Optimal. Leaf size=26 \[ e^{\frac {50 e^9 \left (\frac {e^x}{5}+x\right )^2 \left (5+x^2\right )}{x^2}} \]
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Rubi [F] time = 4.69, 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 {\exp \left (\frac {e^{9+2 x} \left (10+2 x^2\right )+e^{9+x} \left (100 x+20 x^3\right )+e^9 \left (250 x^2+50 x^4\right )}{x^2}\right ) \left (100 e^9 x^4+e^{9+2 x} \left (-20+20 x+4 x^3\right )+e^{9+x} \left (-100 x+100 x^2+20 x^3+20 x^4\right )\right )}{x^3} \, 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 {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} \left (100 e^9 x^4+e^{9+2 x} \left (-20+20 x+4 x^3\right )+e^{9+x} \left (-100 x+100 x^2+20 x^3+20 x^4\right )\right )}{x^3} \, dx\\ &=\int \left (100 e^{9+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} x+\frac {4 \exp \left (9+2 x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}\right ) \left (-5+5 x+x^3\right )}{x^3}+\frac {20 e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} \left (-5+5 x+x^2+x^3\right )}{x^2}\right ) \, dx\\ &=4 \int \frac {\exp \left (9+2 x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}\right ) \left (-5+5 x+x^3\right )}{x^3} \, dx+20 \int \frac {e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} \left (-5+5 x+x^2+x^3\right )}{x^2} \, dx+100 \int e^{9+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} x \, dx\\ &=4 \int \left (\exp \left (9+2 x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}\right )-\frac {5 \exp \left (9+2 x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}\right )}{x^3}+\frac {5 \exp \left (9+2 x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}\right )}{x^2}\right ) \, dx+20 \int \left (e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}}-\frac {5 e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}}}{x^2}+\frac {5 e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}}}{x}+e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} x\right ) \, dx+100 \int e^{9+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} x \, dx\\ &=4 \int \exp \left (9+2 x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}\right ) \, dx+20 \int e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} \, dx-20 \int \frac {\exp \left (9+2 x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}\right )}{x^3} \, dx+20 \int \frac {\exp \left (9+2 x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}\right )}{x^2} \, dx+20 \int e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} x \, dx-100 \int \frac {e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}}}{x^2} \, dx+100 \int \frac {e^{9+x+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}}}{x} \, dx+100 \int e^{9+\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} x \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 24, normalized size = 0.92 \begin {gather*} e^{\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.89, size = 47, normalized size = 1.81 \begin {gather*} e^{\left (\frac {2 \, {\left (25 \, {\left (x^{4} + 5 \, x^{2}\right )} e^{18} + {\left (x^{2} + 5\right )} e^{\left (2 \, x + 18\right )} + 10 \, {\left (x^{3} + 5 \, x\right )} e^{\left (x + 18\right )}\right )} e^{\left (-9\right )}}{x^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.42, size = 48, normalized size = 1.85 \begin {gather*} e^{\left (50 \, x^{2} e^{9} + 20 \, x e^{\left (x + 9\right )} + \frac {100 \, e^{\left (x + 9\right )}}{x} + \frac {10 \, e^{\left (2 \, x + 9\right )}}{x^{2}} + 250 \, e^{9} + 2 \, e^{\left (2 \, x + 9\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 33, normalized size = 1.27
method | result | size |
risch | \({\mathrm e}^{\frac {2 \left (x^{2}+5\right ) \left (25 x^{2} {\mathrm e}^{9}+10 x \,{\mathrm e}^{x +9}+{\mathrm e}^{2 x +9}\right )}{x^{2}}}\) | \(33\) |
norman | \({\mathrm e}^{\frac {\left (2 x^{2}+10\right ) {\mathrm e}^{9} {\mathrm e}^{2 x}+\left (20 x^{3}+100 x \right ) {\mathrm e}^{9} {\mathrm e}^{x}+\left (50 x^{4}+250 x^{2}\right ) {\mathrm e}^{9}}{x^{2}}}\) | \(49\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.07, size = 48, normalized size = 1.85 \begin {gather*} e^{\left (50 \, x^{2} e^{9} + 20 \, x e^{\left (x + 9\right )} + \frac {100 \, e^{\left (x + 9\right )}}{x} + \frac {10 \, e^{\left (2 \, x + 9\right )}}{x^{2}} + 250 \, e^{9} + 2 \, e^{\left (2 \, x + 9\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.22, size = 53, normalized size = 2.04 \begin {gather*} {\mathrm {e}}^{50\,x^2\,{\mathrm {e}}^9}\,{\mathrm {e}}^{250\,{\mathrm {e}}^9}\,{\mathrm {e}}^{\frac {10\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^9}{x^2}}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^9}\,{\mathrm {e}}^{20\,x\,{\mathrm {e}}^9\,{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {100\,{\mathrm {e}}^9\,{\mathrm {e}}^x}{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.44, size = 48, normalized size = 1.85 \begin {gather*} e^{\frac {\left (2 x^{2} + 10\right ) e^{9} e^{2 x} + \left (20 x^{3} + 100 x\right ) e^{9} e^{x} + \left (50 x^{4} + 250 x^{2}\right ) e^{9}}{x^{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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