Optimal. Leaf size=25 \[ 4 \left (x+\frac {1}{4} \left (-x+\left (4-8 x^2\right )^2\right )\right )^{\frac {1}{x^4}} \]
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Rubi [F] time = 3.20, 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 {4^{-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}} \left (12 x-512 x^2+1024 x^4+\left (-256-48 x+1024 x^2-1024 x^4\right ) \log \left (\frac {1}{4} \left (16+3 x-64 x^2+64 x^4\right )\right )\right )}{16 x^5+3 x^6-64 x^7+64 x^9} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4^{-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{-1+\frac {1}{x^4}} \left (12 x-512 x^2+1024 x^4+\left (-256-48 x+1024 x^2-1024 x^4\right ) \log \left (\frac {1}{4} \left (16+3 x-64 x^2+64 x^4\right )\right )\right )}{x^5} \, dx\\ &=\int \left (\frac {4^{1-\frac {1}{x^4}} \left (3-128 x+256 x^3\right ) \left (16+3 x-64 x^2+64 x^4\right )^{-1+\frac {1}{x^4}}}{x^4}-\frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}} \log \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )}{x^5}\right ) \, dx\\ &=\int \frac {4^{1-\frac {1}{x^4}} \left (3-128 x+256 x^3\right ) \left (16+3 x-64 x^2+64 x^4\right )^{-1+\frac {1}{x^4}}}{x^4} \, dx-\int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}} \log \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )}{x^5} \, dx\\ &=-\left (\log \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right ) \int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx\right )+\int \frac {\left (3-128 x+256 x^3\right ) \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )^{-1+\frac {1}{x^4}}}{x^4} \, dx+\int \frac {\left (3-128 x+256 x^3\right ) \int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx}{16+3 x-64 x^2+64 x^4} \, dx\\ &=-\left (\log \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right ) \int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx\right )+\int \left (\frac {3 \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )^{-1+\frac {1}{x^4}}}{x^4}-\frac {128 \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )^{-1+\frac {1}{x^4}}}{x^3}+\frac {256 \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )^{-1+\frac {1}{x^4}}}{x}\right ) \, dx+\int \left (\frac {3 \int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx}{16+3 x-64 x^2+64 x^4}-\frac {128 x \int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx}{16+3 x-64 x^2+64 x^4}+\frac {256 x^3 \int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx}{16+3 x-64 x^2+64 x^4}\right ) \, dx\\ &=3 \int \frac {\left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )^{-1+\frac {1}{x^4}}}{x^4} \, dx+3 \int \frac {\int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx}{16+3 x-64 x^2+64 x^4} \, dx-128 \int \frac {\left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )^{-1+\frac {1}{x^4}}}{x^3} \, dx-128 \int \frac {x \int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx}{16+3 x-64 x^2+64 x^4} \, dx+256 \int \frac {\left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )^{-1+\frac {1}{x^4}}}{x} \, dx+256 \int \frac {x^3 \int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx}{16+3 x-64 x^2+64 x^4} \, dx-\log \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right ) \int \frac {4^{2-\frac {1}{x^4}} \left (16+3 x-64 x^2+64 x^4\right )^{\frac {1}{x^4}}}{x^5} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.28, size = 23, normalized size = 0.92 \begin {gather*} 4 \left (4+\frac {3 x}{4}-16 x^2+16 x^4\right )^{\frac {1}{x^4}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 21, normalized size = 0.84 \begin {gather*} 4 \, {\left (16 \, x^{4} - 16 \, x^{2} + \frac {3}{4} \, x + 4\right )}^{\left (\frac {1}{x^{4}}\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 {4 \, {\left (256 \, x^{4} - 128 \, x^{2} - 4 \, {\left (64 \, x^{4} - 64 \, x^{2} + 3 \, x + 16\right )} \log \left (16 \, x^{4} - 16 \, x^{2} + \frac {3}{4} \, x + 4\right ) + 3 \, x\right )} {\left (16 \, x^{4} - 16 \, x^{2} + \frac {3}{4} \, x + 4\right )}^{\left (\frac {1}{x^{4}}\right )}}{64 \, x^{9} - 64 \, x^{7} + 3 \, x^{6} + 16 \, x^{5}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 22, normalized size = 0.88
method | result | size |
risch | \(4 \left (16 x^{4}-16 x^{2}+\frac {3}{4} x +4\right )^{\frac {1}{x^{4}}}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 31, normalized size = 1.24 \begin {gather*} 4 \, e^{\left (-\frac {2 \, \log \relax (2)}{x^{4}} + \frac {\log \left (64 \, x^{4} - 64 \, x^{2} + 3 \, x + 16\right )}{x^{4}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.42, size = 21, normalized size = 0.84 \begin {gather*} 4\,{\left (16\,x^4-16\,x^2+\frac {3\,x}{4}+4\right )}^{\frac {1}{x^4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.72, size = 24, normalized size = 0.96 \begin {gather*} 4 e^{\frac {\log {\left (16 x^{4} - 16 x^{2} + \frac {3 x}{4} + 4 \right )}}{x^{4}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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