Optimal. Leaf size=34 \[ \frac {x^2}{3+\log ^2\left (2 x+\frac {x}{4-5 \left (\frac {e^5}{x}-x\right )}\right )} \]
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Rubi [F] time = 23.84, 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 {300 e^{10} x+216 x^3+510 x^4+300 x^5+e^5 \left (-510 x^2-600 x^3\right )+\left (-100 e^{10} x-72 x^3-160 x^4-100 x^5+e^5 \left (180 x^2+200 x^3\right )\right ) \log \left (\frac {10 e^5 x-9 x^2-10 x^3}{5 e^5-4 x-5 x^2}\right )+\left (100 e^{10} x+72 x^3+170 x^4+100 x^5+e^5 \left (-170 x^2-200 x^3\right )\right ) \log ^2\left (\frac {10 e^5 x-9 x^2-10 x^3}{5 e^5-4 x-5 x^2}\right )}{450 e^{10}+324 x^2+765 x^3+450 x^4+e^5 \left (-765 x-900 x^2\right )+\left (300 e^{10}+216 x^2+510 x^3+300 x^4+e^5 \left (-510 x-600 x^2\right )\right ) \log ^2\left (\frac {10 e^5 x-9 x^2-10 x^3}{5 e^5-4 x-5 x^2}\right )+\left (50 e^{10}+36 x^2+85 x^3+50 x^4+e^5 \left (-85 x-100 x^2\right )\right ) \log ^4\left (\frac {10 e^5 x-9 x^2-10 x^3}{5 e^5-4 x-5 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} &=\int \frac {2 x \left (3 \left (50 e^{10}-5 e^5 x (17+20 x)+x^2 \left (36+85 x+50 x^2\right )\right )-2 \left (25 e^{10}-5 e^5 x (9+10 x)+x^2 \left (18+40 x+25 x^2\right )\right ) \log \left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )+\left (50 e^{10}-5 e^5 x (17+20 x)+x^2 \left (36+85 x+50 x^2\right )\right ) \log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )}{\left (50 e^{10}-85 e^5 x+4 \left (9-25 e^5\right ) x^2+85 x^3+50 x^4\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx\\ &=2 \int \frac {x \left (3 \left (50 e^{10}-5 e^5 x (17+20 x)+x^2 \left (36+85 x+50 x^2\right )\right )-2 \left (25 e^{10}-5 e^5 x (9+10 x)+x^2 \left (18+40 x+25 x^2\right )\right ) \log \left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )+\left (50 e^{10}-5 e^5 x (17+20 x)+x^2 \left (36+85 x+50 x^2\right )\right ) \log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )}{\left (50 e^{10}-85 e^5 x+4 \left (9-25 e^5\right ) x^2+85 x^3+50 x^4\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx\\ &=2 \int \left (\frac {2 x \left (-25 e^{10}+45 e^5 x-2 \left (9-25 e^5\right ) x^2-40 x^3-25 x^4\right ) \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (10 e^5-9 x-10 x^2\right ) \left (5 e^5-4 x-5 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2}+\frac {x}{3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )}\right ) \, dx\\ &=2 \int \frac {x}{3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )} \, dx+4 \int \frac {x \left (-25 e^{10}+45 e^5 x-2 \left (9-25 e^5\right ) x^2-40 x^3-25 x^4\right ) \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (10 e^5-9 x-10 x^2\right ) \left (5 e^5-4 x-5 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx\\ &=2 \int \frac {x}{3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )} \, dx+4 \int \left (\frac {\log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{20 \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2}-\frac {x \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{2 \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2}+\frac {\left (-90 e^5+\left (81+200 e^5\right ) x\right ) \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{20 \left (10 e^5-9 x-10 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2}+\frac {\left (10 e^5-\left (8+25 e^5\right ) x\right ) \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{5 \left (5 e^5-4 x-5 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {\log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx+\frac {1}{5} \int \frac {\left (-90 e^5+\left (81+200 e^5\right ) x\right ) \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (10 e^5-9 x-10 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx+\frac {4}{5} \int \frac {\left (10 e^5-\left (8+25 e^5\right ) x\right ) \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (5 e^5-4 x-5 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx-2 \int \frac {x \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx+2 \int \frac {x}{3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )} \, dx\\ &=\frac {1}{5} \int \frac {\log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx+\frac {1}{5} \int \left (\frac {\left (81+200 e^5\right ) x \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (10 e^5-9 x-10 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2}+\frac {90 e^5 \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (-10 e^5+9 x+10 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2}\right ) \, dx+\frac {4}{5} \int \left (\frac {10 e^5 \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (5 e^5-4 x-5 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2}+\frac {\left (-8-25 e^5\right ) x \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (5 e^5-4 x-5 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2}\right ) \, dx-2 \int \frac {x \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx+2 \int \frac {x}{3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )} \, dx\\ &=\frac {1}{5} \int \frac {\log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx-2 \int \frac {x \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx+2 \int \frac {x}{3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )} \, dx+\left (8 e^5\right ) \int \frac {\log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (5 e^5-4 x-5 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx+\left (18 e^5\right ) \int \frac {\log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (-10 e^5+9 x+10 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx-\frac {1}{5} \left (4 \left (8+25 e^5\right )\right ) \int \frac {x \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (5 e^5-4 x-5 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx+\frac {1}{5} \left (81+200 e^5\right ) \int \frac {x \log \left (\frac {x \left (-10 e^5+9 x+10 x^2\right )}{-5 e^5+4 x+5 x^2}\right )}{\left (10 e^5-9 x-10 x^2\right ) \left (3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 41, normalized size = 1.21 \begin {gather*} \frac {x^2}{3+\log ^2\left (\frac {x \left (-10 e^5+x (9+10 x)\right )}{-5 e^5+x (4+5 x)}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.89, size = 43, normalized size = 1.26 \begin {gather*} \frac {x^{2}}{\log \left (\frac {10 \, x^{3} + 9 \, x^{2} - 10 \, x e^{5}}{5 \, x^{2} + 4 \, x - 5 \, e^{5}}\right )^{2} + 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 9.17, size = 44, normalized size = 1.29 \begin {gather*} \frac {2 \, x^{2}}{\log \left (\frac {10 \, x^{3} + 9 \, x^{2} - 10 \, x e^{5}}{5 \, x^{2} + 4 \, x - 5 \, e^{5}}\right )^{2} + 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 44, normalized size = 1.29
method | result | size |
risch | \(\frac {x^{2}}{\ln \left (\frac {10 x \,{\mathrm e}^{5}-10 x^{3}-9 x^{2}}{5 \,{\mathrm e}^{5}-5 x^{2}-4 x}\right )^{2}+3}\) | \(44\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.51, size = 104, normalized size = 3.06 \begin {gather*} -\frac {x^{2}}{2 \, {\left (\log \left (5 \, x^{2} + 4 \, x - 5 \, e^{5}\right ) - \log \relax (x)\right )} \log \left (10 \, x^{2} + 9 \, x - 10 \, e^{5}\right ) - \log \left (10 \, x^{2} + 9 \, x - 10 \, e^{5}\right )^{2} - \log \left (5 \, x^{2} + 4 \, x - 5 \, e^{5}\right )^{2} + 2 \, \log \left (5 \, x^{2} + 4 \, x - 5 \, e^{5}\right ) \log \relax (x) - \log \relax (x)^{2} - 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.56, size = 43, normalized size = 1.26 \begin {gather*} \frac {x^2}{{\ln \left (\frac {10\,x^3+9\,x^2-10\,{\mathrm {e}}^5\,x}{5\,x^2+4\,x-5\,{\mathrm {e}}^5}\right )}^2+3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.61, size = 37, normalized size = 1.09 \begin {gather*} \frac {x^{2}}{\log {\left (\frac {- 10 x^{3} - 9 x^{2} + 10 x e^{5}}{- 5 x^{2} - 4 x + 5 e^{5}} \right )}^{2} + 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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