Optimal. Leaf size=26 \[ x \left (-e^{2 x}+5 \left (x+2 e^3 x+x \log (4)\right )\right )^2 \]
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Rubi [B] time = 0.16, antiderivative size = 89, normalized size of antiderivative = 3.42, number of steps used = 23, number of rules used = 6, integrand size = 99, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6, 2176, 2194, 12, 30, 2196} \begin {gather*} 25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+50 \left (1+2 e^3\right ) x^3 \log (4)-10 e^{2 x} x^2-10 e^{2 x+3} x^2 \left (2+\frac {\log (4)}{e^3}\right )-\frac {e^{4 x}}{4}+\frac {1}{4} e^{4 x} (4 x+1) \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 30
Rule 2176
Rule 2194
Rule 2196
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (300 e^6 x^2+\left (75+300 e^3\right ) x^2+e^{4 x} (1+4 x)+\left (150 x^2+300 e^3 x^2\right ) \log (4)+75 x^2 \log ^2(4)+e^{2 x} \left (-20 x-20 x^2+e^3 \left (-40 x-40 x^2\right )+\left (-20 x-20 x^2\right ) \log (4)\right )\right ) \, dx\\ &=\int \left (\left (75+300 e^3+300 e^6\right ) x^2+e^{4 x} (1+4 x)+\left (150 x^2+300 e^3 x^2\right ) \log (4)+75 x^2 \log ^2(4)+e^{2 x} \left (-20 x-20 x^2+e^3 \left (-40 x-40 x^2\right )+\left (-20 x-20 x^2\right ) \log (4)\right )\right ) \, dx\\ &=\int \left (e^{4 x} (1+4 x)+\left (150 x^2+300 e^3 x^2\right ) \log (4)+e^{2 x} \left (-20 x-20 x^2+e^3 \left (-40 x-40 x^2\right )+\left (-20 x-20 x^2\right ) \log (4)\right )+x^2 \left (75+300 e^3+300 e^6+75 \log ^2(4)\right )\right ) \, dx\\ &=25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+\log (4) \int \left (150 x^2+300 e^3 x^2\right ) \, dx+\int e^{4 x} (1+4 x) \, dx+\int e^{2 x} \left (-20 x-20 x^2+e^3 \left (-40 x-40 x^2\right )+\left (-20 x-20 x^2\right ) \log (4)\right ) \, dx\\ &=\frac {1}{4} e^{4 x} (1+4 x)+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+\log (4) \int \left (150+300 e^3\right ) x^2 \, dx-\int e^{4 x} \, dx+\int \left (-20 e^{2 x} x-20 e^{2 x} x^2-40 e^{3+2 x} x (1+x) \left (1+\frac {\log (4)}{2 e^3}\right )\right ) \, dx\\ &=-\frac {e^{4 x}}{4}+\frac {1}{4} e^{4 x} (1+4 x)+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )-20 \int e^{2 x} x \, dx-20 \int e^{2 x} x^2 \, dx+\left (150 \left (1+2 e^3\right ) \log (4)\right ) \int x^2 \, dx-\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} x (1+x) \, dx\\ &=-\frac {e^{4 x}}{4}-10 e^{2 x} x-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+10 \int e^{2 x} \, dx+20 \int e^{2 x} x \, dx-\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int \left (e^{3+2 x} x+e^{3+2 x} x^2\right ) \, dx\\ &=5 e^{2 x}-\frac {e^{4 x}}{4}-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )-10 \int e^{2 x} \, dx-\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} x \, dx-\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} x^2 \, dx\\ &=-\frac {e^{4 x}}{4}-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)-10 e^{3+2 x} x \left (2+\frac {\log (4)}{e^3}\right )-10 e^{3+2 x} x^2 \left (2+\frac {\log (4)}{e^3}\right )+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+\left (10 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} \, dx+\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} x \, dx\\ &=-\frac {e^{4 x}}{4}-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)+5 e^{3+2 x} \left (2+\frac {\log (4)}{e^3}\right )-10 e^{3+2 x} x^2 \left (2+\frac {\log (4)}{e^3}\right )+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )-\left (10 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} \, dx\\ &=-\frac {e^{4 x}}{4}-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)-10 e^{3+2 x} x^2 \left (2+\frac {\log (4)}{e^3}\right )+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 23, normalized size = 0.88 \begin {gather*} x \left (e^{2 x}-10 e^3 x-5 x (1+\log (4))\right )^2 \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.75, size = 74, normalized size = 2.85 \begin {gather*} 100 \, x^{3} \log \relax (2)^{2} + 100 \, x^{3} e^{6} + 100 \, x^{3} e^{3} + 25 \, x^{3} + x e^{\left (4 \, x\right )} - 10 \, {\left (2 \, x^{2} e^{3} + 2 \, x^{2} \log \relax (2) + x^{2}\right )} e^{\left (2 \, x\right )} + 100 \, {\left (2 \, x^{3} e^{3} + x^{3}\right )} \log \relax (2) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.16, size = 78, normalized size = 3.00 \begin {gather*} 100 \, x^{3} \log \relax (2)^{2} + 100 \, x^{3} e^{6} + 100 \, x^{3} e^{3} + 25 \, x^{3} - 20 \, x^{2} e^{\left (2 \, x + 3\right )} + x e^{\left (4 \, x\right )} - 10 \, {\left (2 \, x^{2} \log \relax (2) + x^{2}\right )} e^{\left (2 \, x\right )} + 100 \, {\left (2 \, x^{3} e^{3} + x^{3}\right )} \log \relax (2) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.07, size = 60, normalized size = 2.31
method | result | size |
norman | \(x \,{\mathrm e}^{4 x}+\left (100 \,{\mathrm e}^{6}+200 \,{\mathrm e}^{3} \ln \relax (2)+100 \ln \relax (2)^{2}+100 \,{\mathrm e}^{3}+100 \ln \relax (2)+25\right ) x^{3}+\left (-20 \,{\mathrm e}^{3}-20 \ln \relax (2)-10\right ) x^{2} {\mathrm e}^{2 x}\) | \(60\) |
risch | \(x \,{\mathrm e}^{4 x}-10 \left (2 \,{\mathrm e}^{3}+2 \ln \relax (2)+1\right ) x^{2} {\mathrm e}^{2 x}+100 x^{3} \ln \relax (2)^{2}+200 \ln \relax (2) x^{3} {\mathrm e}^{3}+100 x^{3} \ln \relax (2)+100 x^{3} {\mathrm e}^{6}+100 x^{3} {\mathrm e}^{3}+25 x^{3}\) | \(71\) |
derivativedivides | \(200 \ln \relax (2) x^{3} {\mathrm e}^{3}+100 x^{3} \ln \relax (2)+25 x^{3}+100 x^{3} {\mathrm e}^{3}+100 x^{3} {\mathrm e}^{6}+100 x^{3} \ln \relax (2)^{2}+x \,{\mathrm e}^{4 x}-10 \,{\mathrm e}^{2 x} x^{2}-20 x^{2} {\mathrm e}^{3} {\mathrm e}^{2 x}-20 \,{\mathrm e}^{2 x} \ln \relax (2) x^{2}\) | \(87\) |
default | \(200 \ln \relax (2) x^{3} {\mathrm e}^{3}+100 x^{3} \ln \relax (2)+25 x^{3}+100 x^{3} {\mathrm e}^{3}+100 x^{3} {\mathrm e}^{6}+100 x^{3} \ln \relax (2)^{2}+x \,{\mathrm e}^{4 x}-10 \,{\mathrm e}^{2 x} x^{2}-20 x^{2} {\mathrm e}^{3} {\mathrm e}^{2 x}-20 \,{\mathrm e}^{2 x} \ln \relax (2) x^{2}\) | \(87\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 69, normalized size = 2.65 \begin {gather*} 100 \, x^{3} \log \relax (2)^{2} + 100 \, x^{3} e^{6} + 100 \, x^{3} e^{3} - 10 \, x^{2} {\left (2 \, e^{3} + 2 \, \log \relax (2) + 1\right )} e^{\left (2 \, x\right )} + 25 \, x^{3} + x e^{\left (4 \, x\right )} + 100 \, {\left (2 \, x^{3} e^{3} + x^{3}\right )} \log \relax (2) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.19, size = 24, normalized size = 0.92 \begin {gather*} x\,{\left (5\,x-{\mathrm {e}}^{2\,x}+10\,x\,{\mathrm {e}}^3+10\,x\,\ln \relax (2)\right )}^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.18, size = 70, normalized size = 2.69 \begin {gather*} x^{3} \left (25 + 100 \log {\relax (2 )}^{2} + 100 \log {\relax (2 )} + 100 e^{3} + 200 e^{3} \log {\relax (2 )} + 100 e^{6}\right ) + x e^{4 x} + \left (- 20 x^{2} e^{3} - 20 x^{2} \log {\relax (2 )} - 10 x^{2}\right ) e^{2 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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