Optimal. Leaf size=31 \[ 5+\frac {x}{2 \left (e^2-e^{\frac {1}{-4+x}}-4 \left (25-\frac {2}{x^2}\right )\right )} \]
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Rubi [F] time = 4.98, 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 {384 x^2-192 x^3-1576 x^4+800 x^5-100 x^6+e^{\frac {1}{-4+x}} \left (-16 x^4+7 x^5-x^6\right )+e^2 \left (16 x^4-8 x^5+x^6\right )}{2048-1024 x-51072 x^2+25600 x^3+316800 x^4-160000 x^5+20000 x^6+e^2 \left (512 x^2-256 x^3-6368 x^4+3200 x^5-400 x^6\right )+e^4 \left (32 x^4-16 x^5+2 x^6\right )+e^{\frac {2}{-4+x}} \left (32 x^4-16 x^5+2 x^6\right )+e^{\frac {1}{-4+x}} \left (-512 x^2+256 x^3+6368 x^4-3200 x^5+400 x^6+e^2 \left (-64 x^4+32 x^5-4 x^6\right )\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 {x^2 \left (384-192 x-8 \left (197-2 e^2+2 e^{\frac {1}{-4+x}}\right ) x^2-\left (-800+8 e^2-7 e^{\frac {1}{-4+x}}\right ) x^3-\left (100-e^2+e^{\frac {1}{-4+x}}\right ) x^4\right )}{2 (4-x)^2 \left (8-\left (100-e^2+e^{\frac {1}{-4+x}}\right ) x^2\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {x^2 \left (384-192 x-8 \left (197-2 e^2+2 e^{\frac {1}{-4+x}}\right ) x^2-\left (-800+8 e^2-7 e^{\frac {1}{-4+x}}\right ) x^3-\left (100-e^2+e^{\frac {1}{-4+x}}\right ) x^4\right )}{(4-x)^2 \left (8-\left (100-e^2+e^{\frac {1}{-4+x}}\right ) x^2\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {x^2 \left (16-7 x+x^2\right )}{(4-x)^2 \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )}+\frac {x^2 \left (256-136 x+16 x^2+\left (100-e^2\right ) x^3\right )}{(4-x)^2 \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {x^2 \left (16-7 x+x^2\right )}{(4-x)^2 \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )} \, dx+\frac {1}{2} \int \frac {x^2 \left (256-136 x+16 x^2+\left (100-e^2\right ) x^3\right )}{(4-x)^2 \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {64 \left (399-4 e^2\right )}{\left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2}+\frac {512 \left (199-2 e^2\right )}{(4-x)^2 \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2}+\frac {128 \left (-997+10 e^2\right )}{(4-x) \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2}+\frac {8 \left (599-6 e^2\right ) x}{\left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2}+\frac {8 \left (102-e^2\right ) x^2}{\left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2}+\frac {\left (100-e^2\right ) x^3}{\left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2}\right ) \, dx+\frac {1}{2} \int \left (\frac {8}{8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2}+\frac {64}{(4-x)^2 \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )}+\frac {48}{(-4+x) \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )}+\frac {x}{8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2}+\frac {x^2}{8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {x}{8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2} \, dx+\frac {1}{2} \int \frac {x^2}{8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2} \, dx+4 \int \frac {1}{8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2} \, dx+24 \int \frac {1}{(-4+x) \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )} \, dx+32 \int \frac {1}{(4-x)^2 \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )} \, dx-\left (64 \left (997-10 e^2\right )\right ) \int \frac {1}{(4-x) \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2} \, dx+\left (4 \left (599-6 e^2\right )\right ) \int \frac {x}{\left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2} \, dx+\left (32 \left (399-4 e^2\right )\right ) \int \frac {1}{\left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2} \, dx+\left (256 \left (199-2 e^2\right )\right ) \int \frac {1}{(4-x)^2 \left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2} \, dx+\frac {1}{2} \left (100-e^2\right ) \int \frac {x^3}{\left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2} \, dx+\left (4 \left (102-e^2\right )\right ) \int \frac {x^2}{\left (8-e^{\frac {1}{-4+x}} x^2-100 \left (1-\frac {e^2}{100}\right ) x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 29, normalized size = 0.94 \begin {gather*} -\frac {x^3}{2 \left (-8+\left (100-e^2+e^{\frac {1}{-4+x}}\right ) x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 31, normalized size = 1.00 \begin {gather*} \frac {x^{3}}{2 \, {\left (x^{2} e^{2} - x^{2} e^{\left (\frac {1}{x - 4}\right )} - 100 \, x^{2} + 8\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.29, size = 41, normalized size = 1.32 \begin {gather*} \frac {x^{3} e^{\frac {1}{4}}}{2 \, {\left (x^{2} e^{\frac {9}{4}} - 100 \, x^{2} e^{\frac {1}{4}} - x^{2} e^{\left (\frac {x}{4 \, {\left (x - 4\right )}}\right )} + 8 \, e^{\frac {1}{4}}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.59, size = 32, normalized size = 1.03
method | result | size |
risch | \(\frac {x^{3}}{2 x^{2} {\mathrm e}^{2}-2 x^{2} {\mathrm e}^{\frac {1}{x -4}}-200 x^{2}+16}\) | \(32\) |
norman | \(\frac {-2 x^{3}+\frac {1}{2} x^{4}}{\left (x -4\right ) \left (x^{2} {\mathrm e}^{2}-x^{2} {\mathrm e}^{\frac {1}{x -4}}-100 x^{2}+8\right )}\) | \(44\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 28, normalized size = 0.90 \begin {gather*} \frac {x^{3}}{2 \, {\left (x^{2} {\left (e^{2} - 100\right )} - x^{2} e^{\left (\frac {1}{x - 4}\right )} + 8\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.49, size = 129, normalized size = 4.16 \begin {gather*} -\frac {{\left (x^5-8\,x^4+16\,x^3\right )}^2\,\left (16\,x^2-x^3\,{\mathrm {e}}^2-136\,x+100\,x^3+256\right )}{2\,x^2\,\left ({\mathrm {e}}^{\frac {1}{x-4}}-\frac {x^2\,{\mathrm {e}}^2-100\,x^2+8}{x^2}\right )\,{\left (x-4\right )}^2\,\left (8\,x^7\,{\mathrm {e}}^2-16\,x^6\,{\mathrm {e}}^2-x^8\,{\mathrm {e}}^2+4096\,x^3-4224\,x^4+1600\,x^5+1336\,x^6-784\,x^7+100\,x^8\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.29, size = 31, normalized size = 1.00 \begin {gather*} - \frac {x^{3}}{2 x^{2} e^{\frac {1}{x - 4}} - 2 x^{2} e^{2} + 200 x^{2} - 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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