Optimal. Leaf size=27 \[ 2 x^2+\frac {1}{4+5 \left (\frac {3}{2 x}-x\right )+\log (5 x)} \]
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Rubi [F] time = 0.68, 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 {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (15+448 x+490 x^2-472 x^3-320 x^4+200 x^5+8 x^2 \left (15+8 x-10 x^2\right ) \log (5 x)+8 x^3 \log ^2(5 x)\right )}{\left (15+8 x-10 x^2+2 x \log (5 x)\right )^2} \, dx\\ &=2 \int \frac {15+448 x+490 x^2-472 x^3-320 x^4+200 x^5+8 x^2 \left (15+8 x-10 x^2\right ) \log (5 x)+8 x^3 \log ^2(5 x)}{\left (15+8 x-10 x^2+2 x \log (5 x)\right )^2} \, dx\\ &=2 \int \left (2 x+\frac {15-2 x+10 x^2}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2}\right ) \, dx\\ &=2 x^2+2 \int \frac {15-2 x+10 x^2}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2} \, dx\\ &=2 x^2+2 \int \left (\frac {15}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2}-\frac {2 x}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2}+\frac {10 x^2}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2}\right ) \, dx\\ &=2 x^2-4 \int \frac {x}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2} \, dx+20 \int \frac {x^2}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2} \, dx+30 \int \frac {1}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.30, size = 27, normalized size = 1.00 \begin {gather*} 2 \left (x^2+\frac {x}{15+8 x-10 x^2+2 x \log (5 x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 49, normalized size = 1.81 \begin {gather*} \frac {2 \, {\left (10 \, x^{4} - 2 \, x^{3} \log \left (5 \, x\right ) - 8 \, x^{3} - 15 \, x^{2} - x\right )}}{10 \, x^{2} - 2 \, x \log \left (5 \, x\right ) - 8 \, x - 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 28, normalized size = 1.04 \begin {gather*} 2 \, x^{2} - \frac {2 \, x}{10 \, x^{2} - 2 \, x \log \left (5 \, x\right ) - 8 \, x - 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 29, normalized size = 1.07
method | result | size |
risch | \(2 x^{2}-\frac {2 x}{10 x^{2}-2 x \ln \left (5 x \right )-8 x -15}\) | \(29\) |
norman | \(\frac {-30 x^{2}-2 x -16 x^{3}+20 x^{4}-4 x^{3} \ln \left (5 x \right )}{10 x^{2}-2 x \ln \left (5 x \right )-8 x -15}\) | \(49\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.61, size = 53, normalized size = 1.96 \begin {gather*} \frac {2 \, {\left (10 \, x^{4} - 2 \, x^{3} {\left (\log \relax (5) + 4\right )} - 2 \, x^{3} \log \relax (x) - 15 \, x^{2} - x\right )}}{10 \, x^{2} - 2 \, x {\left (\log \relax (5) + 4\right )} - 2 \, x \log \relax (x) - 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.43, size = 28, normalized size = 1.04 \begin {gather*} \frac {2\,x}{8\,x+2\,x\,\ln \left (5\,x\right )-10\,x^2+15}+2\,x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.20, size = 26, normalized size = 0.96 \begin {gather*} 2 x^{2} + \frac {2 x}{- 10 x^{2} + 2 x \log {\left (5 x \right )} + 8 x + 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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