Optimal. Leaf size=36 \[ \log \left (-2+x^2-\frac {x^2 \left (5-e^x+x\right )}{4-e^{\frac {5}{5+x^2}}}\right ) \]
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Rubi [F]
time = 132.05, 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 {200 x+300 x^2+80 x^3+120 x^4+8 x^5+12 x^6+e^{\frac {10}{5+x^2}} \left (-50 x-20 x^3-2 x^5\right )+e^{\frac {5}{5+x^2}} \left (150 x-75 x^2+10 x^3-40 x^4+6 x^5-3 x^6\right )+e^x \left (-200 x-100 x^2-80 x^3-40 x^4-8 x^5-4 x^6+e^{\frac {5}{5+x^2}} \left (50 x+25 x^2+30 x^3+10 x^4+2 x^5+x^6\right )\right )}{800+420 x^2+100 x^3+72 x^4+40 x^5+4 x^6+4 x^7+e^{\frac {10}{5+x^2}} \left (50-5 x^2-8 x^4-x^6\right )+e^{\frac {5}{5+x^2}} \left (-400-85 x^2-25 x^3+14 x^4-10 x^5+3 x^6-x^7\right )+e^x \left (-100 x^2-40 x^4-4 x^6+e^{\frac {5}{5+x^2}} \left (25 x^2+10 x^4+x^6\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 {x \left (-2 e^{\frac {10}{5+x^2}} \left (5+x^2\right )^2-4 e^x (2+x) \left (5+x^2\right )^2+4 (2+3 x) \left (5+x^2\right )^2+e^{\frac {5}{5+x^2}} \left (150-75 x+10 x^2-40 x^3+6 x^4-3 x^5\right )+e^{x+\frac {5}{5+x^2}} \left (50+25 x+30 x^2+10 x^3+2 x^4+x^5\right )\right )}{\left (4-e^{\frac {5}{5+x^2}}\right ) \left (5+x^2\right )^2 \left (8+x^2-e^x x^2+x^3+e^{\frac {5}{5+x^2}} \left (-2+x^2\right )\right )} \, dx\\ &=\int \left (\frac {2 x}{-2+x^2}+\frac {40 x}{\left (-4+e^{\frac {5}{5+x^2}}\right ) \left (5+x^2\right )^2}-\frac {x \left (660-100 e^x+150 x-50 e^x x+140 x^2-60 e^x x^2+55 x^3+5 e^x x^3+10 x^4+6 e^x x^4-14 x^5+8 e^x x^5-x^7+e^x x^7\right )}{\left (-2+x^2\right ) \left (5+x^2\right )^2 \left (8-2 e^{\frac {5}{5+x^2}}+x^2-e^x x^2+e^{\frac {5}{5+x^2}} x^2+x^3\right )}\right ) \, dx\\ &=2 \int \frac {x}{-2+x^2} \, dx+40 \int \frac {x}{\left (-4+e^{\frac {5}{5+x^2}}\right ) \left (5+x^2\right )^2} \, dx-\int \frac {x \left (660-100 e^x+150 x-50 e^x x+140 x^2-60 e^x x^2+55 x^3+5 e^x x^3+10 x^4+6 e^x x^4-14 x^5+8 e^x x^5-x^7+e^x x^7\right )}{\left (-2+x^2\right ) \left (5+x^2\right )^2 \left (8-2 e^{\frac {5}{5+x^2}}+x^2-e^x x^2+e^{\frac {5}{5+x^2}} x^2+x^3\right )} \, dx\\ &=\log \left (2-x^2\right )+20 \text {Subst}\left (\int \frac {1}{\left (-4+e^{\frac {5}{5+x}}\right ) (5+x)^2} \, dx,x,x^2\right )-\int \left (\frac {x \left (660-100 e^x+150 x-50 e^x x+140 x^2-60 e^x x^2+55 x^3+5 e^x x^3+10 x^4+6 e^x x^4-14 x^5+8 e^x x^5-x^7+e^x x^7\right )}{49 \left (-2+x^2\right ) \left (8-2 e^{\frac {5}{5+x^2}}+x^2-e^x x^2+e^{\frac {5}{5+x^2}} x^2+x^3\right )}-\frac {x \left (660-100 e^x+150 x-50 e^x x+140 x^2-60 e^x x^2+55 x^3+5 e^x x^3+10 x^4+6 e^x x^4-14 x^5+8 e^x x^5-x^7+e^x x^7\right )}{7 \left (5+x^2\right )^2 \left (8-2 e^{\frac {5}{5+x^2}}+x^2-e^x x^2+e^{\frac {5}{5+x^2}} x^2+x^3\right )}-\frac {x \left (660-100 e^x+150 x-50 e^x x+140 x^2-60 e^x x^2+55 x^3+5 e^x x^3+10 x^4+6 e^x x^4-14 x^5+8 e^x x^5-x^7+e^x x^7\right )}{49 \left (5+x^2\right ) \left (8-2 e^{\frac {5}{5+x^2}}+x^2-e^x x^2+e^{\frac {5}{5+x^2}} x^2+x^3\right )}\right ) \, dx\\ &=\log \left (2-x^2\right )-\frac {1}{49} \int \frac {x \left (660-100 e^x+150 x-50 e^x x+140 x^2-60 e^x x^2+55 x^3+5 e^x x^3+10 x^4+6 e^x x^4-14 x^5+8 e^x x^5-x^7+e^x x^7\right )}{\left (-2+x^2\right ) \left (8-2 e^{\frac {5}{5+x^2}}+x^2-e^x x^2+e^{\frac {5}{5+x^2}} x^2+x^3\right )} \, dx+\frac {1}{49} \int \frac {x \left (660-100 e^x+150 x-50 e^x x+140 x^2-60 e^x x^2+55 x^3+5 e^x x^3+10 x^4+6 e^x x^4-14 x^5+8 e^x x^5-x^7+e^x x^7\right )}{\left (5+x^2\right ) \left (8-2 e^{\frac {5}{5+x^2}}+x^2-e^x x^2+e^{\frac {5}{5+x^2}} x^2+x^3\right )} \, dx+\frac {1}{7} \int \frac {x \left (660-100 e^x+150 x-50 e^x x+140 x^2-60 e^x x^2+55 x^3+5 e^x x^3+10 x^4+6 e^x x^4-14 x^5+8 e^x x^5-x^7+e^x x^7\right )}{\left (5+x^2\right )^2 \left (8-2 e^{\frac {5}{5+x^2}}+x^2-e^x x^2+e^{\frac {5}{5+x^2}} x^2+x^3\right )} \, dx+20 \text {Subst}\left (\int \frac {1}{\left (-4+e^{5/x}\right ) x^2} \, dx,x,5+x^2\right )\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.20, size = 64, normalized size = 1.78 \begin {gather*} -\log \left (4-e^{\frac {5}{5+x^2}}\right )+\log \left (8-2 e^{\frac {5}{5+x^2}}+x^2-e^x x^2+e^{\frac {5}{5+x^2}} x^2+x^3\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 58, normalized size = 1.61
method | result | size |
risch | \(\ln \left (x^{2}-2\right )+\ln \left ({\mathrm e}^{\frac {5}{x^{2}+5}}+\frac {x^{3}-{\mathrm e}^{x} x^{2}+x^{2}+8}{x^{2}-2}\right )-\ln \left ({\mathrm e}^{\frac {5}{x^{2}+5}}-4\right )\) | \(58\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.58, size = 62, normalized size = 1.72 \begin {gather*} \log \left (x^{2} - 2\right ) + \log \left (\frac {x^{3} - x^{2} e^{x} + x^{2} + {\left (x^{2} - 2\right )} e^{\left (\frac {5}{x^{2} + 5}\right )} + 8}{x^{2} - 2}\right ) - \log \left (e^{\left (\frac {5}{x^{2} + 5}\right )} - 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 57, normalized size = 1.58 \begin {gather*} 2 \, \log \left (x\right ) + \log \left (-\frac {x^{3} - x^{2} e^{x} + x^{2} + {\left (x^{2} - 2\right )} e^{\left (\frac {5}{x^{2} + 5}\right )} + 8}{x^{2}}\right ) - \log \left (e^{\left (\frac {5}{x^{2} + 5}\right )} - 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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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Mupad [B]
time = 4.82, size = 528, normalized size = 14.67 \begin {gather*} \ln \left (x^7+8\,x^5+6\,x^4+5\,x^3-60\,x^2-50\,x-100\right )-\ln \left (\frac {5280\,x-400\,x^2\,{\mathrm {e}}^x-480\,x^3\,{\mathrm {e}}^x+40\,x^4\,{\mathrm {e}}^x+48\,x^5\,{\mathrm {e}}^x+64\,x^6\,{\mathrm {e}}^x+8\,x^8\,{\mathrm {e}}^x-1320\,x\,{\mathrm {e}}^{\frac {5}{x^2+5}}-300\,x^2\,{\mathrm {e}}^{\frac {5}{x^2+5}}-280\,x^3\,{\mathrm {e}}^{\frac {5}{x^2+5}}-110\,x^4\,{\mathrm {e}}^{\frac {5}{x^2+5}}-20\,x^5\,{\mathrm {e}}^{\frac {5}{x^2+5}}+28\,x^6\,{\mathrm {e}}^{\frac {5}{x^2+5}}+2\,x^8\,{\mathrm {e}}^{\frac {5}{x^2+5}}-800\,x\,{\mathrm {e}}^x+1200\,x^2+1120\,x^3+440\,x^4+80\,x^5-112\,x^6-8\,x^8+200\,x\,{\mathrm {e}}^{\frac {5}{x^2+5}}\,{\mathrm {e}}^x+100\,x^2\,{\mathrm {e}}^{\frac {5}{x^2+5}}\,{\mathrm {e}}^x+120\,x^3\,{\mathrm {e}}^{\frac {5}{x^2+5}}\,{\mathrm {e}}^x-10\,x^4\,{\mathrm {e}}^{\frac {5}{x^2+5}}\,{\mathrm {e}}^x-12\,x^5\,{\mathrm {e}}^{\frac {5}{x^2+5}}\,{\mathrm {e}}^x-16\,x^6\,{\mathrm {e}}^{\frac {5}{x^2+5}}\,{\mathrm {e}}^x-2\,x^8\,{\mathrm {e}}^{\frac {5}{x^2+5}}\,{\mathrm {e}}^x}{x^8+6\,x^6-11\,x^4-60\,x^2+100}\right )-\ln \left (x^2-2\right )+\ln \left (\frac {150\,x-100\,{\mathrm {e}}^x-60\,x^2\,{\mathrm {e}}^x+5\,x^3\,{\mathrm {e}}^x+6\,x^4\,{\mathrm {e}}^x+8\,x^5\,{\mathrm {e}}^x+x^7\,{\mathrm {e}}^x-50\,x\,{\mathrm {e}}^x+140\,x^2+55\,x^3+10\,x^4-14\,x^5-x^7+660}{x^7+8\,x^5+6\,x^4+5\,x^3-60\,x^2-50\,x-100}\right )+\ln \left (\frac {x\,\left (x^2\,{\mathrm {e}}^{\frac {5}{x^2+5}}-x^2\,{\mathrm {e}}^x-2\,{\mathrm {e}}^{\frac {5}{x^2+5}}+x^2+x^3+8\right )}{\left (x^2-2\right )\,{\left (x^2+5\right )}^2}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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