Optimal. Leaf size=27 \[ e^{\frac {2 x+e^{5+x} x}{\left (-1+\frac {4 x}{5}\right )^2+x}} \]
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Rubi [F]
time = 6.20, 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 {e^{\frac {50 x+25 e^{5+x} x}{25-15 x+16 x^2}} \left (1250-800 x^2+e^{5+x} \left (625+625 x-775 x^2+400 x^3\right )\right )}{625-750 x+1025 x^2-480 x^3+256 x^4} \, 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^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} \left (1250-800 x^2+e^{5+x} \left (625+625 x-775 x^2+400 x^3\right )\right )}{625-750 x+1025 x^2-480 x^3+256 x^4} \, dx\\ &=\int \left (-\frac {50 e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} \left (-25+16 x^2\right )}{\left (25-15 x+16 x^2\right )^2}+\frac {25 e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} \left (25+25 x-31 x^2+16 x^3\right )}{\left (25-15 x+16 x^2\right )^2}\right ) \, dx\\ &=25 \int \frac {e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} \left (25+25 x-31 x^2+16 x^3\right )}{\left (25-15 x+16 x^2\right )^2} \, dx-50 \int \frac {e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} \left (-25+16 x^2\right )}{\left (25-15 x+16 x^2\right )^2} \, dx\\ &=25 \int \left (-\frac {5 e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} (-10+3 x)}{\left (25-15 x+16 x^2\right )^2}+\frac {e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} (-1+x)}{25-15 x+16 x^2}\right ) \, dx-50 \int \left (\frac {5 e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} (-10+3 x)}{\left (25-15 x+16 x^2\right )^2}+\frac {e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{25-15 x+16 x^2}\right ) \, dx\\ &=25 \int \frac {e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} (-1+x)}{25-15 x+16 x^2} \, dx-50 \int \frac {e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{25-15 x+16 x^2} \, dx-125 \int \frac {e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} (-10+3 x)}{\left (25-15 x+16 x^2\right )^2} \, dx-250 \int \frac {e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} (-10+3 x)}{\left (25-15 x+16 x^2\right )^2} \, dx\\ &=25 \int \left (\frac {\left (1+\frac {17 i}{5 \sqrt {55}}\right ) e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{-15-5 i \sqrt {55}+32 x}+\frac {\left (1-\frac {17 i}{5 \sqrt {55}}\right ) e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{-15+5 i \sqrt {55}+32 x}\right ) \, dx-50 \int \left (\frac {32 i e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{5 \sqrt {55} \left (15+5 i \sqrt {55}-32 x\right )}+\frac {32 i e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{5 \sqrt {55} \left (-15+5 i \sqrt {55}+32 x\right )}\right ) \, dx-125 \int \left (-\frac {10 e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{\left (25-15 x+16 x^2\right )^2}+\frac {3 e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} x}{\left (25-15 x+16 x^2\right )^2}\right ) \, dx-250 \int \left (-\frac {10 e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{\left (25-15 x+16 x^2\right )^2}+\frac {3 e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} x}{\left (25-15 x+16 x^2\right )^2}\right ) \, dx\\ &=-\left (375 \int \frac {e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} x}{\left (25-15 x+16 x^2\right )^2} \, dx\right )-750 \int \frac {e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} x}{\left (25-15 x+16 x^2\right )^2} \, dx+1250 \int \frac {e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{\left (25-15 x+16 x^2\right )^2} \, dx+2500 \int \frac {e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{\left (25-15 x+16 x^2\right )^2} \, dx-\left (64 i \sqrt {\frac {5}{11}}\right ) \int \frac {e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{15+5 i \sqrt {55}-32 x} \, dx-\left (64 i \sqrt {\frac {5}{11}}\right ) \int \frac {e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{-15+5 i \sqrt {55}+32 x} \, dx+\frac {1}{11} \left (275-17 i \sqrt {55}\right ) \int \frac {e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{-15+5 i \sqrt {55}+32 x} \, dx+\frac {1}{11} \left (275+17 i \sqrt {55}\right ) \int \frac {e^{5+x+\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}}}{-15-5 i \sqrt {55}+32 x} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.31, size = 24, normalized size = 0.89 \begin {gather*} e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.20, size = 23, normalized size = 0.85
method | result | size |
risch | \({\mathrm e}^{\frac {25 x \left ({\mathrm e}^{5+x}+2\right )}{16 x^{2}-15 x +25}}\) | \(23\) |
norman | \(\frac {-15 x \,{\mathrm e}^{\frac {25 x \,{\mathrm e}^{5+x}+50 x}{16 x^{2}-15 x +25}}+16 x^{2} {\mathrm e}^{\frac {25 x \,{\mathrm e}^{5+x}+50 x}{16 x^{2}-15 x +25}}+25 \,{\mathrm e}^{\frac {25 x \,{\mathrm e}^{5+x}+50 x}{16 x^{2}-15 x +25}}}{16 x^{2}-15 x +25}\) | \(100\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.36, size = 36, normalized size = 1.33 \begin {gather*} e^{\left (\frac {25 \, x e^{\left (x + 5\right )}}{16 \, x^{2} - 15 \, x + 25} + \frac {50 \, x}{16 \, x^{2} - 15 \, x + 25}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 25, normalized size = 0.93 \begin {gather*} e^{\left (\frac {25 \, {\left (x e^{\left (x + 5\right )} + 2 \, x\right )}}{16 \, x^{2} - 15 \, x + 25}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.20, size = 22, normalized size = 0.81 \begin {gather*} e^{\frac {25 x e^{x + 5} + 50 x}{16 x^{2} - 15 x + 25}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.43, size = 36, normalized size = 1.33 \begin {gather*} e^{\left (\frac {25 \, x e^{\left (x + 5\right )}}{16 \, x^{2} - 15 \, x + 25} + \frac {50 \, x}{16 \, x^{2} - 15 \, x + 25}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.66, size = 25, normalized size = 0.93 \begin {gather*} {\mathrm {e}}^{\frac {50\,x+25\,x\,{\mathrm {e}}^5\,{\mathrm {e}}^x}{16\,x^2-15\,x+25}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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