Optimal. Leaf size=30 \[ e^x-x-\log \left (-2 x+e^4 x \left (x-\frac {1}{5} (3+x)^2\right )\right ) \]
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Rubi [A]
time = 0.62, antiderivative size = 33, normalized size of antiderivative = 1.10, number of steps
used = 7, number of rules used = 5, integrand size = 139, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6820, 6860,
2225, 1642, 642} \begin {gather*} -\log \left (e^4 x^2+e^4 x+9 e^4+10\right )-x+e^x-\log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 642
Rule 1642
Rule 2225
Rule 6820
Rule 6860
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 e^x x-10 (1+x)+e^{4+x} x \left (9+x+x^2\right )-e^4 \left (9+11 x+4 x^2+x^3\right )}{x \left (10+9 e^4+e^4 x+e^4 x^2\right )} \, dx\\ &=\int \left (e^x+\frac {-10-9 e^4-\left (10+11 e^4\right ) x-4 e^4 x^2-e^4 x^3}{x \left (10+9 e^4+e^4 x+e^4 x^2\right )}\right ) \, dx\\ &=\int e^x \, dx+\int \frac {-10-9 e^4-\left (10+11 e^4\right ) x-4 e^4 x^2-e^4 x^3}{x \left (10+9 e^4+e^4 x+e^4 x^2\right )} \, dx\\ &=e^x+\int \left (-1-\frac {1}{x}-\frac {e^4 (1+2 x)}{10+9 e^4+e^4 x+e^4 x^2}\right ) \, dx\\ &=e^x-x-\log (x)-e^4 \int \frac {1+2 x}{10+9 e^4+e^4 x+e^4 x^2} \, dx\\ &=e^x-x-\log (x)-\log \left (10+9 e^4+e^4 x+e^4 x^2\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 2.51, size = 26, normalized size = 0.87 \begin {gather*} e^x-x-\log (x)-\log \left (10+e^4 \left (9+x+x^2\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.46, size = 30, normalized size = 1.00
method | result | size |
default | \(-x -\ln \left (x \right )-\ln \left (x^{2} {\mathrm e}^{4}+x \,{\mathrm e}^{4}+9 \,{\mathrm e}^{4}+10\right )+{\mathrm e}^{x}\) | \(30\) |
norman | \(\frac {{\mathrm e}^{x} x -x^{2}}{x}-\ln \left (x \right )-\ln \left (x^{2} {\mathrm e}^{4}+x \,{\mathrm e}^{4}+9 \,{\mathrm e}^{4}+10\right )\) | \(39\) |
risch | \(-x -\ln \left (x^{3}+x^{2}+9 x \right )+{\mathrm e}^{x}-\ln \left (5\right )+4+\ln \left (x \right )+\ln \left (x^{2}+x +9\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \left (x^{2}+x +9\right )\right ) \left (-\mathrm {csgn}\left (i x \left (x^{2}+x +9\right )\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \left (x^{2}+x +9\right )\right )+\mathrm {csgn}\left (i \left (x^{2}+x +9\right )\right )\right )}{2}+i \pi \mathrm {csgn}\left (i x \left (x^{2}+x +9\right )\right )^{2} \left (\mathrm {csgn}\left (i x \left (x^{2}+x +9\right )\right )-1\right )-\ln \left (2 x -\frac {x \left (x^{2}+x +9\right ) {\mathrm e}^{4} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \left (x^{2}+x +9\right )\right )^{3}}{2}} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \left (x^{2}+x +9\right )\right )^{2} \mathrm {csgn}\left (i x \right )}{2}} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \left (x^{2}+x +9\right )\right )^{2} \mathrm {csgn}\left (i \left (x^{2}+x +9\right )\right )}{2}} {\mathrm e}^{-\frac {i \pi \,\mathrm {csgn}\left (i x \left (x^{2}+x +9\right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x^{2}+x +9\right )\right )}{2}}}{5}\right )\) | \(240\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 703 vs.
\(2 (29) = 58\).
time = 0.56, size = 703, normalized size = 23.43 \begin {gather*} -\frac {{\left (17 \, e^{4} + 20\right )} \arctan \left (\frac {{\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}}\right ) e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}} + \frac {1}{700} \, {\left (35 \, {\left (17 \, e^{4} + 20\right )} e^{\left (-4\right )} \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right ) + \frac {70 \, {\left (127 \, e^{8} + 320 \, e^{4} + 200\right )} \arctan \left (\frac {{\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}}\right ) e^{\left (-6\right )}}{\sqrt {35 \, e^{4} + 40}} - 700 \, x e^{\left (-4\right )} - 254 \, \sqrt {35} \arctan \left (\frac {1}{35} \, \sqrt {35} {\left (2 \, x + 1\right )}\right ) - 595 \, \log \left (x^{2} + x + 9\right )\right )} e^{4} - \frac {1}{70} \, {\left (35 \, {\left (4 \, e^{4} + 5\right )} e^{\left (-4\right )} \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right ) - \frac {70 \, {\left (13 \, e^{4} + 15\right )} \arctan \left (\frac {{\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}}\right ) e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}} + 26 \, \sqrt {35} \arctan \left (\frac {1}{35} \, \sqrt {35} {\left (2 \, x + 1\right )}\right ) - 140 \, \log \left (x^{2} + x + 9\right )\right )} e^{4} - \frac {6}{175} \, {\left (\frac {70 \, {\left (17 \, e^{4} + 20\right )} \arctan \left (\frac {{\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}}\right ) e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}} - 34 \, \sqrt {35} \arctan \left (\frac {1}{35} \, \sqrt {35} {\left (2 \, x + 1\right )}\right ) + 35 \, \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right ) - 35 \, \log \left (x^{2} + x + 9\right )\right )} e^{4} + \frac {2}{25} \, {\left (2 \, \sqrt {35} \arctan \left (\frac {1}{35} \, \sqrt {35} {\left (2 \, x + 1\right )}\right ) - \frac {70 \, \arctan \left (\frac {{\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}}\right ) e^{2}}{\sqrt {35 \, e^{4} + 40}} + 35 \, \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right ) - 35 \, \log \left (x^{2} + x + 9\right )\right )} e^{4} + \frac {9}{700} \, {\left (2 \, \sqrt {35} \arctan \left (\frac {1}{35} \, \sqrt {35} {\left (2 \, x + 1\right )}\right ) - \frac {315 \, e^{4} \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right )}{9 \, e^{4} + 10} - \frac {630 \, \arctan \left (\frac {{\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}}\right ) e^{6}}{\sqrt {35 \, e^{4} + 40} {\left (9 \, e^{4} + 10\right )}} - \frac {700 \, \log \left (x\right )}{9 \, e^{4} + 10} + 35 \, \log \left (x^{2} + x + 9\right )\right )} e^{4} - \frac {108}{175} \, {\left (\sqrt {35} \arctan \left (\frac {1}{35} \, \sqrt {35} {\left (2 \, x + 1\right )}\right ) - \frac {35 \, \arctan \left (\frac {{\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}}\right ) e^{2}}{\sqrt {35 \, e^{4} + 40}}\right )} e^{4} + \frac {18 \, \arctan \left (\frac {{\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}}\right ) e^{2}}{\sqrt {35 \, e^{4} + 40}} - \frac {9 \, e^{4} \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right )}{2 \, {\left (9 \, e^{4} + 10\right )}} - \frac {9 \, \arctan \left (\frac {{\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2\right )}}{\sqrt {35 \, e^{4} + 40}}\right ) e^{6}}{\sqrt {35 \, e^{4} + 40} {\left (9 \, e^{4} + 10\right )}} - \frac {10 \, \log \left (x\right )}{9 \, e^{4} + 10} + e^{x} + \frac {1}{2} \, \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.41, size = 26, normalized size = 0.87 \begin {gather*} -x + e^{x} - \log \left ({\left (x^{3} + x^{2} + 9 \, x\right )} e^{4} + 10 \, x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.35, size = 27, normalized size = 0.90 \begin {gather*} - x + e^{x} - \log {\left (x^{3} e^{4} + x^{2} e^{4} + x \left (10 + 9 e^{4}\right ) \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.44, size = 40, normalized size = 1.33 \begin {gather*} -{\left (x e^{4} + e^{4} \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right ) + e^{4} \log \left (x\right ) - e^{\left (x + 4\right )}\right )} e^{\left (-4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.38, size = 24, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^x-\ln \left (9\,x+10\,x\,{\mathrm {e}}^{-4}+x^2+x^3\right )-x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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