Optimal. Leaf size=31 \[ 9+\frac {3 x (2+x)}{4+x^2+x \left (\frac {x}{2-e^x}+\log (5)\right )} \]
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Rubi [F] time = 25.72, 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 {96+96 x-36 x^2+12 x^2 \log (5)+e^x \left (-96-96 x+30 x^2-6 x^3-3 x^4-12 x^2 \log (5)\right )+e^{2 x} \left (24+24 x-6 x^2+3 x^2 \log (5)\right )}{64+48 x^2+9 x^4+\left (32 x+12 x^3\right ) \log (5)+4 x^2 \log ^2(5)+e^x \left (-64-40 x^2-6 x^4+\left (-32 x-10 x^3\right ) \log (5)-4 x^2 \log ^2(5)\right )+e^{2 x} \left (16+8 x^2+x^4+\left (8 x+2 x^3\right ) \log (5)+x^2 \log ^2(5)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {96+96 x+x^2 (-36+12 \log (5))+e^x \left (-96-96 x+30 x^2-6 x^3-3 x^4-12 x^2 \log (5)\right )+e^{2 x} \left (24+24 x-6 x^2+3 x^2 \log (5)\right )}{64+48 x^2+9 x^4+\left (32 x+12 x^3\right ) \log (5)+4 x^2 \log ^2(5)+e^x \left (-64-40 x^2-6 x^4+\left (-32 x-10 x^3\right ) \log (5)-4 x^2 \log ^2(5)\right )+e^{2 x} \left (16+8 x^2+x^4+\left (8 x+2 x^3\right ) \log (5)+x^2 \log ^2(5)\right )} \, dx\\ &=\int \frac {96+96 x+x^2 (-36+12 \log (5))+e^x \left (-96-96 x+30 x^2-6 x^3-3 x^4-12 x^2 \log (5)\right )+e^{2 x} \left (24+24 x-6 x^2+3 x^2 \log (5)\right )}{64+9 x^4+\left (32 x+12 x^3\right ) \log (5)+x^2 \left (48+4 \log ^2(5)\right )+e^x \left (-64-40 x^2-6 x^4+\left (-32 x-10 x^3\right ) \log (5)-4 x^2 \log ^2(5)\right )+e^{2 x} \left (16+8 x^2+x^4+\left (8 x+2 x^3\right ) \log (5)+x^2 \log ^2(5)\right )} \, dx\\ &=\int \frac {3 \left (4 \left (8+8 x+x^2 (-3+\log (5))\right )+e^{2 x} \left (8+8 x+x^2 (-2+\log (5))\right )-e^x \left (32+32 x+2 x^3+x^4+2 x^2 (-5+\log (25))\right )\right )}{\left (8+3 x^2-e^x \left (4+x^2+x \log (5)\right )+x \log (25)\right )^2} \, dx\\ &=3 \int \frac {4 \left (8+8 x+x^2 (-3+\log (5))\right )+e^{2 x} \left (8+8 x+x^2 (-2+\log (5))\right )-e^x \left (32+32 x+2 x^3+x^4+2 x^2 (-5+\log (25))\right )}{\left (8+3 x^2-e^x \left (4+x^2+x \log (5)\right )+x \log (25)\right )^2} \, dx\\ &=3 \int \left (\frac {8+8 x-x^2 (2-\log (5))}{\left (4+x^2+x \log (5)\right )^2}+\frac {x^2 \left (-24+6 x^2+x^4+x^3 (2+\log (5))-x (8+\log (25))\right )}{\left (4+x^2+x \log (5)\right )^2 \left (8-4 e^x+3 x^2-e^x x^2-e^x x \log (5)+x \log (25)\right )}+\frac {x^2 \left (-64 x-3 x^6-6 x^5 \left (1+\frac {5 \log (5)}{6}\right )+32 \log ^2(5) \left (1-\frac {\log ^2(25)}{4 \log ^2(5)}\right )-20 x^4 \left (1+\frac {1}{20} \log (5) (9+\log (25))\right )-32 x^3 \left (1+\frac {1}{32} \left (-8 \log ^2(5)+7 \log (25)+6 \log (5) \log (25)\right )\right )-16 x^2 \left (1+\frac {1}{16} \left (12 \log ^2(5)-4 \log ^3(5)+32 \log (25)+\log (5) \left (-32-6 \log (25)+\log ^2(25)\right )\right )\right )\right )}{\left (4+x^2+x \log (5)\right )^2 \left (8-4 e^x+3 x^2-e^x x^2-e^x x \log (5)+x \log (25)\right )^2}\right ) \, dx\\ &=3 \int \frac {8+8 x-x^2 (2-\log (5))}{\left (4+x^2+x \log (5)\right )^2} \, dx+3 \int \frac {x^2 \left (-24+6 x^2+x^4+x^3 (2+\log (5))-x (8+\log (25))\right )}{\left (4+x^2+x \log (5)\right )^2 \left (8-4 e^x+3 x^2-e^x x^2-e^x x \log (5)+x \log (25)\right )} \, dx+3 \int \frac {x^2 \left (-64 x-3 x^6-6 x^5 \left (1+\frac {5 \log (5)}{6}\right )+32 \log ^2(5) \left (1-\frac {\log ^2(25)}{4 \log ^2(5)}\right )-20 x^4 \left (1+\frac {1}{20} \log (5) (9+\log (25))\right )-32 x^3 \left (1+\frac {1}{32} \left (-8 \log ^2(5)+7 \log (25)+6 \log (5) \log (25)\right )\right )-16 x^2 \left (1+\frac {1}{16} \left (12 \log ^2(5)-4 \log ^3(5)+32 \log (25)+\log (5) \left (-32-6 \log (25)+\log ^2(25)\right )\right )\right )\right )}{\left (4+x^2+x \log (5)\right )^2 \left (8-4 e^x+3 x^2-e^x x^2-e^x x \log (5)+x \log (25)\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [B] time = 0.45, size = 207, normalized size = 6.68 \begin {gather*} \frac {3 \left (256+3 x^6+96 x (-2+\log (25))+x^5 (-12+9 \log (5)+\log (25))+4 x^2 \left (56+4 \log (5) (-3+\log (25))-5 \log (25)+\log ^2(25)\right )+x^4 (44-4 \log (25)+\log (5) (-13+6 \log (25)))+x^3 \left (-88-4 \log ^2(5)+20 \log (25)+\log (5) \left (60-3 \log (25)+\log ^2(25)\right )\right )-e^x (4+x (-2+\log (5))) \left (32+3 x^4+x^2 (20+\log (5) (-1+\log (25)))+x^3 \log (3125)+2 x (-4+\log (390625))\right )\right )}{\left (-8-3 x^2+e^x \left (4+x^2+x \log (5)\right )-x \log (25)\right ) \left (32+3 x^4+x^2 (20+\log (5) (-1+\log (25)))+x^3 \log (3125)+2 x (-4+\log (390625))\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.44, size = 55, normalized size = 1.77 \begin {gather*} -\frac {3 \, {\left (x^{2} - {\left (x \log \relax (5) - 2 \, x + 4\right )} e^{x} + 2 \, x \log \relax (5) - 4 \, x + 8\right )}}{3 \, x^{2} - {\left (x^{2} + x \log \relax (5) + 4\right )} e^{x} + 2 \, x \log \relax (5) + 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.53, size = 62, normalized size = 2.00 \begin {gather*} -\frac {3 \, {\left (x e^{x} \log \relax (5) - x^{2} - 2 \, x e^{x} - 2 \, x \log \relax (5) + 4 \, x + 4 \, e^{x} - 8\right )}}{x^{2} e^{x} + x e^{x} \log \relax (5) - 3 \, x^{2} - 2 \, x \log \relax (5) + 4 \, e^{x} - 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.38, size = 61, normalized size = 1.97
method | result | size |
norman | \(\frac {-12 \,{\mathrm e}^{x}+3 x^{2}+\left (-12+6 \ln \relax (5)\right ) x +\left (-3 \ln \relax (5)+6\right ) x \,{\mathrm e}^{x}+24}{x \,{\mathrm e}^{x} \ln \relax (5)+{\mathrm e}^{x} x^{2}-2 x \ln \relax (5)-3 x^{2}+4 \,{\mathrm e}^{x}-8}\) | \(61\) |
risch | \(\frac {\left (-3 \ln \relax (5)+6\right ) x -12}{x \ln \relax (5)+x^{2}+4}+\frac {3 \left (2+x \right ) x^{3}}{\left (x \ln \relax (5)+x^{2}+4\right ) \left (x \,{\mathrm e}^{x} \ln \relax (5)+{\mathrm e}^{x} x^{2}-2 x \ln \relax (5)-3 x^{2}+4 \,{\mathrm e}^{x}-8\right )}\) | \(73\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 53, normalized size = 1.71 \begin {gather*} -\frac {3 \, {\left (x^{2} + 2 \, x {\left (\log \relax (5) - 2\right )} - {\left (x {\left (\log \relax (5) - 2\right )} + 4\right )} e^{x} + 8\right )}}{3 \, x^{2} - {\left (x^{2} + x \log \relax (5) + 4\right )} e^{x} + 2 \, x \log \relax (5) + 8} \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 = -0.03 \begin {gather*} \int \frac {96\,x+12\,x^2\,\ln \relax (5)+{\mathrm {e}}^{2\,x}\,\left (24\,x+3\,x^2\,\ln \relax (5)-6\,x^2+24\right )-{\mathrm {e}}^x\,\left (96\,x+12\,x^2\,\ln \relax (5)-30\,x^2+6\,x^3+3\,x^4+96\right )-36\,x^2+96}{4\,x^2\,{\ln \relax (5)}^2+\ln \relax (5)\,\left (12\,x^3+32\,x\right )+{\mathrm {e}}^{2\,x}\,\left (x^2\,{\ln \relax (5)}^2+\ln \relax (5)\,\left (2\,x^3+8\,x\right )+8\,x^2+x^4+16\right )-{\mathrm {e}}^x\,\left (4\,x^2\,{\ln \relax (5)}^2+\ln \relax (5)\,\left (10\,x^3+32\,x\right )+40\,x^2+6\,x^4+64\right )+48\,x^2+9\,x^4+64} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.83, size = 104, normalized size = 3.35 \begin {gather*} \frac {3 x^{4} + 6 x^{3}}{- 3 x^{4} - 5 x^{3} \log {\relax (5 )} - 20 x^{2} - 2 x^{2} \log {\relax (5 )}^{2} - 16 x \log {\relax (5 )} + \left (x^{4} + 2 x^{3} \log {\relax (5 )} + x^{2} \log {\relax (5 )}^{2} + 8 x^{2} + 8 x \log {\relax (5 )} + 16\right ) e^{x} - 32} - \frac {x \left (-6 + 3 \log {\relax (5 )}\right ) + 12}{x^{2} + x \log {\relax (5 )} + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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