Optimal. Leaf size=432 \[ \frac {\left (1+\sqrt [4]{-1}\right ) \text {ArcTan}\left (\frac {(-1)^{7/8} \sqrt {2+\sqrt {2}} \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}{(-1)^{3/4} \sqrt [4]{-1+2 a} x^2+\sqrt {1+a x^8}}\right )}{8 \sqrt [8]{-1+2 a}}-\frac {i \left (-i \sqrt {2}+\sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \text {ArcTan}\left (\frac {(-1)^{7/8} \left (-2+\sqrt {2}\right ) \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}{(-1)^{3/4} \sqrt {2-\sqrt {2}} \sqrt [4]{-1+2 a} x^2+\sqrt {2-\sqrt {2}} \sqrt {1+a x^8}}\right )}{16 \sqrt [8]{-1+2 a}}+\frac {\left (\sqrt {2}+i \sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {(-1)^{7/8} \sqrt [4]{-1+2 a} x^2-\sqrt [8]{-1} \sqrt {1+a x^8}}{\sqrt {2-\sqrt {2}} \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )}{16 \sqrt [8]{-1+2 a}}+\frac {\left (1+\sqrt [4]{-1}\right ) \tanh ^{-1}\left (\frac {(-1)^{7/8} \sqrt [4]{-1+2 a} x^2-\sqrt [8]{-1} \sqrt {1+a x^8}}{\sqrt {2+\sqrt {2}} \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )}{8 \sqrt [8]{-1+2 a}} \]
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Rubi [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in
optimal.
time = 0.27, antiderivative size = 160, normalized size of antiderivative = 0.37, number of steps
used = 4, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6860, 440}
\begin {gather*} \frac {a \left (1-\frac {2 a+1}{\sqrt {1-4 a^2}}\right ) x F_1\left (\frac {1}{8};-\frac {3}{4},1;\frac {9}{8};-a x^8,-\frac {2 a^2 x^8}{1-\sqrt {1-4 a^2}}\right )}{1-\sqrt {1-4 a^2}}+\frac {a \left (\frac {2 a+1}{\sqrt {1-4 a^2}}+1\right ) x F_1\left (\frac {1}{8};-\frac {3}{4},1;\frac {9}{8};-a x^8,-\frac {2 a^2 x^8}{\sqrt {1-4 a^2}+1}\right )}{\sqrt {1-4 a^2}+1} \end {gather*}
Antiderivative was successfully verified.
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Rule 440
Rule 6860
Rubi steps
\begin {align*} \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx &=\int \left (\frac {\left (a-\frac {a (1+2 a)}{\sqrt {1-4 a^2}}\right ) \left (1+a x^8\right )^{3/4}}{1-\sqrt {1-4 a^2}+2 a^2 x^8}+\frac {\left (a+\frac {a (1+2 a)}{\sqrt {1-4 a^2}}\right ) \left (1+a x^8\right )^{3/4}}{1+\sqrt {1-4 a^2}+2 a^2 x^8}\right ) \, dx\\ &=\left (a \left (1-\frac {1+2 a}{\sqrt {1-4 a^2}}\right )\right ) \int \frac {\left (1+a x^8\right )^{3/4}}{1-\sqrt {1-4 a^2}+2 a^2 x^8} \, dx+\left (a \left (1+\frac {1+2 a}{\sqrt {1-4 a^2}}\right )\right ) \int \frac {\left (1+a x^8\right )^{3/4}}{1+\sqrt {1-4 a^2}+2 a^2 x^8} \, dx\\ &=\frac {a \left (1-\frac {1+2 a}{\sqrt {1-4 a^2}}\right ) x F_1\left (\frac {1}{8};-\frac {3}{4},1;\frac {9}{8};-a x^8,-\frac {2 a^2 x^8}{1-\sqrt {1-4 a^2}}\right )}{1-\sqrt {1-4 a^2}}+\frac {a \left (1+\frac {1+2 a}{\sqrt {1-4 a^2}}\right ) x F_1\left (\frac {1}{8};-\frac {3}{4},1;\frac {9}{8};-a x^8,-\frac {2 a^2 x^8}{1+\sqrt {1-4 a^2}}\right )}{1+\sqrt {1-4 a^2}}\\ \end {align*}
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Mathematica [A]
time = 8.74, size = 385, normalized size = 0.89 \begin {gather*} \frac {-2 \left (i+(-1)^{3/4}\right ) \tanh ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (\left (-i+(-1)^{3/4}\right ) \sqrt [4]{-1+2 a} x^2+\left (1+(-1)^{3/4}\right ) \sqrt {1+a x^8}\right )}{\sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )+\sqrt {2} \left (i \left (i+\sqrt {3+2 \sqrt {2}}\right ) \text {ArcTan}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (-\left (\left (-i+(-1)^{3/4}\right ) \sqrt [4]{-1+2 a} x^2\right )+\left (1+(-1)^{3/4}\right ) \sqrt {1+a x^8}\right )}{\sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )+\left (1+i \sqrt {3-2 \sqrt {2}}\right ) \text {ArcTan}\left (\frac {2 \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}{\left ((1-i)+\sqrt {2}\right ) \sqrt [4]{-1+2 a} x^2-\left ((1+i)+\sqrt {2}\right ) \sqrt {1+a x^8}}\right )+\left (-1-i \sqrt {3-2 \sqrt {2}}\right ) \tanh ^{-1}\left (\frac {\left ((1-i)+\sqrt {2}\right ) \sqrt [4]{-1+2 a} x^2+\left ((1+i)+\sqrt {2}\right ) \sqrt {1+a x^8}}{2 \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )\right )}{16 \sqrt [8]{-1+2 a}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{8}-1\right ) \left (a \,x^{8}+1\right )^{\frac {3}{4}}}{a^{2} x^{16}+x^{8}+1}\, dx\]
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*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
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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Sympy [F(-1)] Timed out
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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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 = -0.00 \begin {gather*} \int \frac {\left (a\,x^8-1\right )\,{\left (a\,x^8+1\right )}^{3/4}}{a^2\,x^{16}+x^8+1} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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