Optimal. Leaf size=19 \[ 25+\log \left (e^{2-\frac {15}{4 \log (x)}}+x^2\right ) \]
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Rubi [F]
time = 1.23, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps
used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {}
\begin {gather*} \int \frac {15+8 e^{\frac {15-8 \log (x)}{4 \log (x)}} x^2 \log ^2(x)}{4 x \log ^2(x)+4 e^{\frac {15-8 \log (x)}{4 \log (x)}} x^3 \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15 e^2+8 e^{\frac {15}{4 \log (x)}} x^2 \log ^2(x)}{4 x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)} \, dx\\ &=\frac {1}{4} \int \frac {15 e^2+8 e^{\frac {15}{4 \log (x)}} x^2 \log ^2(x)}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)} \, dx\\ &=\frac {1}{4} \int \left (\frac {8}{x}-\frac {e^2 \left (-15+8 \log ^2(x)\right )}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)}\right ) \, dx\\ &=2 \log (x)-\frac {1}{4} e^2 \int \frac {-15+8 \log ^2(x)}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)} \, dx\\ &=2 \log (x)-\frac {1}{4} e^2 \int \left (\frac {8}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right )}-\frac {15}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)}\right ) \, dx\\ &=2 \log (x)-\left (2 e^2\right ) \int \frac {1}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right )} \, dx+\frac {1}{4} \left (15 e^2\right ) \int \frac {1}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.13, size = 28, normalized size = 1.47 \begin {gather*} -\frac {15}{4 \log (x)}+\log \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.02, size = 36, normalized size = 1.89
method | result | size |
risch | \(2 \ln \left (x \right )-\frac {-8 \ln \left (x \right )+15}{4 \ln \left (x \right )}+\ln \left ({\mathrm e}^{-\frac {8 \ln \left (x \right )-15}{4 \ln \left (x \right )}}+\frac {1}{x^{2}}\right )\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.30, size = 30, normalized size = 1.58 \begin {gather*} -\frac {15}{4 \, \log \left (x\right )} + 2 \, \log \left (x\right ) + \log \left (\frac {x^{2} e^{\left (\frac {15}{4 \, \log \left (x\right )}\right )} + e^{2}}{x^{2}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 42 vs.
\(2 (16) = 32\).
time = 0.41, size = 42, normalized size = 2.21 \begin {gather*} \frac {8 \, \log \left (x\right )^{2} + 4 \, \log \left (x\right ) \log \left (\frac {x^{2} e^{\left (-\frac {8 \, \log \left (x\right ) - 15}{4 \, \log \left (x\right )}\right )} + 1}{x^{2}}\right ) - 15}{4 \, \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 31 vs.
\(2 (15) = 30\).
time = 0.13, size = 31, normalized size = 1.63 \begin {gather*} 2 \log {\left (x \right )} + \log {\left (e^{\frac {\frac {15}{4} - 2 \log {\left (x \right )}}{\log {\left (x \right )}}} + \frac {1}{x^{2}} \right )} - \frac {15}{4 \log {\left (x \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.48, size = 32, normalized size = 1.68 \begin {gather*} \frac {4 \, \log \left (x^{2} e^{\left (-\frac {8 \, \log \left (x\right ) - 15}{4 \, \log \left (x\right )}\right )} + 1\right ) \log \left (x\right ) - 15}{4 \, \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.20, size = 31, normalized size = 1.63 \begin {gather*} \ln \left (\frac {1}{x^2}\right )+\ln \left (x^2\,{\mathrm {e}}^{\frac {15}{4\,\ln \left (x\right )}}\,{\mathrm {e}}^{-2}+1\right )+2\,\ln \left (x\right )-\frac {15}{4\,\ln \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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