x^2}{\left (1-(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\left (\left (\frac {1}{4}-\frac {i}{4}\right ) \left (1+(-1)^{3/4}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{4}+\frac {i}{4}\right ) (-1)^{3/4} \left (1+(-1)^{3/4}\right )\right ) \int \frac {1+x^2}{\left (1+(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \left ((1+i)-i \sqrt {2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{4} \left (i \left ((-1-i)+\sqrt {2}\right )\right ) \int \frac {1+x^2}{\left (1-\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{4} \left (i \left ((1+i)+\sqrt {2}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (1+\sqrt [4]{-1}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {1+x^4}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \left (1+(-1)^{3/4}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {1+x^4}}-\frac {\left ((1+i)-i \sqrt {2}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {1+x^4}}+\frac {i \left ((1+i)+\sqrt {2}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {1+x^4}}\\ \end {align*}
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Mathematica [C] time = 0.44, size = 106, normalized size = 2.00 \begin {gather*} \frac {1}{2} \sqrt [4]{-1} \left (-2 F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (-\sqrt [4]{-1};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (\sqrt [4]{-1};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (-(-1)^{3/4};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left ((-1)^{3/4};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.28, size = 53, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.56, size = 199, normalized size = 3.75 \begin {gather*} -\frac {1}{4} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )} + 2^{\frac {1}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )}\right )} + 4 \, \sqrt {x^{4} + 1} {\left (2^{\frac {3}{4}} x^{3} + 2^{\frac {1}{4}} {\left (x^{5} + x\right )}\right )}}{2 \, {\left (x^{8} + 1\right )}}\right ) - \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} + 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 1}\right ) + \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} - 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 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 {x^{4} + 1} {\left (x^{4} - 1\right )}}{x^{8} + 1}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 74, normalized size = 1.40 \begin {gather*} \frac {2^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right )}{4}-\frac {2^{\frac {3}{4}} \ln \left (\frac {\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}+\frac {2^{\frac {3}{4}}}{2}}{\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}-\frac {2^{\frac {3}{4}}}{2}}\right )}{8} \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 {x^{4} + 1} {\left (x^{4} - 1\right )}}{x^{8} + 1}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\left (x^4-1\right )\,\sqrt {x^4+1}}{x^8+1} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \sqrt {x^{4} + 1}}{x^{8} + 1}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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