Optimal. Leaf size=30 \[ -4+3 e^{\frac {3}{5 \left (-e^x+5 x\right ) \left (2-x^2\right )}} \]
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Rubi [F]
time = 5.28, 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^{\frac {3}{50 x-25 x^3+e^x \left (-10+5 x^2\right )}} \left (-90+135 x^2+e^x \left (18-18 x-9 x^2\right )\right )}{500 x^2-500 x^4+125 x^6+e^{2 x} \left (20-20 x^2+5 x^4\right )+e^x \left (-200 x+200 x^3-50 x^5\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 {9 e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} \left (-10+15 x^2-e^x \left (-2+2 x+x^2\right )\right )}{5 \left (e^x-5 x\right )^2 \left (2-x^2\right )^2} \, dx\\ &=\frac {9}{5} \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} \left (-10+15 x^2-e^x \left (-2+2 x+x^2\right )\right )}{\left (e^x-5 x\right )^2 \left (2-x^2\right )^2} \, dx\\ &=\frac {9}{5} \int \left (-\frac {5 e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} (-1+x)}{\left (e^x-5 x\right )^2 \left (-2+x^2\right )}-\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} \left (-2+2 x+x^2\right )}{\left (e^x-5 x\right ) \left (-2+x^2\right )^2}\right ) \, dx\\ &=-\left (\frac {9}{5} \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} \left (-2+2 x+x^2\right )}{\left (e^x-5 x\right ) \left (-2+x^2\right )^2} \, dx\right )-9 \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} (-1+x)}{\left (e^x-5 x\right )^2 \left (-2+x^2\right )} \, dx\\ &=-\left (\frac {9}{5} \int \left (\frac {2 e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} x}{\left (e^x-5 x\right ) \left (-2+x^2\right )^2}+\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right ) \left (-2+x^2\right )}\right ) \, dx\right )-9 \int \left (-\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right )^2 \left (-2+x^2\right )}+\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} x}{\left (e^x-5 x\right )^2 \left (-2+x^2\right )}\right ) \, dx\\ &=-\left (\frac {9}{5} \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right ) \left (-2+x^2\right )} \, dx\right )-\frac {18}{5} \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} x}{\left (e^x-5 x\right ) \left (-2+x^2\right )^2} \, dx+9 \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right )^2 \left (-2+x^2\right )} \, dx-9 \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} x}{\left (e^x-5 x\right )^2 \left (-2+x^2\right )} \, dx\\ &=-\left (\frac {9}{5} \int \left (-\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{2 \sqrt {2} \left (e^x-5 x\right ) \left (\sqrt {2}-x\right )}-\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{2 \sqrt {2} \left (e^x-5 x\right ) \left (\sqrt {2}+x\right )}\right ) \, dx\right )-\frac {18}{5} \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} x}{\left (e^x-5 x\right ) \left (-2+x^2\right )^2} \, dx-9 \int \left (-\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{2 \left (e^x-5 x\right )^2 \left (\sqrt {2}-x\right )}+\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{2 \left (e^x-5 x\right )^2 \left (\sqrt {2}+x\right )}\right ) \, dx+9 \int \left (-\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{2 \sqrt {2} \left (e^x-5 x\right )^2 \left (\sqrt {2}-x\right )}-\frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{2 \sqrt {2} \left (e^x-5 x\right )^2 \left (\sqrt {2}+x\right )}\right ) \, dx\\ &=-\left (\frac {18}{5} \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} x}{\left (e^x-5 x\right ) \left (-2+x^2\right )^2} \, dx\right )+\frac {9}{2} \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right )^2 \left (\sqrt {2}-x\right )} \, dx-\frac {9}{2} \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right )^2 \left (\sqrt {2}+x\right )} \, dx+\frac {9 \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right ) \left (\sqrt {2}-x\right )} \, dx}{10 \sqrt {2}}+\frac {9 \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right ) \left (\sqrt {2}+x\right )} \, dx}{10 \sqrt {2}}-\frac {9 \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right )^2 \left (\sqrt {2}-x\right )} \, dx}{2 \sqrt {2}}-\frac {9 \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}}}{\left (e^x-5 x\right )^2 \left (\sqrt {2}+x\right )} \, dx}{2 \sqrt {2}}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.19, size = 24, normalized size = 0.80 \begin {gather*} 3 e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 21, normalized size = 0.70
method | result | size |
risch | \(3 \,{\mathrm e}^{\frac {3}{5 \left (x^{2}-2\right ) \left ({\mathrm e}^{x}-5 x \right )}}\) | \(21\) |
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. 76 vs.
\(2 (24) = 48\).
time = 0.51, size = 76, normalized size = 2.53 \begin {gather*} 3 \, e^{\left (-\frac {3 \, x}{50 \, x^{2} - {\left (x^{2} - 2\right )} e^{\left (2 \, x\right )} - 100} - \frac {3 \, e^{x}}{5 \, {\left (50 \, x^{2} - {\left (x^{2} - 2\right )} e^{\left (2 \, x\right )} - 100\right )}} - \frac {15}{5 \, x e^{\left (2 \, x\right )} - 250 \, x - e^{\left (3 \, x\right )} + 50 \, e^{x}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 25, normalized size = 0.83 \begin {gather*} 3 \, e^{\left (-\frac {3}{5 \, {\left (5 \, x^{3} - {\left (x^{2} - 2\right )} e^{x} - 10 \, x\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.31, size = 22, normalized size = 0.73 \begin {gather*} 3 e^{\frac {3}{- 25 x^{3} + 50 x + \left (5 x^{2} - 10\right ) e^{x}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 27, normalized size = 0.90 \begin {gather*} 3 \, e^{\left (-\frac {3}{5 \, {\left (5 \, x^{3} - x^{2} e^{x} - 10 \, x + 2 \, e^{x}\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.07, size = 27, normalized size = 0.90 \begin {gather*} 3\,{\mathrm {e}}^{\frac {3}{5\,\left (10\,x-2\,{\mathrm {e}}^x+x^2\,{\mathrm {e}}^x-5\,x^3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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