Optimal. Leaf size=29 \[ \frac {\log (16)}{\log (4) \log ^2\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \]
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Rubi [F] time = 0.79, 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 {\left (16-40 x-4 x^2\right ) \log (16)}{\left (-5-6 x+4 x^2+x^3\right ) \log (4) \log ^3\left (\frac {1+2 x-x^2-2 x^3+x^4}{50+20 x+2 x^2}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\log (16) \int \frac {16-40 x-4 x^2}{\left (-5-6 x+4 x^2+x^3\right ) \log ^3\left (\frac {1+2 x-x^2-2 x^3+x^4}{50+20 x+2 x^2}\right )} \, dx}{\log (4)}\\ &=\frac {\log (16) \int \left (\frac {4}{(5+x) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )}-\frac {4 (-1+2 x)}{\left (-1-x+x^2\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )}\right ) \, dx}{\log (4)}\\ &=\frac {(4 \log (16)) \int \frac {1}{(5+x) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\log (4)}-\frac {(4 \log (16)) \int \frac {-1+2 x}{\left (-1-x+x^2\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\log (4)}\\ &=-\frac {(4 \log (16)) \int \left (-\frac {1}{\left (-1-x+x^2\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )}+\frac {2 x}{\left (-1-x+x^2\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )}\right ) \, dx}{\log (4)}+\frac {(4 \log (16)) \int \frac {1}{(5+x) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\log (4)}\\ &=\frac {(4 \log (16)) \int \frac {1}{(5+x) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\log (4)}+\frac {(4 \log (16)) \int \frac {1}{\left (-1-x+x^2\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\log (4)}-\frac {(8 \log (16)) \int \frac {x}{\left (-1-x+x^2\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\log (4)}\\ &=\frac {(4 \log (16)) \int \left (-\frac {2}{\sqrt {5} \left (1+\sqrt {5}-2 x\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )}-\frac {2}{\sqrt {5} \left (-1+\sqrt {5}+2 x\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )}\right ) \, dx}{\log (4)}+\frac {(4 \log (16)) \int \frac {1}{(5+x) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\log (4)}-\frac {(8 \log (16)) \int \left (\frac {1+\frac {1}{\sqrt {5}}}{\left (-1-\sqrt {5}+2 x\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )}+\frac {1-\frac {1}{\sqrt {5}}}{\left (-1+\sqrt {5}+2 x\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )}\right ) \, dx}{\log (4)}\\ &=\frac {(4 \log (16)) \int \frac {1}{(5+x) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\log (4)}-\frac {(8 \log (16)) \int \frac {1}{\left (1+\sqrt {5}-2 x\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\sqrt {5} \log (4)}-\frac {(8 \log (16)) \int \frac {1}{\left (-1+\sqrt {5}+2 x\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{\sqrt {5} \log (4)}-\frac {\left (8 \left (5-\sqrt {5}\right ) \log (16)\right ) \int \frac {1}{\left (-1+\sqrt {5}+2 x\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{5 \log (4)}-\frac {\left (8 \left (5+\sqrt {5}\right ) \log (16)\right ) \int \frac {1}{\left (-1-\sqrt {5}+2 x\right ) \log ^3\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \, dx}{5 \log (4)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.26, size = 29, normalized size = 1.00 \begin {gather*} \frac {\log (16)}{\log (4) \log ^2\left (\frac {\left (1+x-x^2\right )^2}{2 (5+x)^2}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 35, normalized size = 1.21 \begin {gather*} \frac {2}{\log \left (\frac {x^{4} - 2 \, x^{3} - x^{2} + 2 \, x + 1}{2 \, {\left (x^{2} + 10 \, x + 25\right )}}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 121, normalized size = 4.17 \begin {gather*} \frac {2 \, {\left (x^{2} + 10 \, x - 4\right )}}{x^{2} \log \left (\frac {x^{4} - 2 \, x^{3} - x^{2} + 2 \, x + 1}{2 \, {\left (x^{2} + 10 \, x + 25\right )}}\right )^{2} + 10 \, x \log \left (\frac {x^{4} - 2 \, x^{3} - x^{2} + 2 \, x + 1}{2 \, {\left (x^{2} + 10 \, x + 25\right )}}\right )^{2} - 4 \, \log \left (\frac {x^{4} - 2 \, x^{3} - x^{2} + 2 \, x + 1}{2 \, {\left (x^{2} + 10 \, x + 25\right )}}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 37, normalized size = 1.28
method | result | size |
norman | \(\frac {2}{\ln \left (\frac {x^{4}-2 x^{3}-x^{2}+2 x +1}{2 x^{2}+20 x +50}\right )^{2}}\) | \(37\) |
risch | \(\frac {2}{\ln \left (\frac {x^{4}-2 x^{3}-x^{2}+2 x +1}{2 x^{2}+20 x +50}\right )^{2}}\) | \(37\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.49, size = 58, normalized size = 2.00 \begin {gather*} \frac {2}{\log \relax (2)^{2} - 4 \, {\left (\log \relax (2) + 2 \, \log \left (x + 5\right )\right )} \log \left (x^{2} - x - 1\right ) + 4 \, \log \left (x^{2} - x - 1\right )^{2} + 4 \, \log \relax (2) \log \left (x + 5\right ) + 4 \, \log \left (x + 5\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.68, size = 36, normalized size = 1.24 \begin {gather*} \frac {2}{{\ln \left (\frac {x^4-2\,x^3-x^2+2\,x+1}{2\,x^2+20\,x+50}\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.21, size = 31, normalized size = 1.07 \begin {gather*} \frac {2}{\log {\left (\frac {x^{4} - 2 x^{3} - x^{2} + 2 x + 1}{2 x^{2} + 20 x + 50} \right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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