Optimal. Leaf size=29 \[ 2+(1+x) \left (2-e^2-x+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )} \]
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Rubi [A] time = 0.24, antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6688, 2306, 2310, 2178} \begin {gather*} -x^2-\frac {25}{x^2 \log ^2\left (x^3\right )}+x \left (1-e^2+\log (2)\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2178
Rule 2306
Rule 2310
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-e^2-2 x+\log (2)+\frac {150}{x^3 \log ^3\left (x^3\right )}+\frac {50}{x^3 \log ^2\left (x^3\right )}\right ) \, dx\\ &=-x^2+x \left (1-e^2+\log (2)\right )+50 \int \frac {1}{x^3 \log ^2\left (x^3\right )} \, dx+150 \int \frac {1}{x^3 \log ^3\left (x^3\right )} \, dx\\ &=-x^2+x \left (1-e^2+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )}-\frac {50}{3 x^2 \log \left (x^3\right )}-\frac {100}{3} \int \frac {1}{x^3 \log \left (x^3\right )} \, dx-50 \int \frac {1}{x^3 \log ^2\left (x^3\right )} \, dx\\ &=-x^2+x \left (1-e^2+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )}+\frac {100}{3} \int \frac {1}{x^3 \log \left (x^3\right )} \, dx-\frac {\left (100 \left (x^3\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {e^{-2 x/3}}{x} \, dx,x,\log \left (x^3\right )\right )}{9 x^2}\\ &=-x^2-\frac {100 \left (x^3\right )^{2/3} \text {Ei}\left (-\frac {2}{3} \log \left (x^3\right )\right )}{9 x^2}+x \left (1-e^2+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )}+\frac {\left (100 \left (x^3\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {e^{-2 x/3}}{x} \, dx,x,\log \left (x^3\right )\right )}{9 x^2}\\ &=-x^2+x \left (1-e^2+\log (2)\right )-\frac {25}{x^2 \log ^2\left (x^3\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 28, normalized size = 0.97 \begin {gather*} x-e^2 x-x^2+x \log (2)-\frac {25}{x^2 \log ^2\left (x^3\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.81, size = 42, normalized size = 1.45 \begin {gather*} -\frac {{\left (x^{4} + x^{3} e^{2} - x^{3} \log \relax (2) - x^{3}\right )} \log \left (x^{3}\right )^{2} + 25}{x^{2} \log \left (x^{3}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.18, size = 59, normalized size = 2.03 \begin {gather*} -\frac {x^{4} \log \left (x^{3}\right )^{2} + x^{3} e^{2} \log \left (x^{3}\right )^{2} - x^{3} \log \relax (2) \log \left (x^{3}\right )^{2} - x^{3} \log \left (x^{3}\right )^{2} + 25}{x^{2} \log \left (x^{3}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 28, normalized size = 0.97
method | result | size |
risch | \(-{\mathrm e}^{2} x +x \ln \relax (2)-x^{2}+x -\frac {25}{\ln \left (x^{3}\right )^{2} x^{2}}\) | \(28\) |
norman | \(\frac {-25+\left (-{\mathrm e}^{2}+\ln \relax (2)+1\right ) x^{3} \ln \left (x^{3}\right )^{2}-x^{4} \ln \left (x^{3}\right )^{2}}{x^{2} \ln \left (x^{3}\right )^{2}}\) | \(42\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 25, normalized size = 0.86 \begin {gather*} -x^{2} - x e^{2} + x \log \relax (2) + x - \frac {25}{9 \, x^{2} \log \relax (x)^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.34, size = 27, normalized size = 0.93 \begin {gather*} x-x\,{\mathrm {e}}^2+x\,\ln \relax (2)-x^2-\frac {25}{x^2\,{\ln \left (x^3\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.11, size = 24, normalized size = 0.83 \begin {gather*} - x^{2} + x \left (- e^{2} + \log {\relax (2 )} + 1\right ) - \frac {25}{x^{2} \log {\left (x^{3} \right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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