Optimal. Leaf size=119 \[ -\frac {x}{4 \sqrt {x^4+1}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}} \]
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Rubi [C] time = 1.05, antiderivative size = 404, normalized size of antiderivative = 3.39, number of steps used = 37, number of rules used = 16, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {6725, 220, 2073, 1222, 1179, 1196, 1211, 1699, 207, 1198, 203, 21, 1429, 409, 1217, 1707} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}}-\frac {x \left (1-x^2\right )}{8 \sqrt {x^4+1}}-\frac {x \left (x^2+1\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 )}{16 \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 )}{16 \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 )}{16 \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 )}{16 \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 )}{4 \sqrt {x^4+1}} \end {gather*}
Antiderivative was successfully verified.
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Rule 21
Rule 203
Rule 207
Rule 220
Rule 409
Rule 1179
Rule 1196
Rule 1198
Rule 1211
Rule 1217
Rule 1222
Rule 1429
Rule 1699
Rule 1707
Rule 2073
Rule 6725
Rubi steps
\begin {align*} \int \frac {1+x^{16}}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx &=\int \left (\frac {1}{\sqrt {1+x^4}}+\frac {2}{\sqrt {1+x^4} \left (-1+x^{16}\right )}\right ) \, dx\\ &=2 \int \frac {1}{\sqrt {1+x^4} \left (-1+x^{16}\right )} \, dx+\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}}+2 \int \left (\frac {1}{4 \left (-1+x^2\right ) \left (1+x^4\right )^{3/2}}-\frac {1}{4 \left (1+x^2\right ) \left (1+x^4\right )^{3/2}}+\frac {-1-x^4}{2 \left (1+x^4\right )^{3/2} \left (1+x^8\right )}\right ) \, 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+x^2\right ) \left (1+x^4\right )^{3/2}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+x^2\right ) \left (1+x^4\right )^{3/2}} \, dx+\int \frac {-1-x^4}{\left (1+x^4\right )^{3/2} \left (1+x^8\right )} \, 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}{4} \int \frac {-1-x^2}{\left (1+x^4\right )^{3/2}} \, dx-\frac {1}{4} \int \frac {1-x^2}{\left (1+x^4\right )^{3/2}} \, dx+\frac {1}{4} \int \frac {1}{\left (-1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-\int \frac {1}{\sqrt {1+x^4} \left (1+x^8\right )} \, dx\\ &=-\frac {x \left (1-x^2\right )}{8 \sqrt {1+x^4}}-\frac {x \left (1+x^2\right )}{8 \sqrt {1+x^4}}+\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} i \int \frac {1}{\left (i-x^4\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} i \int \frac {1}{\left (i+x^4\right ) \sqrt {1+x^4}} \, dx-2 \left (\frac {1}{8} \int \frac {1}{\sqrt {1+x^4}} \, dx\right )+\frac {1}{8} \int \frac {-1-x^2}{\sqrt {1+x^4}} \, dx-\frac {1}{8} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx-\frac {1}{8} \int \frac {-1-x^2}{\left (-1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{8} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {x \left (1-x^2\right )}{8 \sqrt {1+x^4}}-\frac {x \left (1+x^2\right )}{8 \sqrt {1+x^4}}+\frac {x \sqrt {1+x^4}}{8 \left (1+x^2\right )}-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {1+x^4}}+\frac {3 \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 {1}{8} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{-1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )-\frac {1}{4} \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1-\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1+\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1-(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1}{\left (1+(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {x \left (1-x^2\right )}{8 \sqrt {1+x^4}}-\frac {x \left (1+x^2\right )}{8 \sqrt {1+x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {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 )}{4 \sqrt {1+x^4}}-\left (\left (\frac {1}{8}+\frac {i}{8}\right ) \left (1+\sqrt [4]{-1}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{8}-\frac {i}{8}\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}{8}-\frac {i}{8}\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}{8}-\frac {i}{8}\right ) \left (1+(-1)^{3/4}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{8}+\frac {i}{8}\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}{8} \left ((1+i)-i \sqrt {2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{8} \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}{8} \left (i \left ((1+i)+\sqrt {2}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x \left (1-x^2\right )}{8 \sqrt {1+x^4}}-\frac {x \left (1+x^2\right )}{8 \sqrt {1+x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{8 \sqrt {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 )}{4 \sqrt {1+x^4}}-\frac {\left (\frac {1}{16}+\frac {i}{16}\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 ((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 )}{16 \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 )}{16 \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 )}{16 \sqrt {1+x^4}}\\ \end {align*}
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Mathematica [C] time = 1.18, size = 227, normalized size = 1.91 \begin {gather*} \left (\frac {1}{8}+\frac {i}{8}\right ) \left (-\frac {(1-i) x}{\sqrt {x^4+1}}-3 \sqrt {2} F\left (\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (-i;\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (i;\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (-\frac {1+i}{\sqrt {2}};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (-\frac {1-i}{\sqrt {2}};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (\frac {1-i}{\sqrt {2}};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\sqrt {2} \Pi \left (\frac {1+i}{\sqrt {2}};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.62, size = 119, normalized size = 1.00 \begin {gather*} -\frac {x}{4 \sqrt {x^4+1}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{8 \sqrt {2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.59, size = 302, normalized size = 2.54 \begin {gather*} -\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \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 ) + 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \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 ) - 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \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 ) + 2 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) - \sqrt {2} {\left (x^{4} + 1\right )} \log \left (\frac {x^{4} - 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) + 8 \, \sqrt {x^{4} + 1} x}{32 \, {\left (x^{4} + 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 {x^{16} + 1}{{\left (x^{16} - 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.04, size = 272, normalized size = 2.29 \begin {gather*} \frac {3 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{4 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {x}{4 \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{4 \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{4 \sqrt {x^{4}+1}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{8}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{6}+x^{2}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}\, \sqrt {x^{4}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}}-\frac {2 \left (-1\right )^{\frac {3}{4}} \underline {\hspace {1.25 ex}}\alpha ^{7} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \underline {\hspace {1.25 ex}}\alpha ^{6}, i\right )}{\sqrt {x^{4}+1}}\right )\right )}{16} \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 {x^{16} + 1}{{\left (x^{16} - 1\right )} \sqrt {x^{4} + 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 = -0.01 \begin {gather*} \int \frac {x^{16}+1}{\sqrt {x^4+1}\,\left (x^{16}-1\right )} \,d x \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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