Optimal. Leaf size=25 \[ -e^x-\frac {4}{6+e^x+\log (2)}+\log ^2(x)+\log (2 x) \]
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Rubi [A] time = 1.58, antiderivative size = 29, normalized size of antiderivative = 1.16, number of steps used = 14, number of rules used = 10, integrand size = 128, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6, 6741, 6742, 2194, 2282, 36, 29, 31, 44, 2301} \begin {gather*} -e^x+\frac {1}{4} (2 \log (x)+1)^2-\frac {4}{e^x+6+\log (2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 29
Rule 31
Rule 36
Rule 44
Rule 2194
Rule 2282
Rule 2301
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {36-e^{3 x} x+12 \log (2)+\log ^2(2)+e^{2 x} (1-12 x-2 x \log (2))+e^x \left (12-32 x+(2-12 x) \log (2)-x \log ^2(2)\right )+\left (72+2 e^{2 x}+24 \log (2)+2 \log ^2(2)+e^x (24+4 \log (2))\right ) \log (x)}{e^{2 x} x+x \log ^2(2)+x (36+12 \log (2))+e^x (12 x+2 x \log (2))} \, dx\\ &=\int \frac {36-e^{3 x} x+12 \log (2)+\log ^2(2)+e^{2 x} (1-12 x-2 x \log (2))+e^x \left (12-32 x+(2-12 x) \log (2)-x \log ^2(2)\right )+\left (72+2 e^{2 x}+24 \log (2)+2 \log ^2(2)+e^x (24+4 \log (2))\right ) \log (x)}{e^{2 x} x+e^x (12 x+2 x \log (2))+x \left (36+12 \log (2)+\log ^2(2)\right )} \, dx\\ &=\int \frac {-e^{3 x} x+e^{2 x} (1-12 x-2 x \log (2))+e^x \left (12-32 x+(2-12 x) \log (2)-x \log ^2(2)\right )+36 \left (1+\frac {1}{36} \log (2) (12+\log (2))\right )+\left (72+2 e^{2 x}+24 \log (2)+2 \log ^2(2)+e^x (24+4 \log (2))\right ) \log (x)}{x \left (e^x+6 \left (1+\frac {\log (2)}{6}\right )\right )^2} \, dx\\ &=\int \left (-e^x+\frac {4}{e^x+6 \left (1+\frac {\log (2)}{6}\right )}+\frac {4 (-6-\log (2))}{\left (e^x+6 \left (1+\frac {\log (2)}{6}\right )\right )^2}+\frac {1+2 \log (x)}{x}\right ) \, dx\\ &=4 \int \frac {1}{e^x+6 \left (1+\frac {\log (2)}{6}\right )} \, dx-(4 (6+\log (2))) \int \frac {1}{\left (e^x+6 \left (1+\frac {\log (2)}{6}\right )\right )^2} \, dx-\int e^x \, dx+\int \frac {1+2 \log (x)}{x} \, dx\\ &=-e^x+\frac {1}{4} (1+2 \log (x))^2+4 \operatorname {Subst}\left (\int \frac {1}{x (6+x+\log (2))} \, dx,x,e^x\right )-(4 (6+\log (2))) \operatorname {Subst}\left (\int \frac {1}{x (6+x+\log (2))^2} \, dx,x,e^x\right )\\ &=-e^x+\frac {1}{4} (1+2 \log (x))^2+\frac {4 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{6+\log (2)}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{6+x+\log (2)} \, dx,x,e^x\right )}{6+\log (2)}-(4 (6+\log (2))) \operatorname {Subst}\left (\int \left (\frac {1}{x (6+\log (2))^2}-\frac {1}{(6+\log (2)) (6+x+\log (2))^2}-\frac {1}{(6+\log (2))^2 (6+x+\log (2))}\right ) \, dx,x,e^x\right )\\ &=-e^x-\frac {4}{6+e^x+\log (2)}+\frac {1}{4} (1+2 \log (x))^2\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.36, size = 59, normalized size = 2.36 \begin {gather*} -e^x-\frac {24-2 \log ^3(2)+\log ^2(2) (-12+\log (4))+\log (16)-\log (2) \log (4096)+\log (4) \log (4096)}{(6+\log (2)) \left (6+e^x+\log (2)\right )}+\log (x)+\log ^2(x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 45, normalized size = 1.80 \begin {gather*} \frac {{\left (e^{x} + \log \relax (2) + 6\right )} \log \relax (x)^{2} - {\left (\log \relax (2) + 6\right )} e^{x} + {\left (e^{x} + \log \relax (2) + 6\right )} \log \relax (x) - e^{\left (2 \, x\right )} - 4}{e^{x} + \log \relax (2) + 6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.23, size = 61, normalized size = 2.44 \begin {gather*} \frac {e^{x} \log \relax (x)^{2} + \log \relax (2) \log \relax (x)^{2} - e^{x} \log \relax (2) + e^{x} \log \relax (x) + \log \relax (2) \log \relax (x) + 6 \, \log \relax (x)^{2} - e^{\left (2 \, x\right )} - 6 \, e^{x} + 6 \, \log \relax (x) - 4}{e^{x} + \log \relax (2) + 6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.20, size = 47, normalized size = 1.88
method | result | size |
risch | \(\ln \relax (x )^{2}+\frac {\ln \relax (2) \ln \relax (x )+{\mathrm e}^{x} \ln \relax (x )-{\mathrm e}^{x} \ln \relax (2)-{\mathrm e}^{2 x}+6 \ln \relax (x )-6 \,{\mathrm e}^{x}-4}{6+\ln \relax (2)+{\mathrm e}^{x}}\) | \(47\) |
norman | \(\frac {\left (\ln \relax (2)+6\right ) \ln \relax (x )^{2}+\left (\ln \relax (2)+6\right ) \ln \relax (x )+{\mathrm e}^{x} \ln \relax (x )^{2}+{\mathrm e}^{x} \ln \relax (x )-{\mathrm e}^{2 x}+\ln \relax (2)^{2}+12 \ln \relax (2)+32}{6+\ln \relax (2)+{\mathrm e}^{x}}\) | \(54\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.50, size = 48, normalized size = 1.92 \begin {gather*} \frac {{\left (\log \relax (2) + 6\right )} \log \relax (x)^{2} + {\left (\log \relax (x)^{2} - \log \relax (2) + \log \relax (x) - 6\right )} e^{x} + {\left (\log \relax (2) + 6\right )} \log \relax (x) - e^{\left (2 \, x\right )} - 4}{e^{x} + \log \relax (2) + 6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {12\,\ln \relax (2)-x\,{\mathrm {e}}^{3\,x}-{\mathrm {e}}^{2\,x}\,\left (12\,x+2\,x\,\ln \relax (2)-1\right )-{\mathrm {e}}^x\,\left (32\,x+\ln \relax (2)\,\left (12\,x-2\right )+x\,{\ln \relax (2)}^2-12\right )+\ln \relax (x)\,\left (2\,{\mathrm {e}}^{2\,x}+24\,\ln \relax (2)+{\mathrm {e}}^x\,\left (4\,\ln \relax (2)+24\right )+2\,{\ln \relax (2)}^2+72\right )+{\ln \relax (2)}^2+36}{36\,x+x\,{\mathrm {e}}^{2\,x}+12\,x\,\ln \relax (2)+x\,{\ln \relax (2)}^2+{\mathrm {e}}^x\,\left (12\,x+2\,x\,\ln \relax (2)\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.41, size = 20, normalized size = 0.80 \begin {gather*} - e^{x} + \log {\relax (x )}^{2} + \log {\relax (x )} - \frac {4}{e^{x} + \log {\relax (2 )} + 6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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