Optimal. Leaf size=18 \[ -3-3 x+\frac {4+e^x}{x^3 \log (100)} \]
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Rubi [A] time = 0.05, antiderivative size = 24, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14, 2197} \begin {gather*} \frac {e^x}{x^3 \log (100)}+\frac {4}{x^3 \log (100)}-3 x \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2197
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-12+e^x (-3+x)-3 x^4 \log (100)}{x^4} \, dx}{\log (100)}\\ &=\frac {\int \left (\frac {e^x (-3+x)}{x^4}-\frac {3 \left (4+x^4 \log (100)\right )}{x^4}\right ) \, dx}{\log (100)}\\ &=\frac {\int \frac {e^x (-3+x)}{x^4} \, dx}{\log (100)}-\frac {3 \int \frac {4+x^4 \log (100)}{x^4} \, dx}{\log (100)}\\ &=\frac {e^x}{x^3 \log (100)}-\frac {3 \int \left (\frac {4}{x^4}+\log (100)\right ) \, dx}{\log (100)}\\ &=-3 x+\frac {4}{x^3 \log (100)}+\frac {e^x}{x^3 \log (100)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.03, size = 23, normalized size = 1.28 \begin {gather*} \frac {\frac {4}{x^3}+\frac {e^x}{x^3}-3 x \log (100)}{\log (100)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 22, normalized size = 1.22 \begin {gather*} -\frac {6 \, x^{4} \log \left (10\right ) - e^{x} - 4}{2 \, x^{3} \log \left (10\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 22, normalized size = 1.22 \begin {gather*} -\frac {6 \, x^{4} \log \left (10\right ) - e^{x} - 4}{2 \, x^{3} \log \left (10\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 24, normalized size = 1.33
method | result | size |
default | \(\frac {\frac {4}{x^{3}}+\frac {{\mathrm e}^{x}}{x^{3}}-6 \ln \left (10\right ) x}{2 \ln \left (10\right )}\) | \(24\) |
norman | \(\frac {-3 x^{4}+\frac {2}{\ln \left (10\right )}+\frac {{\mathrm e}^{x}}{2 \ln \left (10\right )}}{x^{3}}\) | \(25\) |
risch | \(-\frac {3 x \ln \relax (2)}{\ln \relax (2)+\ln \relax (5)}-\frac {3 x \ln \relax (5)}{\ln \relax (2)+\ln \relax (5)}+\frac {2}{\left (\ln \relax (2)+\ln \relax (5)\right ) x^{3}}+\frac {{\mathrm e}^{x}}{2 \left (\ln \relax (2)+\ln \relax (5)\right ) x^{3}}\) | \(52\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.38, size = 29, normalized size = 1.61 \begin {gather*} -\frac {6 \, x \log \left (10\right ) - \frac {4}{x^{3}} + \Gamma \left (-2, -x\right ) + 3 \, \Gamma \left (-3, -x\right )}{2 \, \log \left (10\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.86, size = 18, normalized size = 1.00 \begin {gather*} \frac {\frac {{\mathrm {e}}^x}{2}+2}{x^3\,\ln \left (10\right )}-3\,x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 26, normalized size = 1.44 \begin {gather*} \frac {- 3 x \log {\left (10 \right )} + \frac {2}{x^{3}}}{\log {\left (10 \right )}} + \frac {e^{x}}{2 x^{3} \log {\left (10 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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