Optimal. Leaf size=27 \[ -\frac {x^3}{48 \left (2-e^{2-2 x}\right ) (4+5 x)} \]
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Rubi [F] time = 3.37, 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 {-12 x^2-10 x^3+e^{2-2 x} \left (6 x^2+9 x^3+5 x^4\right )}{1536+3840 x+2400 x^2+e^{2-2 x} \left (-1536-3840 x-2400 x^2\right )+e^{4-4 x} \left (384+960 x+600 x^2\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 {e^{2 x} x^2 \left (-2 e^{2 x} (6+5 x)+e^2 \left (6+9 x+5 x^2\right )\right )}{24 \left (e^2-2 e^{2 x}\right )^2 (4+5 x)^2} \, dx\\ &=\frac {1}{24} \int \frac {e^{2 x} x^2 \left (-2 e^{2 x} (6+5 x)+e^2 \left (6+9 x+5 x^2\right )\right )}{\left (e^2-2 e^{2 x}\right )^2 (4+5 x)^2} \, dx\\ &=\frac {1}{24} \int \left (\frac {e^{2+2 x} x^3}{\left (e^2-2 e^{2 x}\right )^2 (4+5 x)}-\frac {e^{2 x} x^2 (6+5 x)}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2}\right ) \, dx\\ &=\frac {1}{24} \int \frac {e^{2+2 x} x^3}{\left (e^2-2 e^{2 x}\right )^2 (4+5 x)} \, dx-\frac {1}{24} \int \frac {e^{2 x} x^2 (6+5 x)}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx\\ &=-\left (\frac {1}{24} \int \left (-\frac {2 e^{2 x}}{25 \left (-e^2+2 e^{2 x}\right )}+\frac {e^{2 x} x}{5 \left (-e^2+2 e^{2 x}\right )}+\frac {32 e^{2 x}}{25 \left (-e^2+2 e^{2 x}\right ) (4+5 x)^2}\right ) \, dx\right )+\frac {1}{24} \int \left (\frac {16 e^{2+2 x}}{125 \left (-e^2+2 e^{2 x}\right )^2}-\frac {4 e^{2+2 x} x}{25 \left (-e^2+2 e^{2 x}\right )^2}+\frac {e^{2+2 x} x^2}{5 \left (-e^2+2 e^{2 x}\right )^2}-\frac {64 e^{2+2 x}}{125 \left (-e^2+2 e^{2 x}\right )^2 (4+5 x)}\right ) \, dx\\ &=\frac {1}{300} \int \frac {e^{2 x}}{-e^2+2 e^{2 x}} \, dx+\frac {2}{375} \int \frac {e^{2+2 x}}{\left (-e^2+2 e^{2 x}\right )^2} \, dx-\frac {1}{150} \int \frac {e^{2+2 x} x}{\left (-e^2+2 e^{2 x}\right )^2} \, dx-\frac {1}{120} \int \frac {e^{2 x} x}{-e^2+2 e^{2 x}} \, dx+\frac {1}{120} \int \frac {e^{2+2 x} x^2}{\left (-e^2+2 e^{2 x}\right )^2} \, dx-\frac {8}{375} \int \frac {e^{2+2 x}}{\left (-e^2+2 e^{2 x}\right )^2 (4+5 x)} \, dx-\frac {4}{75} \int \frac {e^{2 x}}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx\\ &=-\frac {1}{480} x \log \left (1-2 e^{-2+2 x}\right )+\frac {1}{600} \operatorname {Subst}\left (\int \frac {1}{-e^2+2 x} \, dx,x,e^{2 x}\right )+\frac {1}{480} \int \log \left (1-2 e^{-2+2 x}\right ) \, dx-\frac {4}{75} \int \frac {e^{2 x}}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx+\frac {1}{375} \left (2 e^2\right ) \int \frac {e^{2 x}}{\left (-e^2+2 e^{2 x}\right )^2} \, dx-\frac {1}{150} e^2 \int \frac {e^{2 x} x}{\left (-e^2+2 e^{2 x}\right )^2} \, dx+\frac {1}{120} e^2 \int \frac {e^{2 x} x^2}{\left (-e^2+2 e^{2 x}\right )^2} \, dx-\frac {1}{375} \left (8 e^2\right ) \int \frac {e^{2 x}}{\left (-e^2+2 e^{2 x}\right )^2 (4+5 x)} \, dx\\ &=-\frac {e^2 x}{600 \left (e^2-2 e^{2 x}\right )}+\frac {e^2 x^2}{480 \left (e^2-2 e^{2 x}\right )}-\frac {2 e^2}{375 \left (e^2-2 e^{2 x}\right ) (4+5 x)}+\frac {\log \left (e^2-2 e^{2 x}\right )}{1200}-\frac {1}{480} x \log \left (1-2 e^{-2+2 x}\right )+\frac {1}{960} \operatorname {Subst}\left (\int \frac {\log (1-2 x)}{x} \, dx,x,e^{-2+2 x}\right )-\frac {4}{75} \int \frac {e^{2 x}}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx-\frac {1}{600} e^2 \int \frac {1}{-e^2+2 e^{2 x}} \, dx+\frac {1}{375} e^2 \operatorname {Subst}\left (\int \frac {1}{\left (-e^2+2 x\right )^2} \, dx,x,e^{2 x}\right )+\frac {1}{240} e^2 \int \frac {x}{-e^2+2 e^{2 x}} \, dx+\frac {1}{75} \left (2 e^2\right ) \int \frac {1}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx\\ &=\frac {e^2}{750 \left (e^2-2 e^{2 x}\right )}-\frac {e^2 x}{600 \left (e^2-2 e^{2 x}\right )}-\frac {x^2}{480}+\frac {e^2 x^2}{480 \left (e^2-2 e^{2 x}\right )}-\frac {2 e^2}{375 \left (e^2-2 e^{2 x}\right ) (4+5 x)}+\frac {\log \left (e^2-2 e^{2 x}\right )}{1200}-\frac {1}{480} x \log \left (1-2 e^{-2+2 x}\right )-\frac {1}{960} \text {Li}_2\left (2 e^{-2+2 x}\right )+\frac {1}{120} \int \frac {e^{2 x} x}{-e^2+2 e^{2 x}} \, dx-\frac {4}{75} \int \frac {e^{2 x}}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx-\frac {e^2 \operatorname {Subst}\left (\int \frac {1}{x \left (-e^2+2 x\right )} \, dx,x,e^{2 x}\right )}{1200}+\frac {1}{75} \left (2 e^2\right ) \int \frac {1}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx\\ &=\frac {e^2}{750 \left (e^2-2 e^{2 x}\right )}-\frac {e^2 x}{600 \left (e^2-2 e^{2 x}\right )}-\frac {x^2}{480}+\frac {e^2 x^2}{480 \left (e^2-2 e^{2 x}\right )}-\frac {2 e^2}{375 \left (e^2-2 e^{2 x}\right ) (4+5 x)}+\frac {\log \left (e^2-2 e^{2 x}\right )}{1200}-\frac {1}{960} \text {Li}_2\left (2 e^{-2+2 x}\right )+\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x}\right )}{1200}-\frac {1}{600} \operatorname {Subst}\left (\int \frac {1}{-e^2+2 x} \, dx,x,e^{2 x}\right )-\frac {1}{480} \int \log \left (1-2 e^{-2+2 x}\right ) \, dx-\frac {4}{75} \int \frac {e^{2 x}}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx+\frac {1}{75} \left (2 e^2\right ) \int \frac {1}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx\\ &=\frac {e^2}{750 \left (e^2-2 e^{2 x}\right )}+\frac {x}{600}-\frac {e^2 x}{600 \left (e^2-2 e^{2 x}\right )}-\frac {x^2}{480}+\frac {e^2 x^2}{480 \left (e^2-2 e^{2 x}\right )}-\frac {2 e^2}{375 \left (e^2-2 e^{2 x}\right ) (4+5 x)}-\frac {1}{960} \text {Li}_2\left (2 e^{-2+2 x}\right )-\frac {1}{960} \operatorname {Subst}\left (\int \frac {\log (1-2 x)}{x} \, dx,x,e^{-2+2 x}\right )-\frac {4}{75} \int \frac {e^{2 x}}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx+\frac {1}{75} \left (2 e^2\right ) \int \frac {1}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx\\ &=\frac {e^2}{750 \left (e^2-2 e^{2 x}\right )}+\frac {x}{600}-\frac {e^2 x}{600 \left (e^2-2 e^{2 x}\right )}-\frac {x^2}{480}+\frac {e^2 x^2}{480 \left (e^2-2 e^{2 x}\right )}-\frac {2 e^2}{375 \left (e^2-2 e^{2 x}\right ) (4+5 x)}-\frac {4}{75} \int \frac {e^{2 x}}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx+\frac {1}{75} \left (2 e^2\right ) \int \frac {1}{\left (-e^2+2 e^{2 x}\right ) (4+5 x)^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.42, size = 53, normalized size = 1.96 \begin {gather*} \frac {-24 e^2 (4+5 x)+e^{2 x} \left (192+240 x-125 x^3\right )}{6000 \left (-e^2+2 e^{2 x}\right ) (4+5 x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.81, size = 24, normalized size = 0.89 \begin {gather*} \frac {x^{3}}{48 \, {\left ({\left (5 \, x + 4\right )} e^{\left (-2 \, x + 2\right )} - 10 \, x - 8\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.15, size = 170, normalized size = 6.30 \begin {gather*} -\frac {625 \, x^{4} e^{2} - 1250 \, x^{4} e^{\left (2 \, x\right )} + 500 \, x^{3} e^{2} - 1000 \, x^{3} e^{\left (2 \, x\right )} - 300 \, x^{2} e^{2} + 300 \, x^{2} e^{\left (2 \, x\right )} + 75 \, x^{2} e^{\left (-2 \, x + 4\right )} - 480 \, x e^{2} + 480 \, x e^{\left (2 \, x\right )} + 120 \, x e^{\left (-2 \, x + 4\right )} - 192 \, e^{2} + 192 \, e^{\left (2 \, x\right )} + 48 \, e^{\left (-2 \, x + 4\right )}}{6000 \, {\left (100 \, x^{2} e^{2} - 100 \, x^{2} e^{\left (2 \, x\right )} - 25 \, x^{2} e^{\left (-2 \, x + 4\right )} + 160 \, x e^{2} - 160 \, x e^{\left (2 \, x\right )} - 40 \, x e^{\left (-2 \, x + 4\right )} + 64 \, e^{2} - 64 \, e^{\left (2 \, x\right )} - 16 \, e^{\left (-2 \, x + 4\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 23, normalized size = 0.85
method | result | size |
norman | \(\frac {x^{3}}{48 \left ({\mathrm e}^{-2 x +2}-2\right ) \left (4+5 x \right )}\) | \(23\) |
risch | \(\frac {x^{3}}{48 \left ({\mathrm e}^{-2 x +2}-2\right ) \left (4+5 x \right )}\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.47, size = 50, normalized size = 1.85 \begin {gather*} \frac {40 \, x e^{2} + {\left (125 \, x^{3} - 80 \, x - 64\right )} e^{\left (2 \, x\right )} + 32 \, e^{2}}{6000 \, {\left (5 \, x e^{2} - 2 \, {\left (5 \, x + 4\right )} e^{\left (2 \, x\right )} + 4 \, e^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.20, size = 30, normalized size = 1.11 \begin {gather*} \frac {5\,x^4+4\,x^3}{48\,{\left (5\,x+4\right )}^2\,\left ({\mathrm {e}}^{2-2\,x}-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 19, normalized size = 0.70 \begin {gather*} \frac {x^{3}}{- 480 x + \left (240 x + 192\right ) e^{2 - 2 x} - 384} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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