Optimal. Leaf size=32 \[ \frac {\left (2-\frac {-1+x}{-\frac {e^x}{5-4 x}+x}\right )^2}{16 x^8} \]
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Rubi [F] time = 3.18, 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 {-16 e^{3 x}+625 x-375 x^2-1000 x^3+640 x^4+384 x^5-256 x^6+e^{2 x} \left (80+124 x-162 x^2+8 x^3\right )+e^x \left (-100-560 x+547 x^2+437 x^3-392 x^4+16 x^5\right )}{8 e^{3 x} x^9-1000 x^{12}+2400 x^{13}-1920 x^{14}+512 x^{15}+e^{2 x} \left (-120 x^{10}+96 x^{11}\right )+e^x \left (600 x^{11}-960 x^{12}+384 x^{13}\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 {-16 e^{3 x}-x (-5+4 x)^3 \left (5+9 x+4 x^2\right )+2 e^{2 x} \left (40+62 x-81 x^2+4 x^3\right )+e^x \left (-100-560 x+547 x^2+437 x^3-392 x^4+16 x^5\right )}{8 x^9 \left (e^x+x (-5+4 x)\right )^3} \, dx\\ &=\frac {1}{8} \int \frac {-16 e^{3 x}-x (-5+4 x)^3 \left (5+9 x+4 x^2\right )+2 e^{2 x} \left (40+62 x-81 x^2+4 x^3\right )+e^x \left (-100-560 x+547 x^2+437 x^3-392 x^4+16 x^5\right )}{x^9 \left (e^x+x (-5+4 x)\right )^3} \, dx\\ &=\frac {1}{8} \int \left (-\frac {16}{x^9}+\frac {\left (5-13 x+4 x^2\right ) \left (5-9 x+4 x^2\right )^2}{x^8 \left (e^x-5 x+4 x^2\right )^3}+\frac {2 \left (40-58 x+15 x^2+4 x^3\right )}{x^9 \left (e^x-5 x+4 x^2\right )}-\frac {100-240 x+53 x^2+255 x^3-216 x^4+48 x^5}{x^9 \left (e^x-5 x+4 x^2\right )^2}\right ) \, dx\\ &=\frac {1}{4 x^8}+\frac {1}{8} \int \frac {\left (5-13 x+4 x^2\right ) \left (5-9 x+4 x^2\right )^2}{x^8 \left (e^x-5 x+4 x^2\right )^3} \, dx-\frac {1}{8} \int \frac {100-240 x+53 x^2+255 x^3-216 x^4+48 x^5}{x^9 \left (e^x-5 x+4 x^2\right )^2} \, dx+\frac {1}{4} \int \frac {40-58 x+15 x^2+4 x^3}{x^9 \left (e^x-5 x+4 x^2\right )} \, dx\\ &=\frac {1}{4 x^8}+\frac {1}{8} \int \left (\frac {125}{x^8 \left (e^x-5 x+4 x^2\right )^3}-\frac {775}{x^7 \left (e^x-5 x+4 x^2\right )^3}+\frac {1875}{x^6 \left (e^x-5 x+4 x^2\right )^3}-\frac {2293}{x^5 \left (e^x-5 x+4 x^2\right )^3}+\frac {1500}{x^4 \left (e^x-5 x+4 x^2\right )^3}-\frac {496}{x^3 \left (e^x-5 x+4 x^2\right )^3}+\frac {64}{x^2 \left (e^x-5 x+4 x^2\right )^3}\right ) \, dx-\frac {1}{8} \int \left (\frac {100}{x^9 \left (e^x-5 x+4 x^2\right )^2}-\frac {240}{x^8 \left (e^x-5 x+4 x^2\right )^2}+\frac {53}{x^7 \left (e^x-5 x+4 x^2\right )^2}+\frac {255}{x^6 \left (e^x-5 x+4 x^2\right )^2}-\frac {216}{x^5 \left (e^x-5 x+4 x^2\right )^2}+\frac {48}{x^4 \left (e^x-5 x+4 x^2\right )^2}\right ) \, dx+\frac {1}{4} \int \left (\frac {40}{x^9 \left (e^x-5 x+4 x^2\right )}-\frac {58}{x^8 \left (e^x-5 x+4 x^2\right )}+\frac {15}{x^7 \left (e^x-5 x+4 x^2\right )}+\frac {4}{x^6 \left (e^x-5 x+4 x^2\right )}\right ) \, dx\\ &=\frac {1}{4 x^8}+\frac {15}{4} \int \frac {1}{x^7 \left (e^x-5 x+4 x^2\right )} \, dx-6 \int \frac {1}{x^4 \left (e^x-5 x+4 x^2\right )^2} \, dx-\frac {53}{8} \int \frac {1}{x^7 \left (e^x-5 x+4 x^2\right )^2} \, dx+8 \int \frac {1}{x^2 \left (e^x-5 x+4 x^2\right )^3} \, dx+10 \int \frac {1}{x^9 \left (e^x-5 x+4 x^2\right )} \, dx-\frac {25}{2} \int \frac {1}{x^9 \left (e^x-5 x+4 x^2\right )^2} \, dx-\frac {29}{2} \int \frac {1}{x^8 \left (e^x-5 x+4 x^2\right )} \, dx+\frac {125}{8} \int \frac {1}{x^8 \left (e^x-5 x+4 x^2\right )^3} \, dx+27 \int \frac {1}{x^5 \left (e^x-5 x+4 x^2\right )^2} \, dx+30 \int \frac {1}{x^8 \left (e^x-5 x+4 x^2\right )^2} \, dx-\frac {255}{8} \int \frac {1}{x^6 \left (e^x-5 x+4 x^2\right )^2} \, dx-62 \int \frac {1}{x^3 \left (e^x-5 x+4 x^2\right )^3} \, dx-\frac {775}{8} \int \frac {1}{x^7 \left (e^x-5 x+4 x^2\right )^3} \, dx+\frac {375}{2} \int \frac {1}{x^4 \left (e^x-5 x+4 x^2\right )^3} \, dx+\frac {1875}{8} \int \frac {1}{x^6 \left (e^x-5 x+4 x^2\right )^3} \, dx-\frac {2293}{8} \int \frac {1}{x^5 \left (e^x-5 x+4 x^2\right )^3} \, dx+\int \frac {1}{x^6 \left (e^x-5 x+4 x^2\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.07, size = 35, normalized size = 1.09 \begin {gather*} \frac {\left (5-2 e^x+x-4 x^2\right )^2}{16 x^8 \left (e^x+x (-5+4 x)\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.80, size = 83, normalized size = 2.59 \begin {gather*} \frac {16 \, x^{4} - 8 \, x^{3} - 39 \, x^{2} + 4 \, {\left (4 \, x^{2} - x - 5\right )} e^{x} + 10 \, x + 4 \, e^{\left (2 \, x\right )} + 25}{16 \, {\left (16 \, x^{12} - 40 \, x^{11} + 25 \, x^{10} + x^{8} e^{\left (2 \, x\right )} + 2 \, {\left (4 \, x^{10} - 5 \, x^{9}\right )} e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.15, size = 84, normalized size = 2.62 \begin {gather*} \frac {16 \, x^{4} - 8 \, x^{3} + 16 \, x^{2} e^{x} - 39 \, x^{2} - 4 \, x e^{x} + 10 \, x + 4 \, e^{\left (2 \, x\right )} - 20 \, e^{x} + 25}{16 \, {\left (16 \, x^{12} - 40 \, x^{11} + 8 \, x^{10} e^{x} + 25 \, x^{10} - 10 \, x^{9} e^{x} + x^{8} e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.04, size = 61, normalized size = 1.91
method | result | size |
risch | \(\frac {1}{4 x^{8}}-\frac {48 x^{4}-152 x^{3}+16 \,{\mathrm e}^{x} x^{2}+139 x^{2}-36 \,{\mathrm e}^{x} x -10 x +20 \,{\mathrm e}^{x}-25}{16 x^{8} \left (4 x^{2}+{\mathrm e}^{x}-5 x \right )^{2}}\) | \(61\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 83, normalized size = 2.59 \begin {gather*} \frac {16 \, x^{4} - 8 \, x^{3} - 39 \, x^{2} + 4 \, {\left (4 \, x^{2} - x - 5\right )} e^{x} + 10 \, x + 4 \, e^{\left (2 \, x\right )} + 25}{16 \, {\left (16 \, x^{12} - 40 \, x^{11} + 25 \, x^{10} + x^{8} e^{\left (2 \, x\right )} + 2 \, {\left (4 \, x^{10} - 5 \, x^{9}\right )} e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {16\,{\mathrm {e}}^{3\,x}-625\,x-{\mathrm {e}}^{2\,x}\,\left (8\,x^3-162\,x^2+124\,x+80\right )+{\mathrm {e}}^x\,\left (-16\,x^5+392\,x^4-437\,x^3-547\,x^2+560\,x+100\right )+375\,x^2+1000\,x^3-640\,x^4-384\,x^5+256\,x^6}{{\mathrm {e}}^x\,\left (384\,x^{13}-960\,x^{12}+600\,x^{11}\right )-{\mathrm {e}}^{2\,x}\,\left (120\,x^{10}-96\,x^{11}\right )+8\,x^9\,{\mathrm {e}}^{3\,x}-1000\,x^{12}+2400\,x^{13}-1920\,x^{14}+512\,x^{15}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.26, size = 78, normalized size = 2.44 \begin {gather*} \frac {- 48 x^{4} + 152 x^{3} - 139 x^{2} + 10 x + \left (- 16 x^{2} + 36 x - 20\right ) e^{x} + 25}{256 x^{12} - 640 x^{11} + 400 x^{10} + 16 x^{8} e^{2 x} + \left (128 x^{10} - 160 x^{9}\right ) e^{x}} + \frac {1}{4 x^{8}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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