Optimal. Leaf size=31 \[ \frac {3 e^{5+e^{2 x}} x}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )} \]
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Rubi [F]
time = 3.90, 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 {e^{5+e^{2 x}} \left (6+e^{16+x} \left (-9+9 x-18 e^{2 x} x\right )+e^{16+x} \left (3-3 x+6 e^{2 x} x\right ) \log \left (x^2\right )\right )+e^{5+e^{2 x}} \left (9+18 e^{2 x} x+\left (-3-6 e^{2 x} x\right ) \log \left (x^2\right )\right ) \log \left (3-\log \left (x^2\right )\right )}{-3 e^{32+2 x}+e^{32+2 x} \log \left (x^2\right )+\left (6 e^{16+x}-2 e^{16+x} \log \left (x^2\right )\right ) \log \left (3-\log \left (x^2\right )\right )+\left (-3+\log \left (x^2\right )\right ) \log ^2\left (3-\log \left (x^2\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^{5+e^{2 x}} \left (6+e^{16+x} \left (-9+9 x-18 e^{2 x} x\right )+e^{16+x} \left (3-3 x+6 e^{2 x} x\right ) \log \left (x^2\right )\right )-e^{5+e^{2 x}} \left (9+18 e^{2 x} x+\left (-3-6 e^{2 x} x\right ) \log \left (x^2\right )\right ) \log \left (3-\log \left (x^2\right )\right )}{\left (3-\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx\\ &=\int \left (6 e^{-11+e^{2 x}+x} x+6 e^{-27+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right )-\frac {3 e^{5+e^{2 x}} \left (-2-3 x \log \left (3-\log \left (x^2\right )\right )+x \log \left (x^2\right ) \log \left (3-\log \left (x^2\right )\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2}-\frac {3 e^{e^{2 x}} \left (-e^{32}+e^{32} x-2 x \log ^2\left (3-\log \left (x^2\right )\right )\right )}{e^{43+x}-e^{27} \log \left (3-\log \left (x^2\right )\right )}\right ) \, dx\\ &=-\left (3 \int \frac {e^{5+e^{2 x}} \left (-2-3 x \log \left (3-\log \left (x^2\right )\right )+x \log \left (x^2\right ) \log \left (3-\log \left (x^2\right )\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx\right )-3 \int \frac {e^{e^{2 x}} \left (-e^{32}+e^{32} x-2 x \log ^2\left (3-\log \left (x^2\right )\right )\right )}{e^{43+x}-e^{27} \log \left (3-\log \left (x^2\right )\right )} \, dx+6 \int e^{-11+e^{2 x}+x} x \, dx+6 \int e^{-27+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right ) \, dx\\ &=-\left (3 \int \left (-\frac {2 e^{5+e^{2 x}}}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2}-\frac {3 e^{5+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2}+\frac {e^{5+e^{2 x}} x \log \left (x^2\right ) \log \left (3-\log \left (x^2\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2}\right ) \, dx\right )-3 \int \left (-\frac {e^{5+e^{2 x}}}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )}+\frac {e^{5+e^{2 x}} x}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )}+\frac {2 e^{e^{2 x}} x \log ^2\left (3-\log \left (x^2\right )\right )}{-e^{43+x}+e^{27} \log \left (3-\log \left (x^2\right )\right )}\right ) \, dx+6 \int e^{-11+e^{2 x}+x} x \, dx+6 \int e^{-27+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right ) \, dx\\ &=3 \int \frac {e^{5+e^{2 x}}}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )} \, dx-3 \int \frac {e^{5+e^{2 x}} x}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )} \, dx-3 \int \frac {e^{5+e^{2 x}} x \log \left (x^2\right ) \log \left (3-\log \left (x^2\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx+6 \int e^{-11+e^{2 x}+x} x \, dx+6 \int \frac {e^{5+e^{2 x}}}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx+6 \int e^{-27+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right ) \, dx-6 \int \frac {e^{e^{2 x}} x \log ^2\left (3-\log \left (x^2\right )\right )}{-e^{43+x}+e^{27} \log \left (3-\log \left (x^2\right )\right )} \, dx+9 \int \frac {e^{5+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.22, size = 31, normalized size = 1.00 \begin {gather*} -\frac {3 e^{5+e^{2 x}} x}{-e^{16+x}+\log \left (3-\log \left (x^2\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 4.20, size = 55, normalized size = 1.77
method | result | size |
risch | \(\frac {3 x \,{\mathrm e}^{5+{\mathrm e}^{2 x}}}{{\mathrm e}^{x +16}-\ln \left (3-2 \ln \left (x \right )+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )}\) | \(55\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.35, size = 26, normalized size = 0.84 \begin {gather*} \frac {3 \, x e^{\left (e^{\left (2 \, x\right )} + 5\right )}}{e^{\left (x + 16\right )} - \log \left (-2 \, \log \left (x\right ) + 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.41, size = 36, normalized size = 1.16 \begin {gather*} \frac {3 \, x e^{\left ({\left (5 \, e^{32} + e^{\left (2 \, x + 32\right )}\right )} e^{\left (-32\right )}\right )}}{e^{\left (x + 16\right )} - \log \left (-\log \left (x^{2}\right ) + 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 127 vs.
\(2 (28) = 56\).
time = 0.51, size = 127, normalized size = 4.10 \begin {gather*} -\frac {3 \, {\left (2 \, x^{2} e^{\left (x + e^{\left (2 \, x\right )} + 21\right )} \log \left (x \mathrm {sgn}\left (x\right )\right ) - 3 \, x^{2} e^{\left (x + e^{\left (2 \, x\right )} + 21\right )} - 2 \, x e^{\left (e^{\left (2 \, x\right )} + 5\right )}\right )}}{2 \, x e^{\left (x + 16\right )} \log \left (x \mathrm {sgn}\left (x\right )\right ) \log \left (-2 \, \log \left (x \mathrm {sgn}\left (x\right )\right ) + 3\right ) - 2 \, x e^{\left (2 \, x + 32\right )} \log \left (x \mathrm {sgn}\left (x\right )\right ) - 3 \, x e^{\left (x + 16\right )} \log \left (-2 \, \log \left (x \mathrm {sgn}\left (x\right )\right ) + 3\right ) + 3 \, x e^{\left (2 \, x + 32\right )} + 2 \, e^{\left (x + 16\right )} - 2 \, \log \left (-2 \, \log \left (x \mathrm {sgn}\left (x\right )\right ) + 3\right )} \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 = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,x}+5}\,\left (\ln \left (x^2\right )\,{\mathrm {e}}^{x+16}\,\left (6\,x\,{\mathrm {e}}^{2\,x}-3\,x+3\right )-{\mathrm {e}}^{x+16}\,\left (18\,x\,{\mathrm {e}}^{2\,x}-9\,x+9\right )+6\right )+{\mathrm {e}}^{{\mathrm {e}}^{2\,x}+5}\,\ln \left (3-\ln \left (x^2\right )\right )\,\left (18\,x\,{\mathrm {e}}^{2\,x}-\ln \left (x^2\right )\,\left (6\,x\,{\mathrm {e}}^{2\,x}+3\right )+9\right )}{\left (\ln \left (x^2\right )-3\right )\,{\ln \left (3-\ln \left (x^2\right )\right )}^2+\left (6\,{\mathrm {e}}^{x+16}-2\,\ln \left (x^2\right )\,{\mathrm {e}}^{x+16}\right )\,\ln \left (3-\ln \left (x^2\right )\right )-3\,{\mathrm {e}}^{2\,x+32}+\ln \left (x^2\right )\,{\mathrm {e}}^{2\,x+32}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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