Optimal. Leaf size=28 \[ e^{-2 x} x^2 \left (x+\frac {x^3}{\left (-8+\frac {e^{2 x}}{5}\right )^2}\right ) \]
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Rubi [B] time = 3.90, antiderivative size = 58, normalized size of antiderivative = 2.07, number of steps used = 143, number of rules used = 14, integrand size = 112, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6688, 6742, 2254, 2176, 2194, 2185, 2184, 2190, 2531, 6609, 2282, 6589, 2191, 2196} \begin {gather*} \frac {1}{64} e^{-2 x} x^5+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+e^{-2 x} x^3 \end {gather*}
Antiderivative was successfully verified.
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Rule 2176
Rule 2184
Rule 2185
Rule 2190
Rule 2191
Rule 2194
Rule 2196
Rule 2254
Rule 2282
Rule 2531
Rule 6589
Rule 6609
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-2 x} x^2 \left (-120 e^{4 x} (-3+2 x)+e^{6 x} (-3+2 x)-1000 \left (-192+128 x-5 x^2+2 x^3\right )+25 e^{2 x} \left (-576+384 x-5 x^2+6 x^3\right )\right )}{\left (40-e^{2 x}\right )^3} \, dx\\ &=\int \left (-\frac {4000 e^{-2 x} x^5}{\left (-40+e^{2 x}\right )^3}-e^{-2 x} x^2 (-3+2 x)-\frac {25 e^{-2 x} x^4 (-5+6 x)}{\left (-40+e^{2 x}\right )^2}\right ) \, dx\\ &=-\left (25 \int \frac {e^{-2 x} x^4 (-5+6 x)}{\left (-40+e^{2 x}\right )^2} \, dx\right )-4000 \int \frac {e^{-2 x} x^5}{\left (-40+e^{2 x}\right )^3} \, dx-\int e^{-2 x} x^2 (-3+2 x) \, dx\\ &=-\left (25 \int \left (-\frac {5 e^{-2 x} x^4}{\left (-40+e^{2 x}\right )^2}+\frac {6 e^{-2 x} x^5}{\left (-40+e^{2 x}\right )^2}\right ) \, dx\right )-4000 \int \left (-\frac {e^{-2 x} x^5}{64000}+\frac {x^5}{40 \left (-40+e^{2 x}\right )^3}-\frac {x^5}{1600 \left (-40+e^{2 x}\right )^2}+\frac {x^5}{64000 \left (-40+e^{2 x}\right )}\right ) \, dx-\int \left (-3 e^{-2 x} x^2+2 e^{-2 x} x^3\right ) \, dx\\ &=\frac {1}{16} \int e^{-2 x} x^5 \, dx-\frac {1}{16} \int \frac {x^5}{-40+e^{2 x}} \, dx-2 \int e^{-2 x} x^3 \, dx+\frac {5}{2} \int \frac {x^5}{\left (-40+e^{2 x}\right )^2} \, dx+3 \int e^{-2 x} x^2 \, dx-100 \int \frac {x^5}{\left (-40+e^{2 x}\right )^3} \, dx+125 \int \frac {e^{-2 x} x^4}{\left (-40+e^{2 x}\right )^2} \, dx-150 \int \frac {e^{-2 x} x^5}{\left (-40+e^{2 x}\right )^2} \, dx\\ &=-\frac {3}{2} e^{-2 x} x^2+e^{-2 x} x^3-\frac {1}{32} e^{-2 x} x^5+\frac {x^6}{3840}-\frac {1}{640} \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx+\frac {1}{16} \int \frac {e^{2 x} x^5}{\left (-40+e^{2 x}\right )^2} \, dx-\frac {1}{16} \int \frac {x^5}{-40+e^{2 x}} \, dx+\frac {5}{32} \int e^{-2 x} x^4 \, dx-\frac {5}{2} \int \frac {e^{2 x} x^5}{\left (-40+e^{2 x}\right )^3} \, dx+\frac {5}{2} \int \frac {x^5}{\left (-40+e^{2 x}\right )^2} \, dx+3 \int e^{-2 x} x \, dx-3 \int e^{-2 x} x^2 \, dx+125 \int \left (\frac {e^{-2 x} x^4}{1600}+\frac {x^4}{40 \left (-40+e^{2 x}\right )^2}-\frac {x^4}{1600 \left (-40+e^{2 x}\right )}\right ) \, dx-150 \int \left (\frac {e^{-2 x} x^5}{1600}+\frac {x^5}{40 \left (-40+e^{2 x}\right )^2}-\frac {x^5}{1600 \left (-40+e^{2 x}\right )}\right ) \, dx\\ &=-\frac {3}{2} e^{-2 x} x+e^{-2 x} x^3-\frac {5}{64} e^{-2 x} x^4-\frac {1}{32} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{32 \left (40-e^{2 x}\right )}+\frac {x^6}{1920}-\frac {x^5 \log \left (1-\frac {e^{2 x}}{40}\right )}{1280}-\frac {1}{640} \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx+\frac {1}{256} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{16} \int \frac {e^{2 x} x^5}{\left (-40+e^{2 x}\right )^2} \, dx-\frac {1}{16} \int \frac {x^5}{-40+e^{2 x}} \, dx+\frac {5}{64} \int e^{-2 x} x^4 \, dx-\frac {5}{64} \int \frac {x^4}{-40+e^{2 x}} \, dx-\frac {3}{32} \int e^{-2 x} x^5 \, dx+\frac {3}{32} \int \frac {x^5}{-40+e^{2 x}} \, dx+\frac {5}{32} \int \frac {x^4}{-40+e^{2 x}} \, dx+\frac {5}{16} \int e^{-2 x} x^3 \, dx+\frac {3}{2} \int e^{-2 x} \, dx-3 \int e^{-2 x} x \, dx-\frac {15}{4} \int \frac {x^5}{\left (-40+e^{2 x}\right )^2} \, dx\\ &=-\frac {3}{4} e^{-2 x}+\frac {27}{32} e^{-2 x} x^3-\frac {15}{128} e^{-2 x} x^4-\frac {x^5}{2560}+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{16 \left (40-e^{2 x}\right )}+\frac {x^6}{2560}-\frac {1}{640} x^5 \log \left (1-\frac {e^{2 x}}{40}\right )-\frac {1}{512} x^4 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )-\frac {1}{640} \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx-\frac {1}{512} \int \frac {e^{2 x} x^4}{-40+e^{2 x}} \, dx+\frac {3 \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx}{1280}+\frac {1}{256} \int \frac {e^{2 x} x^4}{-40+e^{2 x}} \, dx+\frac {1}{256} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{128} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{32} \int \frac {e^{2 x} x^5}{\left (-40+e^{2 x}\right )^2} \, dx+\frac {3}{32} \int \frac {x^5}{-40+e^{2 x}} \, dx+\frac {5}{32} \int e^{-2 x} x^3 \, dx+\frac {5}{32} \int \frac {x^4}{-40+e^{2 x}} \, dx-\frac {15}{64} \int e^{-2 x} x^4 \, dx+\frac {15}{32} \int e^{-2 x} x^2 \, dx-\frac {3}{2} \int e^{-2 x} \, dx\\ &=-\frac {15}{64} e^{-2 x} x^2+\frac {49}{64} e^{-2 x} x^3-\frac {3 x^5}{2560}+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {x^4 \log \left (1-\frac {e^{2 x}}{40}\right )}{1024}-\frac {3 x^5 \log \left (1-\frac {e^{2 x}}{40}\right )}{2560}-\frac {1}{256} x^4 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )+\frac {1}{256} x^3 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )+\frac {3 \int \frac {e^{2 x} x^5}{-40+e^{2 x}} \, dx}{1280}+\frac {1}{256} \int \frac {e^{2 x} x^4}{-40+e^{2 x}} \, dx+\frac {1}{256} \int x^3 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{256} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{512} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx-\frac {1}{128} \int x^3 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{128} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {15}{64} \int e^{-2 x} x^2 \, dx-\frac {15}{64} \int \frac {x^4}{-40+e^{2 x}} \, dx+\frac {15}{32} \int e^{-2 x} x \, dx-\frac {15}{32} \int e^{-2 x} x^3 \, dx\\ &=-\frac {15}{64} e^{-2 x} x-\frac {45}{128} e^{-2 x} x^2+e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {3 x^4 \log \left (1-\frac {e^{2 x}}{40}\right )}{1024}+\frac {1}{512} x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )-\frac {3 x^4 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {1}{128} x^3 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )-\frac {3}{512} x^2 \text {Li}_4\left (\frac {e^{2 x}}{40}\right )-\frac {3}{512} \int \frac {e^{2 x} x^4}{-40+e^{2 x}} \, dx-\frac {3}{512} \int x^4 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{512} \int x^2 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {1}{128} \int x^3 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx+\frac {1}{128} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {15}{64} \int e^{-2 x} \, dx+\frac {15}{64} \int e^{-2 x} x \, dx-\frac {45}{64} \int e^{-2 x} x^2 \, dx\\ &=-\frac {15}{128} e^{-2 x}-\frac {45}{128} e^{-2 x} x+e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {3}{512} x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right )-\frac {3 x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {3}{512} x^3 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )-\frac {3}{256} x^2 \text {Li}_4\left (\frac {e^{2 x}}{40}\right )+\frac {3}{512} x \text {Li}_5\left (\frac {e^{2 x}}{40}\right )-\frac {3}{512} \int x \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{512} \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x^3 \log \left (1-\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^3 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{256} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {9}{512} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {15}{128} \int e^{-2 x} \, dx-\frac {45}{64} \int e^{-2 x} x \, dx\\ &=-\frac {45}{256} e^{-2 x}+e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}-\frac {9 x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {3 x \text {Li}_4\left (\frac {e^{2 x}}{40}\right )}{1024}-\frac {9 x^2 \text {Li}_4\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {3}{256} x \text {Li}_5\left (\frac {e^{2 x}}{40}\right )+\frac {3 \int \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx}{1024}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}-\frac {3}{512} \int \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{512} \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {3}{256} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {9}{512} \int x^2 \text {Li}_2\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {9}{512} \int x^2 \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {9}{512} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {45}{128} \int e^{-2 x} \, dx\\ &=e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}+\frac {9 x \text {Li}_4\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {9 x \text {Li}_5\left (\frac {e^{2 x}}{40}\right )}{1024}-\frac {3 \text {Li}_6\left (\frac {e^{2 x}}{40}\right )}{1024}+\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{2048}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}-\frac {3}{512} \int \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {3}{512} \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx+\frac {9 \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx}{1024}-\frac {9}{512} \int x \text {Li}_3\left (\frac {e^{2 x}}{40}\right ) \, dx-\frac {9}{512} \int x \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx\\ &=e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}-\frac {3 \text {Li}_5\left (\frac {e^{2 x}}{40}\right )}{2048}-\frac {3}{512} \text {Li}_6\left (\frac {e^{2 x}}{40}\right )-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{1024}+\frac {9 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{2048}+\frac {9 \int \text {Li}_4\left (\frac {e^{2 x}}{40}\right ) \, dx}{1024}+\frac {9 \int \text {Li}_5\left (\frac {e^{2 x}}{40}\right ) \, dx}{1024}\\ &=e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}-\frac {9 \text {Li}_5\left (\frac {e^{2 x}}{40}\right )}{2048}-\frac {9 \text {Li}_6\left (\frac {e^{2 x}}{40}\right )}{2048}+\frac {9 \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{2048}+\frac {9 \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (\frac {x}{40}\right )}{x} \, dx,x,e^{2 x}\right )}{2048}\\ &=e^{-2 x} x^3+\frac {1}{64} e^{-2 x} x^5+\frac {5 x^5}{8 \left (40-e^{2 x}\right )^2}+\frac {x^5}{64 \left (40-e^{2 x}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.06, size = 38, normalized size = 1.36 \begin {gather*} \frac {e^{-2 x} x^3 \left (-80 e^{2 x}+e^{4 x}+25 \left (64+x^2\right )\right )}{\left (-40+e^{2 x}\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.72, size = 48, normalized size = 1.71 \begin {gather*} \frac {25 \, x^{5} + x^{3} e^{\left (4 \, x\right )} - 80 \, x^{3} e^{\left (2 \, x\right )} + 1600 \, x^{3}}{e^{\left (6 \, x\right )} - 80 \, e^{\left (4 \, x\right )} + 1600 \, e^{\left (2 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.14, size = 48, normalized size = 1.71 \begin {gather*} \frac {25 \, x^{5} + x^{3} e^{\left (4 \, x\right )} - 80 \, x^{3} e^{\left (2 \, x\right )} + 1600 \, x^{3}}{e^{\left (6 \, x\right )} - 80 \, e^{\left (4 \, x\right )} + 1600 \, e^{\left (2 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.20, size = 35, normalized size = 1.25
method | result | size |
risch | \(\left (x^{3}+\frac {1}{64} x^{5}\right ) {\mathrm e}^{-2 x}-\frac {x^{5} \left ({\mathrm e}^{2 x}-80\right )}{64 \left ({\mathrm e}^{2 x}-40\right )^{2}}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 44, normalized size = 1.57 \begin {gather*} \frac {x^{3} e^{\left (2 \, x\right )} - 80 \, x^{3} + 25 \, {\left (x^{5} + 64 \, x^{3}\right )} e^{\left (-2 \, x\right )}}{e^{\left (4 \, x\right )} - 80 \, e^{\left (2 \, x\right )} + 1600} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.21, size = 36, normalized size = 1.29 \begin {gather*} -\frac {80\,x^3-x^3\,{\mathrm {e}}^{-2\,x}\,\left ({\mathrm {e}}^{4\,x}+25\,x^2+1600\right )}{{\left ({\mathrm {e}}^{2\,x}-40\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.19, size = 42, normalized size = 1.50 \begin {gather*} \frac {\left (x^{5} + 64 x^{3}\right ) e^{- 2 x}}{64} + \frac {- x^{5} e^{2 x} + 80 x^{5}}{64 e^{4 x} - 5120 e^{2 x} + 102400} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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