Optimal. Leaf size=27 \[ x^2 \left (-3-e^x+\frac {x}{4 \left (5-\frac {16}{x^2}+\log (3)\right )}\right ) \]
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Rubi [B] time = 1.28, antiderivative size = 71, normalized size of antiderivative = 2.63, number of steps used = 27, number of rules used = 14, integrand size = 163, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6, 6688, 12, 6742, 2196, 2176, 2194, 261, 288, 321, 206, 266, 43, 302} \begin {gather*} \frac {5 x^3}{8 (5+\log (3))}-e^x x^2-3 x^2+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 43
Rule 206
Rule 261
Rule 266
Rule 288
Rule 302
Rule 321
Rule 2176
Rule 2194
Rule 2196
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-6144 x+3840 x^3-80 x^4-600 x^5+15 x^6+\left (768 x^3-240 x^5+3 x^6\right ) \log (3)-24 x^5 \log ^2(3)+e^x \left (-2048 x-1024 x^2+1280 x^3+640 x^4-200 x^5-100 x^6+\left (256 x^3+128 x^4-80 x^5-40 x^6\right ) \log (3)+\left (-8 x^5-4 x^6\right ) \log ^2(3)\right )}{1024-640 x^2+\left (-128 x^2+40 x^4\right ) \log (3)+x^4 \left (100+4 \log ^2(3)\right )} \, dx\\ &=\int \frac {-6144 x+3840 x^3-80 x^4+15 x^6+\left (768 x^3-240 x^5+3 x^6\right ) \log (3)+x^5 \left (-600-24 \log ^2(3)\right )+e^x \left (-2048 x-1024 x^2+1280 x^3+640 x^4-200 x^5-100 x^6+\left (256 x^3+128 x^4-80 x^5-40 x^6\right ) \log (3)+\left (-8 x^5-4 x^6\right ) \log ^2(3)\right )}{1024-640 x^2+\left (-128 x^2+40 x^4\right ) \log (3)+x^4 \left (100+4 \log ^2(3)\right )} \, dx\\ &=\int \frac {x \left (-6144-80 x^3+768 x^2 (5+\log (3))+3 x^5 (5+\log (3))-24 x^4 (5+\log (3))^2-4 e^x (2+x) \left (-16+x^2 (5+\log (3))\right )^2\right )}{4 \left (16-x^2 (5+\log (3))\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {x \left (-6144-80 x^3+768 x^2 (5+\log (3))+3 x^5 (5+\log (3))-24 x^4 (5+\log (3))^2-4 e^x (2+x) \left (-16+x^2 (5+\log (3))\right )^2\right )}{\left (16-x^2 (5+\log (3))\right )^2} \, dx\\ &=\frac {1}{4} \int \left (-4 e^x x (2+x)-\frac {6144 x}{\left (16-x^2 (5+\log (3))\right )^2}-\frac {80 x^4}{\left (16-x^2 (5+\log (3))\right )^2}+\frac {768 x^3 (5+\log (3))}{\left (16-x^2 (5+\log (3))\right )^2}+\frac {3 x^6 (5+\log (3))}{\left (16-x^2 (5+\log (3))\right )^2}-\frac {24 x^5 (5+\log (3))^2}{\left (16-x^2 (5+\log (3))\right )^2}\right ) \, dx\\ &=-\left (20 \int \frac {x^4}{\left (16-x^2 (5+\log (3))\right )^2} \, dx\right )-1536 \int \frac {x}{\left (16-x^2 (5+\log (3))\right )^2} \, dx+\frac {1}{4} (3 (5+\log (3))) \int \frac {x^6}{\left (16-x^2 (5+\log (3))\right )^2} \, dx+(192 (5+\log (3))) \int \frac {x^3}{\left (16-x^2 (5+\log (3))\right )^2} \, dx-\left (6 (5+\log (3))^2\right ) \int \frac {x^5}{\left (16-x^2 (5+\log (3))\right )^2} \, dx-\int e^x x (2+x) \, dx\\ &=\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {768}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-\frac {15}{8} \int \frac {x^4}{16+x^2 (-5-\log (3))} \, dx+\frac {30 \int \frac {x^2}{16+x^2 (-5-\log (3))} \, dx}{5+\log (3)}+(96 (5+\log (3))) \operatorname {Subst}\left (\int \frac {x}{(16+x (-5-\log (3)))^2} \, dx,x,x^2\right )-\left (3 (5+\log (3))^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{(16+x (-5-\log (3)))^2} \, dx,x,x^2\right )-\int \left (2 e^x x+e^x x^2\right ) \, dx\\ &=-\frac {30 x}{(5+\log (3))^2}+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {768}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-\frac {15}{8} \int \left (-\frac {16}{(5+\log (3))^2}-\frac {x^2}{5+\log (3)}+\frac {256}{(5+\log (3))^2 \left (16-x^2 (5+\log (3))\right )}\right ) \, dx-2 \int e^x x \, dx+\frac {480 \int \frac {1}{16+x^2 (-5-\log (3))} \, dx}{(5+\log (3))^2}+(96 (5+\log (3))) \operatorname {Subst}\left (\int \left (\frac {16}{(5+\log (3)) (16-x (5+\log (3)))^2}+\frac {1}{(-5-\log (3)) (16-x (5+\log (3)))}\right ) \, dx,x,x^2\right )-\left (3 (5+\log (3))^2\right ) \operatorname {Subst}\left (\int \left (\frac {1}{(5+\log (3))^2}+\frac {256}{(5+\log (3))^2 (16-x (5+\log (3)))^2}+\frac {32}{(5+\log (3))^2 (-16+x (5+\log (3)))}\right ) \, dx,x,x^2\right )-\int e^x x^2 \, dx\\ &=-2 e^x x-3 x^2-e^x x^2+\frac {120 \tanh ^{-1}\left (\frac {1}{4} x \sqrt {5+\log (3)}\right )}{(5+\log (3))^{5/2}}+\frac {5 x^3}{8 (5+\log (3))}+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}+2 \int e^x \, dx+2 \int e^x x \, dx-\frac {480 \int \frac {1}{16-x^2 (5+\log (3))} \, dx}{(5+\log (3))^2}\\ &=2 e^x-3 x^2-e^x x^2+\frac {5 x^3}{8 (5+\log (3))}+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-2 \int e^x \, dx\\ &=-3 x^2-e^x x^2+\frac {5 x^3}{8 (5+\log (3))}+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.35, size = 48, normalized size = 1.78 \begin {gather*} \frac {x^2 \left (192+x^3-12 x^2 (5+\log (3))-4 e^x \left (-16+x^2 (5+\log (3))\right )\right )}{4 \left (-16+x^2 (5+\log (3))\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 2.02, size = 59, normalized size = 2.19 \begin {gather*} \frac {x^{5} - 12 \, x^{4} \log \relax (3) - 60 \, x^{4} + 192 \, x^{2} - 4 \, {\left (x^{4} \log \relax (3) + 5 \, x^{4} - 16 \, x^{2}\right )} e^{x}}{4 \, {\left (x^{2} \log \relax (3) + 5 \, x^{2} - 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.23, size = 63, normalized size = 2.33 \begin {gather*} -\frac {4 \, x^{4} e^{x} \log \relax (3) - x^{5} + 20 \, x^{4} e^{x} + 12 \, x^{4} \log \relax (3) + 60 \, x^{4} - 64 \, x^{2} e^{x} - 192 \, x^{2}}{4 \, {\left (x^{2} \log \relax (3) + 5 \, x^{2} - 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.32, size = 57, normalized size = 2.11
method | result | size |
norman | \(\frac {\left (-3 \ln \relax (3)-15\right ) x^{4}+48 x^{2}+\left (-5-\ln \relax (3)\right ) x^{4} {\mathrm e}^{x}+\frac {x^{5}}{4}+16 \,{\mathrm e}^{x} x^{2}}{x^{2} \ln \relax (3)+5 x^{2}-16}\) | \(57\) |
risch | \(-\frac {12 \ln \relax (3)^{2} x^{2}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}+\frac {\ln \relax (3) x^{3}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}-\frac {120 \ln \relax (3) x^{2}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}+\frac {5 x^{3}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}-\frac {300 x^{2}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}+\frac {16 x}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}+\frac {256 x}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right ) \left (x^{2} \ln \relax (3)+5 x^{2}-16\right )}-{\mathrm e}^{x} x^{2}\) | \(160\) |
default | \(\text {Expression too large to display}\) | \(3045\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 848, normalized size = 31.41 result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.01, size = 139, normalized size = 5.15 \begin {gather*} \frac {x^3\,\left (\ln \left (27\right )+15\right )}{3\,\left (40\,\ln \relax (3)+4\,{\ln \relax (3)}^2+100\right )}-x\,\left (\frac {80}{40\,\ln \relax (3)+4\,{\ln \relax (3)}^2+100}-\frac {\left (128\,\ln \relax (3)+640\right )\,\left (\ln \left (27\right )+15\right )}{{\left (40\,\ln \relax (3)+4\,{\ln \relax (3)}^2+100\right )}^2}\right )-\frac {64\,x}{\left (-75\,\ln \relax (3)-15\,{\ln \relax (3)}^2-{\ln \relax (3)}^3-125\right )\,x^2+160\,\ln \relax (3)+16\,{\ln \relax (3)}^2+400}-\frac {12\,x^2\,{\left (\ln \relax (3)+5\right )}^2}{40\,\ln \relax (3)+4\,{\ln \relax (3)}^2+100}-x^2\,{\mathrm {e}}^x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.94, size = 75, normalized size = 2.78 \begin {gather*} \frac {x^{3}}{4 \log {\relax (3 )} + 20} - x^{2} e^{x} - 3 x^{2} + \frac {4 x}{\log {\relax (3 )}^{2} + 10 \log {\relax (3 )} + 25} + \frac {64 x}{x^{2} \left (\log {\relax (3 )}^{3} + 15 \log {\relax (3 )}^{2} + 75 \log {\relax (3 )} + 125\right ) - 400 - 160 \log {\relax (3 )} - 16 \log {\relax (3 )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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