Optimal. Leaf size=28 \[ -2+\frac {2 e^2 (4+x)^2}{25 \left (5-e^{-3+x}\right )^2 x} \]
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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 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, 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^8 (4+x) \left (5 e^3 (-4+x)+e^x \left (4+7 x+2 x^2\right )\right )}{25 \left (5 e^3-e^x\right )^3 x^2} \, dx\\ &=\frac {1}{25} \left (2 e^8\right ) \int \frac {(4+x) \left (5 e^3 (-4+x)+e^x \left (4+7 x+2 x^2\right )\right )}{\left (5 e^3-e^x\right )^3 x^2} \, dx\\ &=\frac {1}{25} \left (2 e^8\right ) \int \left (\frac {10 e^3 (4+x)^2}{\left (5 e^3-e^x\right )^3 x}-\frac {16+32 x+15 x^2+2 x^3}{\left (-5 e^3+e^x\right )^2 x^2}\right ) \, dx\\ &=-\left (\frac {1}{25} \left (2 e^8\right ) \int \frac {16+32 x+15 x^2+2 x^3}{\left (-5 e^3+e^x\right )^2 x^2} \, dx\right )+\frac {1}{5} \left (4 e^{11}\right ) \int \frac {(4+x)^2}{\left (5 e^3-e^x\right )^3 x} \, dx\\ &=-\left (\frac {1}{25} \left (2 e^8\right ) \int \left (\frac {15}{\left (-5 e^3+e^x\right )^2}+\frac {16}{\left (-5 e^3+e^x\right )^2 x^2}+\frac {32}{\left (-5 e^3+e^x\right )^2 x}+\frac {2 x}{\left (-5 e^3+e^x\right )^2}\right ) \, dx\right )+\frac {1}{5} \left (4 e^{11}\right ) \int \left (-\frac {8}{\left (-5 e^3+e^x\right )^3}-\frac {16}{\left (-5 e^3+e^x\right )^3 x}-\frac {x}{\left (-5 e^3+e^x\right )^3}\right ) \, dx\\ &=-\left (\frac {1}{25} \left (4 e^8\right ) \int \frac {x}{\left (-5 e^3+e^x\right )^2} \, dx\right )-\frac {1}{5} \left (6 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2} \, dx-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (4 e^{11}\right ) \int \frac {x}{\left (-5 e^3+e^x\right )^3} \, dx-\frac {1}{5} \left (32 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx\\ &=-\left (\frac {1}{125} \left (4 e^5\right ) \int \frac {e^x x}{\left (-5 e^3+e^x\right )^2} \, dx\right )+\frac {1}{125} \left (4 e^5\right ) \int \frac {x}{-5 e^3+e^x} \, dx-\frac {1}{25} \left (4 e^8\right ) \int \frac {e^x x}{\left (-5 e^3+e^x\right )^3} \, dx+\frac {1}{25} \left (4 e^8\right ) \int \frac {x}{\left (-5 e^3+e^x\right )^2} \, dx-\frac {1}{5} \left (6 e^8\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )^2} \, dx,x,e^x\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (32 e^{11}\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )^3} \, dx,x,e^x\right )-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx\\ &=\frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}-\frac {4 e^5 x}{125 \left (5 e^3-e^x\right )}-\frac {2 e^2 x^2}{625}+\frac {1}{625} \left (4 e^2\right ) \int \frac {e^x x}{-5 e^3+e^x} \, dx-\frac {1}{125} \left (4 e^5\right ) \int \frac {1}{-5 e^3+e^x} \, dx+\frac {1}{125} \left (4 e^5\right ) \int \frac {e^x x}{\left (-5 e^3+e^x\right )^2} \, dx-\frac {1}{125} \left (4 e^5\right ) \int \frac {x}{-5 e^3+e^x} \, dx-\frac {1}{25} \left (2 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2} \, dx-\frac {1}{5} \left (6 e^8\right ) \operatorname {Subst}\left (\int \left (\frac {1}{5 e^3 \left (5 e^3-x\right )^2}+\frac {1}{25 e^6 \left (5 e^3-x\right )}+\frac {1}{25 e^6 x}\right ) \, dx,x,e^x\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (32 e^{11}\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{5 e^3 \left (5 e^3-x\right )^3}-\frac {1}{25 e^6 \left (5 e^3-x\right )^2}-\frac {1}{125 e^9 \left (5 e^3-x\right )}-\frac {1}{125 e^9 x}\right ) \, dx,x,e^x\right )-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx\\ &=\frac {16 e^8}{25 \left (5 e^3-e^x\right )^2}+\frac {2 e^5}{125 \left (5 e^3-e^x\right )}+\frac {2 e^2 x}{625}+\frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}+\frac {4}{625} e^2 x \log \left (1-\frac {e^{-3+x}}{5}\right )-\frac {2}{625} e^2 \log \left (5 e^3-e^x\right )-\frac {1}{625} \left (4 e^2\right ) \int \frac {e^x x}{-5 e^3+e^x} \, dx-\frac {1}{625} \left (4 e^2\right ) \int \log \left (1-\frac {e^{-3+x}}{5}\right ) \, dx+\frac {1}{125} \left (4 e^5\right ) \int \frac {1}{-5 e^3+e^x} \, dx-\frac {1}{125} \left (4 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )} \, dx,x,e^x\right )-\frac {1}{25} \left (2 e^8\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )^2} \, dx,x,e^x\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx\\ &=\frac {16 e^8}{25 \left (5 e^3-e^x\right )^2}+\frac {2 e^5}{125 \left (5 e^3-e^x\right )}+\frac {2 e^2 x}{625}+\frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}-\frac {2}{625} e^2 \log \left (5 e^3-e^x\right )+\frac {1}{625} \left (4 e^2\right ) \int \log \left (1-\frac {e^{-3+x}}{5}\right ) \, dx+\frac {1}{625} \left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )-\frac {1}{625} \left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{-5 e^3+x} \, dx,x,e^x\right )-\frac {1}{625} \left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{5}\right )}{x} \, dx,x,e^{-3+x}\right )+\frac {1}{125} \left (4 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )} \, dx,x,e^x\right )-\frac {1}{25} \left (2 e^8\right ) \operatorname {Subst}\left (\int \left (\frac {1}{5 e^3 \left (5 e^3-x\right )^2}+\frac {1}{25 e^6 \left (5 e^3-x\right )}+\frac {1}{25 e^6 x}\right ) \, dx,x,e^x\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx\\ &=\frac {16 e^8}{25 \left (5 e^3-e^x\right )^2}+\frac {4 e^2 x}{625}+\frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}-\frac {4}{625} e^2 \log \left (5 e^3-e^x\right )+\frac {4}{625} e^2 \text {Li}_2\left (\frac {e^{-3+x}}{5}\right )-\frac {1}{625} \left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\frac {1}{625} \left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{-5 e^3+x} \, dx,x,e^x\right )+\frac {1}{625} \left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{5}\right )}{x} \, dx,x,e^{-3+x}\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx\\ &=\frac {16 e^8}{25 \left (5 e^3-e^x\right )^2}+\frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.27, size = 26, normalized size = 0.93 \begin {gather*} \frac {2 e^8 (4+x)^2}{25 \left (-5 e^3+e^x\right )^2 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 35, normalized size = 1.25 \begin {gather*} \frac {2 \, {\left (x^{2} + 8 \, x + 16\right )} e^{6}}{25 \, {\left (25 \, x e^{4} + x e^{\left (2 \, x - 2\right )} - 10 \, x e^{\left (x + 1\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 39, normalized size = 1.39 \begin {gather*} \frac {2 \, {\left (x^{2} e^{8} + 8 \, x e^{8} + 16 \, e^{8}\right )}}{25 \, {\left (25 \, x e^{6} + x e^{\left (2 \, x\right )} - 10 \, x e^{\left (x + 3\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 24, normalized size = 0.86
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{2} \left (x^{2}+8 x +16\right )}{25 x \left ({\mathrm e}^{x -3}-5\right )^{2}}\) | \(24\) |
norman | \(\frac {\frac {16 \,{\mathrm e}^{2} x}{25}+\frac {32 \,{\mathrm e}^{2}}{25}+\frac {2 x^{2} {\mathrm e}^{2}}{25}}{x \left ({\mathrm e}^{x -3}-5\right )^{2}}\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 39, normalized size = 1.39 \begin {gather*} \frac {2 \, {\left (x^{2} e^{8} + 8 \, x e^{8} + 16 \, e^{8}\right )}}{25 \, {\left (25 \, x e^{6} + x e^{\left (2 \, x\right )} - 10 \, x e^{\left (x + 3\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.17, size = 40, normalized size = 1.43 \begin {gather*} \frac {2\,\left ({\mathrm {e}}^2\,x^3+8\,{\mathrm {e}}^2\,x^2+16\,{\mathrm {e}}^2\,x\right )}{25\,x^2\,\left ({\mathrm {e}}^{2\,x-6}-10\,{\mathrm {e}}^{x-3}+25\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.17, size = 41, normalized size = 1.46 \begin {gather*} \frac {2 x^{2} e^{2} + 16 x e^{2} + 32 e^{2}}{- 250 x e^{x - 3} + 25 x e^{2 x - 6} + 625 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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