Optimal. Leaf size=27 \[ \frac {4-e^4+x}{-\frac {2 e^{-x}}{5 x^3}+x^2} \]
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Rubi [F]
time = 1.73, 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^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^x x^2 \left (10 e^{4+x} x^5+2 e^4 (3+x)-5 e^x x^5 (8+x)-2 \left (12+8 x+x^2\right )\right )}{\left (2-5 e^x x^5\right )^2} \, dx\\ &=5 \int \frac {e^x x^2 \left (10 e^{4+x} x^5+2 e^4 (3+x)-5 e^x x^5 (8+x)-2 \left (12+8 x+x^2\right )\right )}{\left (2-5 e^x x^5\right )^2} \, dx\\ &=5 \int \left (\frac {2 e^x x^2 \left (-5 \left (4-e^4\right )-\left (9-e^4\right ) x-x^2\right )}{\left (2-5 e^x x^5\right )^2}-\frac {e^x x^2 \left (8-2 e^4+x\right )}{-2+5 e^x x^5}\right ) \, dx\\ &=-\left (5 \int \frac {e^x x^2 \left (8-2 e^4+x\right )}{-2+5 e^x x^5} \, dx\right )+10 \int \frac {e^x x^2 \left (-5 \left (4-e^4\right )-\left (9-e^4\right ) x-x^2\right )}{\left (2-5 e^x x^5\right )^2} \, dx\\ &=-\left (5 \int \left (-\frac {2 e^x \left (-4+e^4\right ) x^2}{-2+5 e^x x^5}+\frac {e^x x^3}{-2+5 e^x x^5}\right ) \, dx\right )+10 \int \left (\frac {5 e^x \left (-2+e^2\right ) \left (2+e^2\right ) x^2}{\left (-2+5 e^x x^5\right )^2}+\frac {e^x \left (-9+e^4\right ) x^3}{\left (-2+5 e^x x^5\right )^2}-\frac {e^x x^4}{\left (-2+5 e^x x^5\right )^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^x x^3}{-2+5 e^x x^5} \, dx\right )-10 \int \frac {e^x x^4}{\left (-2+5 e^x x^5\right )^2} \, dx-\left (10 \left (4-e^4\right )\right ) \int \frac {e^x x^2}{-2+5 e^x x^5} \, dx-\left (50 \left (4-e^4\right )\right ) \int \frac {e^x x^2}{\left (-2+5 e^x x^5\right )^2} \, dx-\left (10 \left (9-e^4\right )\right ) \int \frac {e^x x^3}{\left (-2+5 e^x x^5\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 5.54, size = 28, normalized size = 1.04 \begin {gather*} -\frac {5 e^x \left (-4+e^4-x\right ) x^3}{-2+5 e^x x^5} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.14, size = 33, normalized size = 1.22
method | result | size |
norman | \(\frac {\left (20-5 \,{\mathrm e}^{4}\right ) x^{3} {\mathrm e}^{x}+5 \,{\mathrm e}^{x} x^{4}}{5 x^{5} {\mathrm e}^{x}-2}\) | \(33\) |
risch | \(\frac {4-{\mathrm e}^{4}+x}{x^{2}}-\frac {2 \left ({\mathrm e}^{4}-x -4\right )}{x^{2} \left (5 x^{5} {\mathrm e}^{x}-2\right )}\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 28, normalized size = 1.04 \begin {gather*} \frac {5 \, {\left (x^{4} - x^{3} {\left (e^{4} - 4\right )}\right )} e^{x}}{5 \, x^{5} e^{x} - 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 31, normalized size = 1.15 \begin {gather*} \frac {5 \, {\left (x^{4} - x^{3} e^{4} + 4 \, x^{3}\right )} e^{x}}{5 \, x^{5} e^{x} - 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.12, size = 32, normalized size = 1.19 \begin {gather*} \frac {2 x - 2 e^{4} + 8}{5 x^{7} e^{x} - 2 x^{2}} - \frac {- x - 4 + e^{4}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 36, normalized size = 1.33 \begin {gather*} \frac {5 \, {\left (x^{4} e^{x} - x^{3} e^{\left (x + 4\right )} + 4 \, x^{3} e^{x}\right )}}{5 \, x^{5} e^{x} - 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.38, size = 25, normalized size = 0.93 \begin {gather*} \frac {5\,x^3\,{\mathrm {e}}^x\,\left (x-{\mathrm {e}}^4+4\right )}{5\,x^5\,{\mathrm {e}}^x-2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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