Optimal. Leaf size=24 \[ \frac {5 (-3+x)}{2 \left (3+e^2\right ) x \log ^2(15)}+\log (x) \]
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Rubi [A]
time = 0.03, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps
used = 5, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6, 12, 192, 45}
\begin {gather*} \log (x)-\frac {15}{2 \left (3+e^2\right ) x \log ^2(15)} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 45
Rule 192
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15+\left (6 x+2 e^2 x\right ) \log ^2(15)}{\left (6+2 e^2\right ) x^2 \log ^2(15)} \, dx\\ &=\frac {\int \frac {15+\left (6 x+2 e^2 x\right ) \log ^2(15)}{x^2} \, dx}{2 \left (3+e^2\right ) \log ^2(15)}\\ &=\frac {\int \frac {15+2 \left (3+e^2\right ) x \log ^2(15)}{x^2} \, dx}{2 \left (3+e^2\right ) \log ^2(15)}\\ &=\frac {\int \left (\frac {15}{x^2}+\frac {2 \left (3+e^2\right ) \log ^2(15)}{x}\right ) \, dx}{2 \left (3+e^2\right ) \log ^2(15)}\\ &=-\frac {15}{2 \left (3+e^2\right ) x \log ^2(15)}+\log (x)\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.01, size = 21, normalized size = 0.88 \begin {gather*} -\frac {15}{2 \left (3+e^2\right ) x \log ^2(15)}+\log (x) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.09, size = 31, normalized size = 1.29
method | result | size |
norman | \(-\frac {15}{2 \ln \left (15\right )^{2} \left ({\mathrm e}^{2}+3\right ) x}+\ln \left (x \right )\) | \(19\) |
default | \(\frac {2 \ln \left (15\right )^{2} \left ({\mathrm e}^{2}+3\right ) \ln \left (x \right )-\frac {15}{x}}{2 \ln \left (15\right )^{2} \left ({\mathrm e}^{2}+3\right )}\) | \(31\) |
risch | \(-\frac {15}{2 \left (\ln \left (3\right )+\ln \left (5\right )\right )^{2} x \left ({\mathrm e}^{2}+3\right )}+\frac {\ln \left (x \right ) \ln \left (3\right )^{2}}{\left (\ln \left (3\right )+\ln \left (5\right )\right )^{2}}+\frac {2 \ln \left (x \right ) \ln \left (3\right ) \ln \left (5\right )}{\left (\ln \left (3\right )+\ln \left (5\right )\right )^{2}}+\frac {\ln \left (x \right ) \ln \left (5\right )^{2}}{\left (\ln \left (3\right )+\ln \left (5\right )\right )^{2}}\) | \(63\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.26, size = 26, normalized size = 1.08 \begin {gather*} \frac {2 \, \log \left (15\right )^{2} \log \left (x\right ) - \frac {15}{x {\left (e^{2} + 3\right )}}}{2 \, \log \left (15\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 34, normalized size = 1.42 \begin {gather*} \frac {2 \, {\left (x e^{2} + 3 \, x\right )} \log \left (15\right )^{2} \log \left (x\right ) - 15}{2 \, {\left (x e^{2} + 3 \, x\right )} \log \left (15\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.07, size = 34, normalized size = 1.42 \begin {gather*} \frac {2 \cdot \left (3 + e^{2}\right ) \log {\left (15 \right )}^{2} \log {\left (x \right )} - \frac {15}{x}}{6 \log {\left (15 \right )}^{2} + 2 e^{2} \log {\left (15 \right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 27, normalized size = 1.12 \begin {gather*} \frac {2 \, \log \left (15\right )^{2} \log \left ({\left | x \right |}\right ) - \frac {15}{x {\left (e^{2} + 3\right )}}}{2 \, \log \left (15\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.10, size = 25, normalized size = 1.04 \begin {gather*} \ln \left (x\right )-\frac {15}{x\,\left (2\,{\mathrm {e}}^2\,{\ln \left (15\right )}^2+6\,{\ln \left (15\right )}^2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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