Optimal. Leaf size=60 \[ \frac {\sqrt {1-2 x^8} \left (2 x^8-3 x^4-1\right )}{6 x^6}-\frac {1}{2} \tanh ^{-1}\left (\frac {x^2 \sqrt {1-2 x^8}}{2 x^8-1}\right ) \]
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Rubi [C] time = 0.89, antiderivative size = 335, normalized size of antiderivative = 5.58, number of steps used = 47, number of rules used = 17, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.415, Rules used = {21, 6728, 275, 277, 195, 221, 279, 307, 1181, 424, 1491, 1209, 1177, 524, 248, 1213, 537} \begin {gather*} \frac {1}{15} \sqrt [4]{2} \left (3+5 \sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {\left (1+\sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2^{3/4}}+\frac {\left (1-\sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {1}{15} 2^{3/4} \left (5-3 \sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {6}{5} \sqrt [4]{2} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {\Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}-\frac {6}{5} \sqrt {1-2 x^8} x^6-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {2}{3} \sqrt {1-2 x^8} x^2-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}+\frac {1}{15} \left (5-3 x^4\right ) \sqrt {1-2 x^8} x^2+\frac {1}{15} \left (6 x^4+5\right ) \sqrt {1-2 x^8} x^2 \end {gather*}
Antiderivative was successfully verified.
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Rule 21
Rule 195
Rule 221
Rule 248
Rule 275
Rule 277
Rule 279
Rule 307
Rule 424
Rule 524
Rule 537
Rule 1177
Rule 1181
Rule 1209
Rule 1213
Rule 1491
Rule 6728
Rubi steps
\begin {align*} \int \frac {\sqrt {1-2 x^8} \left (-1+2 x^8\right ) \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx &=-\int \frac {\left (1-2 x^8\right )^{3/2} \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx\\ &=-\int \left (-\frac {\left (1-2 x^8\right )^{3/2}}{x^7}-\frac {\left (1-2 x^8\right )^{3/2}}{x^3}-\frac {x \left (1-2 x^8\right )^{3/2}}{1+x^4}+\frac {4 x \left (1-2 x^8\right )^{3/2}}{-1+2 x^4}\right ) \, dx\\ &=-\left (4 \int \frac {x \left (1-2 x^8\right )^{3/2}}{-1+2 x^4} \, dx\right )+\int \frac {\left (1-2 x^8\right )^{3/2}}{x^7} \, dx+\int \frac {\left (1-2 x^8\right )^{3/2}}{x^3} \, dx+\int \frac {x \left (1-2 x^8\right )^{3/2}}{1+x^4} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (1-2 x^4\right )^{3/2}}{x^4} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (1-2 x^4\right )^{3/2}}{x^2} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (1-2 x^4\right )^{3/2}}{1+x^2} \, dx,x,x^2\right )-2 \operatorname {Subst}\left (\int \frac {\left (1-2 x^4\right )^{3/2}}{-1+2 x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {1-2 x^4}}{1+x^2} \, dx,x,x^2\right )-\frac {1}{2} \operatorname {Subst}\left (\int \left (-2+2 x^2\right ) \sqrt {1-2 x^4} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \left (2+4 x^2\right ) \sqrt {1-2 x^4} \, dx,x,x^2\right )-2 \operatorname {Subst}\left (\int \sqrt {1-2 x^4} \, dx,x,x^2\right )-6 \operatorname {Subst}\left (\int x^2 \sqrt {1-2 x^4} \, dx,x,x^2\right )-\operatorname {Subst}\left (\int \frac {\sqrt {1-2 x^4}}{-1+2 x^2} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}-\frac {1}{30} \operatorname {Subst}\left (\int \frac {-20+12 x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )+\frac {1}{30} \operatorname {Subst}\left (\int \frac {20+24 x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {2+4 x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {1-2 x^4}} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {-2+2 x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\left (-1+2 x^2\right ) \sqrt {1-2 x^4}} \, dx,x,x^2\right )-\frac {4}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )-\frac {12}{5} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {\operatorname {Subst}\left (\int \frac {-20+12 x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{15 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {20+24 x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{15 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {2+4 x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{2 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \left (1+x^2\right ) \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{\sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {-2+2 x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{\sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \left (-1+2 x^2\right ) \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{\sqrt {2}}+\frac {1}{5} \left (6 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )-\frac {1}{5} \left (6 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1+\sqrt {2} x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}+\frac {6}{5} \sqrt [4]{2} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {\Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}-\frac {12}{5} \operatorname {Subst}\left (\int \frac {1+\sqrt {2} x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )+2 \frac {\operatorname {Subst}\left (\int \frac {\sqrt {\sqrt {2}+2 x^2}}{\sqrt {\sqrt {2}-2 x^2}} \, dx,x,x^2\right )}{\sqrt {2}}-\frac {1}{5} \sqrt {2} \operatorname {Subst}\left (\int \frac {\sqrt {\sqrt {2}+2 x^2}}{\sqrt {\sqrt {2}-2 x^2}} \, dx,x,x^2\right )+\frac {1}{5} \left (2 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {\sqrt {2}+2 x^2}}{\sqrt {\sqrt {2}-2 x^2}} \, dx,x,x^2\right )-\frac {1}{15} \left (2 \left (6-5 \sqrt {2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )+\frac {1}{2} \left (-2+\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )-\left (1+\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )+\frac {1}{15} \left (2 \left (3+5 \sqrt {2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}+\frac {6}{5} \sqrt [4]{2} E\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {6}{5} \sqrt [4]{2} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {\Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}-\frac {1}{5} \left (6 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {\sqrt {2}+2 x^2}}{\sqrt {\sqrt {2}-2 x^2}} \, dx,x,x^2\right )-\frac {1}{15} \left (2 \left (6-5 \sqrt {2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-4 x^4}} \, dx,x,x^2\right )+\frac {1}{2} \left (-2+\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-4 x^4}} \, dx,x,x^2\right )-\left (1+\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-4 x^4}} \, dx,x,x^2\right )+\frac {1}{15} \left (2 \left (3+5 \sqrt {2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-4 x^4}} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}+\frac {6}{5} \sqrt [4]{2} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {1}{15} 2^{3/4} \left (5-3 \sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\left (1-\sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}-\frac {\left (1+\sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2^{3/4}}+\frac {1}{15} \sqrt [4]{2} \left (3+5 \sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {\Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}\\ \end {align*}
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Mathematica [C] time = 0.30, size = 149, normalized size = 2.48 \begin {gather*} -\frac {6 \sqrt {1-2 x^8} x^8 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};2 x^8\right )+8 x^{16}-12 x^{12}-8 x^8+6 x^4-3\ 2^{3/4} \sqrt {1-2 x^8} x^6 \Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-3\ 2^{3/4} \sqrt {1-2 x^8} x^6 \Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+2}{12 x^6 \sqrt {1-2 x^8}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 22.25, size = 60, normalized size = 1.00 \begin {gather*} \frac {\sqrt {1-2 x^8} \left (2 x^8-3 x^4-1\right )}{6 x^6}-\frac {1}{2} \tanh ^{-1}\left (\frac {x^2 \sqrt {1-2 x^8}}{2 x^8-1}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 75, normalized size = 1.25 \begin {gather*} \frac {3 \, x^{6} \log \left (-\frac {2 \, x^{8} - x^{4} - 2 \, \sqrt {-2 \, x^{8} + 1} x^{2} - 1}{2 \, x^{8} + x^{4} - 1}\right ) + 2 \, {\left (2 \, x^{8} - 3 \, x^{4} - 1\right )} \sqrt {-2 \, x^{8} + 1}}{12 \, x^{6}} \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 {{\left (2 \, x^{8} + 1\right )} {\left (2 \, x^{8} - 1\right )} \sqrt {-2 \, x^{8} + 1}}{{\left (2 \, x^{8} + x^{4} - 1\right )} x^{7}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.54, size = 85, normalized size = 1.42 \begin {gather*} -\frac {4 x^{16}-6 x^{12}-4 x^{8}+3 x^{4}+1}{6 x^{6} \sqrt {-2 x^{8}+1}}-\frac {\ln \left (-\frac {2 x^{8}-x^{4}+2 \sqrt {-2 x^{8}+1}\, x^{2}-1}{\left (2 x^{4}-1\right ) \left (x^{4}+1\right )}\right )}{4} \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 {{\left (2 \, x^{8} + 1\right )} {\left (2 \, x^{8} - 1\right )} \sqrt {-2 \, x^{8} + 1}}{{\left (2 \, x^{8} + x^{4} - 1\right )} x^{7}}\,{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 (1-2\,x^8\right )}^{3/2}\,\left (2\,x^8+1\right )}{x^7\,\left (2\,x^8+x^4-1\right )} \,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 {\sqrt {1 - 2 x^{8}} \left (2 x^{8} - 1\right ) \left (2 x^{8} + 1\right )}{x^{7} \left (x^{4} + 1\right ) \left (2 x^{4} - 1\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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