Optimal. Leaf size=36 \[ e^2 x-\frac {x}{\left (e^{e^x}-(5+2 x)^2\right ) \left (1-\log ^2(4)\right )} \]
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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 {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\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 {25-e^{e^x}+e^{e^x+x} x-4 x^2-e^{2+2 e^x} \left (-1+\log ^2(4)\right )+2 e^{2+e^x} (5+2 x)^2 \left (-1+\log ^2(4)\right )-e^2 (5+2 x)^4 \left (-1+\log ^2(4)\right )}{\left (e^{e^x}-(5+2 x)^2\right )^2 \left (1-\log ^2(4)\right )} \, dx\\ &=\frac {\int \frac {25-e^{e^x}+e^{e^x+x} x-4 x^2-e^{2+2 e^x} \left (-1+\log ^2(4)\right )+2 e^{2+e^x} (5+2 x)^2 \left (-1+\log ^2(4)\right )-e^2 (5+2 x)^4 \left (-1+\log ^2(4)\right )}{\left (e^{e^x}-(5+2 x)^2\right )^2} \, dx}{1-\log ^2(4)}\\ &=\frac {\int \left (\frac {25}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}-\frac {e^{e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}+\frac {e^{e^x+x} x}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}-\frac {4 x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {e^{2 \left (1+e^x\right )} (1-\log (4)) (1+\log (4))}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {2 e^{2+e^x} (5+2 x)^2 (-1+\log (4)) (1+\log (4))}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}-\frac {e^2 (5+2 x)^4 (-1+\log (4)) (1+\log (4))}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}\right ) \, dx}{1-\log ^2(4)}\\ &=-\left (2 \int \frac {e^{2+e^x} (5+2 x)^2}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx\right )+e^2 \int \frac {(5+2 x)^4}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx-\frac {\int \frac {e^{e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {\int \frac {e^{e^x+x} x}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}-\frac {4 \int \frac {x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {25 \int \frac {1}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\int \frac {e^{2 \left (1+e^x\right )}}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx\\ &=-\left (2 \int \left (\frac {25 e^{2+e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}+\frac {20 e^{2+e^x} x}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {4 e^{2+e^x} x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}\right ) \, dx\right )+e^2 \int \left (\frac {625}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}+\frac {1000 x}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {600 x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {160 x^3}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {16 x^4}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}\right ) \, dx-\frac {\int \frac {e^{e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {\int \frac {e^{e^x+x} x}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}-\frac {4 \int \frac {x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {25 \int \frac {1}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\int \frac {e^{2 \left (1+e^x\right )}}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx\\ &=-\left (8 \int \frac {e^{2+e^x} x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx\right )-40 \int \frac {e^{2+e^x} x}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx-50 \int \frac {e^{2+e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx+\left (16 e^2\right ) \int \frac {x^4}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx+\left (160 e^2\right ) \int \frac {x^3}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx+\left (600 e^2\right ) \int \frac {x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx+\left (625 e^2\right ) \int \frac {1}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx+\left (1000 e^2\right ) \int \frac {x}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx-\frac {\int \frac {e^{e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {\int \frac {e^{e^x+x} x}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}-\frac {4 \int \frac {x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {25 \int \frac {1}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\int \frac {e^{2 \left (1+e^x\right )}}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 40, normalized size = 1.11 \begin {gather*} \frac {\frac {x}{-25+e^{e^x}-20 x-4 x^2}+e^2 x \left (-1+\log ^2(4)\right )}{-1+\log ^2(4)} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.63, size = 106, normalized size = 2.94 \begin {gather*} \frac {4 \, {\left (4 \, x^{3} + 20 \, x^{2} + 25 \, x\right )} e^{2} \log \relax (2)^{2} - {\left (4 \, x^{3} + 20 \, x^{2} + 25 \, x\right )} e^{2} - {\left (4 \, x e^{2} \log \relax (2)^{2} - x e^{2}\right )} e^{\left (e^{x}\right )} - x}{4 \, {\left (4 \, x^{2} + 20 \, x + 25\right )} \log \relax (2)^{2} - 4 \, x^{2} - {\left (4 \, \log \relax (2)^{2} - 1\right )} e^{\left (e^{x}\right )} - 20 \, x - 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.25, size = 120, normalized size = 3.33 \begin {gather*} \frac {16 \, x^{3} e^{2} \log \relax (2)^{2} + 80 \, x^{2} e^{2} \log \relax (2)^{2} - 4 \, x^{3} e^{2} + 100 \, x e^{2} \log \relax (2)^{2} - 4 \, x e^{\left (e^{x} + 2\right )} \log \relax (2)^{2} - 20 \, x^{2} e^{2} - 25 \, x e^{2} + x e^{\left (e^{x} + 2\right )} - x}{16 \, x^{2} \log \relax (2)^{2} + 80 \, x \log \relax (2)^{2} - 4 \, e^{\left (e^{x}\right )} \log \relax (2)^{2} - 4 \, x^{2} + 100 \, \log \relax (2)^{2} - 20 \, x + e^{\left (e^{x}\right )} - 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.66, size = 36, normalized size = 1.00
method | result | size |
risch | \({\mathrm e}^{2} x -\frac {x}{\left (4 \ln \relax (2)^{2}-1\right ) \left (4 x^{2}+20 x -{\mathrm e}^{{\mathrm e}^{x}}+25\right )}\) | \(36\) |
norman | \(\frac {\frac {\left (100 \,{\mathrm e}^{2} \ln \relax (2)^{2}-25 \,{\mathrm e}^{2}-1\right ) x}{4 \ln \relax (2)^{2}-1}+20 x^{2} {\mathrm e}^{2}+4 x^{3} {\mathrm e}^{2}-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{x}} x}{4 x^{2}+20 x -{\mathrm e}^{{\mathrm e}^{x}}+25}\) | \(68\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.57, size = 111, normalized size = 3.08 \begin {gather*} \frac {4 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} x^{3} e^{2} + 20 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} x^{2} e^{2} - {\left (4 \, \log \relax (2)^{2} - 1\right )} x e^{\left (e^{x} + 2\right )} + {\left (25 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} e^{2} - 1\right )} x}{4 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} x^{2} + 20 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} x - {\left (4 \, \log \relax (2)^{2} - 1\right )} e^{\left (e^{x}\right )} + 100 \, \log \relax (2)^{2} - 25} \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 {{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left ({\mathrm {e}}^2-4\,{\mathrm {e}}^2\,{\ln \relax (2)}^2\right )+{\mathrm {e}}^2\,\left (16\,x^4+160\,x^3+600\,x^2+1000\,x+625\right )-4\,x^2-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^2\,\left (8\,x^2+40\,x+50\right )-x\,{\mathrm {e}}^x-4\,{\mathrm {e}}^2\,{\ln \relax (2)}^2\,\left (8\,x^2+40\,x+50\right )+1\right )-4\,{\mathrm {e}}^2\,{\ln \relax (2)}^2\,\left (16\,x^4+160\,x^3+600\,x^2+1000\,x+625\right )+25}{1000\,x-4\,{\ln \relax (2)}^2\,\left (16\,x^4+160\,x^3+600\,x^2+1000\,x+625\right )-{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (4\,{\ln \relax (2)}^2-1\right )-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (40\,x-4\,{\ln \relax (2)}^2\,\left (8\,x^2+40\,x+50\right )+8\,x^2+50\right )+600\,x^2+160\,x^3+16\,x^4+625} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.27, size = 54, normalized size = 1.50 \begin {gather*} x e^{2} + \frac {x}{- 16 x^{2} \log {\relax (2 )}^{2} + 4 x^{2} - 80 x \log {\relax (2 )}^{2} + 20 x + \left (-1 + 4 \log {\relax (2 )}^{2}\right ) e^{e^{x}} - 100 \log {\relax (2 )}^{2} + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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