Optimal. Leaf size=27 \[ \frac {e^{e^x}+\frac {4}{(7-4 x)^2 \left (5+e^x+x\right )}}{x} \]
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Rubi [F] time = 4.10, 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 {140-184 x-64 x^2+e^x \left (28-20 x-16 x^2\right )+e^{e^x} \left (8575-11270 x+2863 x^2+1172 x^3-304 x^4-64 x^5+e^{3 x} \left (-343 x+588 x^2-336 x^3+64 x^4\right )+e^{2 x} \left (343-4018 x+5530 x^2-2248 x^3-32 x^4+128 x^5\right )+e^x \left (3430-13769 x+13454 x^2-2831 x^3-1300 x^4+304 x^5+64 x^6\right )\right )}{-8575 x^2+11270 x^3-2863 x^4-1172 x^5+304 x^6+64 x^7+e^{2 x} \left (-343 x^2+588 x^3-336 x^4+64 x^5\right )+e^x \left (-3430 x^2+5194 x^3-2184 x^4-32 x^5+128 x^6\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^{e^x+3 x} x (-7+4 x)^3+e^{e^x} (5+x)^2 (-7+4 x)^3-e^{e^x+2 x} (-7+4 x)^3 \left (-1+10 x+2 x^2\right )+4 e^x \left (-7+5 x+4 x^2\right )+4 \left (-35+46 x+16 x^2\right )-e^{e^x+x} (-7+4 x)^3 \left (-10+23 x+10 x^2+x^3\right )}{(7-4 x)^3 x^2 \left (5+e^x+x\right )^2} \, dx\\ &=\int \left (-\frac {e^{e^x}}{x^2}+\frac {e^{e^x+x}}{x}+\frac {4 (4+x)}{x \left (5+e^x+x\right )^2 (-7+4 x)^2}-\frac {4 \left (-7+5 x+4 x^2\right )}{x^2 \left (5+e^x+x\right ) (-7+4 x)^3}\right ) \, dx\\ &=4 \int \frac {4+x}{x \left (5+e^x+x\right )^2 (-7+4 x)^2} \, dx-4 \int \frac {-7+5 x+4 x^2}{x^2 \left (5+e^x+x\right ) (-7+4 x)^3} \, dx-\int \frac {e^{e^x}}{x^2} \, dx+\int \frac {e^{e^x+x}}{x} \, dx\\ &=4 \int \left (\frac {4}{49 x \left (5+e^x+x\right )^2}+\frac {23}{7 \left (5+e^x+x\right )^2 (-7+4 x)^2}-\frac {16}{49 \left (5+e^x+x\right )^2 (-7+4 x)}\right ) \, dx-4 \int \left (\frac {1}{49 x^2 \left (5+e^x+x\right )}+\frac {1}{49 x \left (5+e^x+x\right )}+\frac {32}{7 \left (5+e^x+x\right ) (-7+4 x)^3}+\frac {12}{49 \left (5+e^x+x\right ) (-7+4 x)^2}-\frac {4}{49 \left (5+e^x+x\right ) (-7+4 x)}\right ) \, dx-\int \frac {e^{e^x}}{x^2} \, dx+\int \frac {e^{e^x+x}}{x} \, dx\\ &=-\left (\frac {4}{49} \int \frac {1}{x^2 \left (5+e^x+x\right )} \, dx\right )-\frac {4}{49} \int \frac {1}{x \left (5+e^x+x\right )} \, dx+\frac {16}{49} \int \frac {1}{x \left (5+e^x+x\right )^2} \, dx+\frac {16}{49} \int \frac {1}{\left (5+e^x+x\right ) (-7+4 x)} \, dx-\frac {48}{49} \int \frac {1}{\left (5+e^x+x\right ) (-7+4 x)^2} \, dx-\frac {64}{49} \int \frac {1}{\left (5+e^x+x\right )^2 (-7+4 x)} \, dx+\frac {92}{7} \int \frac {1}{\left (5+e^x+x\right )^2 (-7+4 x)^2} \, dx-\frac {128}{7} \int \frac {1}{\left (5+e^x+x\right ) (-7+4 x)^3} \, dx-\int \frac {e^{e^x}}{x^2} \, dx+\int \frac {e^{e^x+x}}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 27, normalized size = 1.00 \begin {gather*} \frac {e^{e^x}+\frac {4}{(7-4 x)^2 \left (5+e^x+x\right )}}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.92, size = 73, normalized size = 2.70 \begin {gather*} \frac {{\left (16 \, x^{3} + 24 \, x^{2} + {\left (16 \, x^{2} - 56 \, x + 49\right )} e^{x} - 231 \, x + 245\right )} e^{\left (e^{x}\right )} + 4}{16 \, x^{4} + 24 \, x^{3} - 231 \, x^{2} + {\left (16 \, x^{3} - 56 \, x^{2} + 49 \, x\right )} e^{x} + 245 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.47, size = 126, normalized size = 4.67 \begin {gather*} \frac {16 \, x^{3} e^{\left (x + e^{x}\right )} + 16 \, x^{2} e^{\left (2 \, x + e^{x}\right )} + 24 \, x^{2} e^{\left (x + e^{x}\right )} - 56 \, x e^{\left (2 \, x + e^{x}\right )} - 231 \, x e^{\left (x + e^{x}\right )} + 49 \, e^{\left (2 \, x + e^{x}\right )} + 245 \, e^{\left (x + e^{x}\right )} + 4 \, e^{x}}{16 \, x^{4} e^{x} + 16 \, x^{3} e^{\left (2 \, x\right )} + 24 \, x^{3} e^{x} - 56 \, x^{2} e^{\left (2 \, x\right )} - 231 \, x^{2} e^{x} + 49 \, x e^{\left (2 \, x\right )} + 245 \, x e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 33, normalized size = 1.22
method | result | size |
risch | \(\frac {4}{x \left (16 x^{2}-56 x +49\right ) \left ({\mathrm e}^{x}+5+x \right )}+\frac {{\mathrm e}^{{\mathrm e}^{x}}}{x}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.61, size = 73, normalized size = 2.70 \begin {gather*} \frac {{\left (16 \, x^{3} + 24 \, x^{2} + {\left (16 \, x^{2} - 56 \, x + 49\right )} e^{x} - 231 \, x + 245\right )} e^{\left (e^{x}\right )} + 4}{16 \, x^{4} + 24 \, x^{3} - 231 \, x^{2} + {\left (16 \, x^{3} - 56 \, x^{2} + 49 \, x\right )} e^{x} + 245 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.95, size = 46, normalized size = 1.70 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{x}+\frac {4\,\left (4\,x^3+9\,x^2-28\,x\right )}{x^2\,{\left (4\,x-7\right )}^3\,\left (x+4\right )\,\left (x+{\mathrm {e}}^x+5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.38, size = 42, normalized size = 1.56 \begin {gather*} \frac {4}{16 x^{4} + 24 x^{3} - 231 x^{2} + 245 x + \left (16 x^{3} - 56 x^{2} + 49 x\right ) e^{x}} + \frac {e^{e^{x}}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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