Optimal. Leaf size=31 \[ 5 e^{-2 x}-\frac {4}{x+\frac {5}{258+\frac {2}{x}+\log (1-x)}} \]
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Rubi [F] time = 6.78, 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 {490 x^2+35630 x^3+629520 x^4-665640 x^5+e^{2 x} \left (-56-4072 x-262148 x^2+266256 x^3\right )+\left (140 x^3+5020 x^4-5160 x^5+e^{2 x} \left (-16 x-2048 x^2+2064 x^3\right )\right ) \log (1-x)+\left (10 x^4-10 x^5+e^{2 x} \left (-4 x^2+4 x^3\right )\right ) \log ^2(1-x)}{e^{2 x} \left (-49 x^2-3563 x^3-62952 x^4+66564 x^5\right )+e^{2 x} \left (-14 x^3-502 x^4+516 x^5\right ) \log (1-x)+e^{2 x} \left (-x^4+x^5\right ) \log ^2(1-x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{-2 x} \left (5 (-1+x) x^2 (7+258 x)^2-2 e^{2 x} \left (-14-1018 x-65537 x^2+66564 x^3\right )-2 (-1+x) x \left (-5 x^2 (7+258 x)+e^{2 x} (4+516 x)\right ) \log (1-x)-(-1+x) x^2 \left (2 e^{2 x}-5 x^2\right ) \log ^2(1-x)\right )}{(1-x) x^2 (7+258 x+x \log (1-x))^2} \, dx\\ &=2 \int \frac {e^{-2 x} \left (5 (-1+x) x^2 (7+258 x)^2-2 e^{2 x} \left (-14-1018 x-65537 x^2+66564 x^3\right )-2 (-1+x) x \left (-5 x^2 (7+258 x)+e^{2 x} (4+516 x)\right ) \log (1-x)-(-1+x) x^2 \left (2 e^{2 x}-5 x^2\right ) \log ^2(1-x)\right )}{(1-x) x^2 (7+258 x+x \log (1-x))^2} \, dx\\ &=2 \int \left (-5 e^{-2 x}+\frac {2 \left (-14-1018 x-65537 x^2+66564 x^3-4 x \log (1-x)-512 x^2 \log (1-x)+516 x^3 \log (1-x)-x^2 \log ^2(1-x)+x^3 \log ^2(1-x)\right )}{(-1+x) x^2 (7+258 x+x \log (1-x))^2}\right ) \, dx\\ &=4 \int \frac {-14-1018 x-65537 x^2+66564 x^3-4 x \log (1-x)-512 x^2 \log (1-x)+516 x^3 \log (1-x)-x^2 \log ^2(1-x)+x^3 \log ^2(1-x)}{(-1+x) x^2 (7+258 x+x \log (1-x))^2} \, dx-10 \int e^{-2 x} \, dx\\ &=5 e^{-2 x}+4 \int \left (\frac {1}{x^2}-\frac {5 \left (7-7 x+x^2\right )}{(-1+x) x^2 (7+258 x+x \log (1-x))^2}-\frac {10}{x^2 (7+258 x+x \log (1-x))}\right ) \, dx\\ &=5 e^{-2 x}-\frac {4}{x}-20 \int \frac {7-7 x+x^2}{(-1+x) x^2 (7+258 x+x \log (1-x))^2} \, dx-40 \int \frac {1}{x^2 (7+258 x+x \log (1-x))} \, dx\\ &=5 e^{-2 x}-\frac {4}{x}-20 \int \left (\frac {1}{(-1+x) (7+258 x+x \log (1-x))^2}-\frac {7}{x^2 (7+258 x+x \log (1-x))^2}\right ) \, dx-40 \int \frac {1}{x^2 (7+258 x+x \log (1-x))} \, dx\\ &=5 e^{-2 x}-\frac {4}{x}-20 \int \frac {1}{(-1+x) (7+258 x+x \log (1-x))^2} \, dx-40 \int \frac {1}{x^2 (7+258 x+x \log (1-x))} \, dx+140 \int \frac {1}{x^2 (7+258 x+x \log (1-x))^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 37, normalized size = 1.19 \begin {gather*} 2 \left (\frac {5 e^{-2 x}}{2}-\frac {2}{x}+\frac {10}{x (7+258 x+x \log (1-x))}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.85, size = 72, normalized size = 2.32 \begin {gather*} \frac {1290 \, x^{2} - 8 \, {\left (129 \, x + 1\right )} e^{\left (2 \, x\right )} + {\left (5 \, x^{2} - 4 \, x e^{\left (2 \, x\right )}\right )} \log \left (-x + 1\right ) + 35 \, x}{x^{2} e^{\left (2 \, x\right )} \log \left (-x + 1\right ) + {\left (258 \, x^{2} + 7 \, x\right )} e^{\left (2 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.66, size = 67, normalized size = 2.16 \begin {gather*} \frac {5 \, x^{2} e^{\left (-2 \, x\right )} \log \left (-x + 1\right ) + 1290 \, x^{2} e^{\left (-2 \, x\right )} + 35 \, x e^{\left (-2 \, x\right )} - 4 \, x \log \left (-x + 1\right ) - 1032 \, x - 8}{x^{2} \log \left (-x + 1\right ) + 258 \, x^{2} + 7 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 40, normalized size = 1.29
method | result | size |
risch | \(\frac {\left (-4 \,{\mathrm e}^{2 x}+5 x \right ) {\mathrm e}^{-2 x}}{x}+\frac {20}{x \left (x \ln \left (1-x \right )+258 x +7\right )}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.77, size = 72, normalized size = 2.32 \begin {gather*} \frac {1290 \, x^{2} - 8 \, {\left (129 \, x + 1\right )} e^{\left (2 \, x\right )} + {\left (5 \, x^{2} - 4 \, x e^{\left (2 \, x\right )}\right )} \log \left (-x + 1\right ) + 35 \, x}{x^{2} e^{\left (2 \, x\right )} \log \left (-x + 1\right ) + {\left (258 \, x^{2} + 7 \, x\right )} e^{\left (2 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.84, size = 32, normalized size = 1.03 \begin {gather*} 5\,{\mathrm {e}}^{-2\,x}+\frac {20}{x\,\left (258\,x+x\,\ln \left (1-x\right )+7\right )}-\frac {4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.55, size = 27, normalized size = 0.87 \begin {gather*} 5 e^{- 2 x} + \frac {20}{x^{2} \log {\left (1 - x \right )} + 258 x^{2} + 7 x} - \frac {4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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