Optimal. Leaf size=22 \[ e^x \left (5+\log ^2(2)\right )-\frac {5}{16} x^8 \log ^8(x) \]
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Rubi [A] time = 0.23, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 2194, 2305, 2304} \begin {gather*} e^x \left (5+\log ^2(2)\right )-\frac {5}{16} x^8 \log ^8(x) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2194
Rule 2304
Rule 2305
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (e^x \left (10+2 \log ^2(2)\right )-5 x^7 \log ^7(x)-5 x^7 \log ^8(x)\right ) \, dx\\ &=-\left (\frac {5}{2} \int x^7 \log ^7(x) \, dx\right )-\frac {5}{2} \int x^7 \log ^8(x) \, dx+\left (5+\log ^2(2)\right ) \int e^x \, dx\\ &=e^x \left (5+\log ^2(2)\right )-\frac {5}{16} x^8 \log ^7(x)-\frac {5}{16} x^8 \log ^8(x)+\frac {35}{16} \int x^7 \log ^6(x) \, dx+\frac {5}{2} \int x^7 \log ^7(x) \, dx\\ &=e^x \left (5+\log ^2(2)\right )+\frac {35}{128} x^8 \log ^6(x)-\frac {5}{16} x^8 \log ^8(x)-\frac {105}{64} \int x^7 \log ^5(x) \, dx-\frac {35}{16} \int x^7 \log ^6(x) \, dx\\ &=e^x \left (5+\log ^2(2)\right )-\frac {105}{512} x^8 \log ^5(x)-\frac {5}{16} x^8 \log ^8(x)+\frac {525}{512} \int x^7 \log ^4(x) \, dx+\frac {105}{64} \int x^7 \log ^5(x) \, dx\\ &=e^x \left (5+\log ^2(2)\right )+\frac {525 x^8 \log ^4(x)}{4096}-\frac {5}{16} x^8 \log ^8(x)-\frac {525 \int x^7 \log ^3(x) \, dx}{1024}-\frac {525}{512} \int x^7 \log ^4(x) \, dx\\ &=e^x \left (5+\log ^2(2)\right )-\frac {525 x^8 \log ^3(x)}{8192}-\frac {5}{16} x^8 \log ^8(x)+\frac {1575 \int x^7 \log ^2(x) \, dx}{8192}+\frac {525 \int x^7 \log ^3(x) \, dx}{1024}\\ &=e^x \left (5+\log ^2(2)\right )+\frac {1575 x^8 \log ^2(x)}{65536}-\frac {5}{16} x^8 \log ^8(x)-\frac {1575 \int x^7 \log (x) \, dx}{32768}-\frac {1575 \int x^7 \log ^2(x) \, dx}{8192}\\ &=\frac {1575 x^8}{2097152}+e^x \left (5+\log ^2(2)\right )-\frac {1575 x^8 \log (x)}{262144}-\frac {5}{16} x^8 \log ^8(x)+\frac {1575 \int x^7 \log (x) \, dx}{32768}\\ &=e^x \left (5+\log ^2(2)\right )-\frac {5}{16} x^8 \log ^8(x)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 25, normalized size = 1.14 \begin {gather*} 5 e^x+e^x \log ^2(2)-\frac {5}{16} x^8 \log ^8(x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.85, size = 19, normalized size = 0.86 \begin {gather*} -\frac {5}{16} \, x^{8} \log \relax (x)^{8} + {\left (\log \relax (2)^{2} + 5\right )} e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 19, normalized size = 0.86 \begin {gather*} -\frac {5}{16} \, x^{8} \log \relax (x)^{8} + {\left (\log \relax (2)^{2} + 5\right )} e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 22, normalized size = 1.00
method | result | size |
default | \(-\frac {5 x^{8} \ln \relax (x )^{8}}{16}+\ln \relax (2)^{2} {\mathrm e}^{x}+5 \,{\mathrm e}^{x}\) | \(22\) |
risch | \(-\frac {5 x^{8} \ln \relax (x )^{8}}{16}+\ln \relax (2)^{2} {\mathrm e}^{x}+5 \,{\mathrm e}^{x}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 110, normalized size = 5.00 \begin {gather*} -\frac {5}{2097152} \, {\left (131072 \, \log \relax (x)^{8} - 131072 \, \log \relax (x)^{7} + 114688 \, \log \relax (x)^{6} - 86016 \, \log \relax (x)^{5} + 53760 \, \log \relax (x)^{4} - 26880 \, \log \relax (x)^{3} + 10080 \, \log \relax (x)^{2} - 2520 \, \log \relax (x) + 315\right )} x^{8} - \frac {5}{2097152} \, {\left (131072 \, \log \relax (x)^{7} - 114688 \, \log \relax (x)^{6} + 86016 \, \log \relax (x)^{5} - 53760 \, \log \relax (x)^{4} + 26880 \, \log \relax (x)^{3} - 10080 \, \log \relax (x)^{2} + 2520 \, \log \relax (x) - 315\right )} x^{8} + {\left (\log \relax (2)^{2} + 5\right )} e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.30, size = 19, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^x\,\left ({\ln \relax (2)}^2+5\right )-\frac {5\,x^8\,{\ln \relax (x)}^8}{16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.27, size = 20, normalized size = 0.91 \begin {gather*} - \frac {5 x^{8} \log {\relax (x )}^{8}}{16} + \left (\log {\relax (2 )}^{2} + 5\right ) e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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