Optimal. Leaf size=22 \[ \frac {e^{8 e^5} x^2}{\left (-e^3+5 x\right )^2} \]
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Rubi [B] time = 0.04, antiderivative size = 45, normalized size of antiderivative = 2.05, number of steps used = 3, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 2074} \begin {gather*} \frac {e^{6+8 e^5}}{25 \left (e^3-5 x\right )^2}-\frac {2 e^{3+8 e^5}}{25 \left (e^3-5 x\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2074
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\left (2 e^{3+8 e^5}\right ) \int \frac {x}{e^9-15 e^6 x+75 e^3 x^2-125 x^3} \, dx\\ &=\left (2 e^{3+8 e^5}\right ) \int \left (\frac {e^3}{5 \left (e^3-5 x\right )^3}-\frac {1}{5 \left (e^3-5 x\right )^2}\right ) \, dx\\ &=\frac {e^{6+8 e^5}}{25 \left (e^3-5 x\right )^2}-\frac {2 e^{3+8 e^5}}{25 \left (e^3-5 x\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.01, size = 29, normalized size = 1.32 \begin {gather*} -\frac {e^{3+8 e^5} \left (e^3-10 x\right )}{25 \left (e^3-5 x\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.03, size = 32, normalized size = 1.45 \begin {gather*} \frac {{\left (10 \, x - e^{3}\right )} e^{\left (8 \, e^{5} + 3\right )}}{25 \, {\left (25 \, x^{2} - 10 \, x e^{3} + e^{6}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, x e^{\left (8 \, e^{5} + 3\right )}}{125 \, x^{3} - 75 \, x^{2} e^{3} + 15 \, x e^{6} - e^{9}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 32, normalized size = 1.45
method | result | size |
norman | \(\frac {\frac {2 \,{\mathrm e}^{3} {\mathrm e}^{8 \,{\mathrm e}^{5}} x}{5}-\frac {{\mathrm e}^{6} {\mathrm e}^{8 \,{\mathrm e}^{5}}}{25}}{\left ({\mathrm e}^{3}-5 x \right )^{2}}\) | \(32\) |
risch | \(\frac {2 \left (\frac {x}{5}-\frac {{\mathrm e}^{3}}{50}\right ) {\mathrm e}^{3+8 \,{\mathrm e}^{5}}}{{\mathrm e}^{6}-10 x \,{\mathrm e}^{3}+25 x^{2}}\) | \(33\) |
gosper | \(-\frac {{\mathrm e}^{8 \,{\mathrm e}^{5}} {\mathrm e}^{3} \left ({\mathrm e}^{3}-10 x \right )}{25 \left ({\mathrm e}^{6}-10 x \,{\mathrm e}^{3}+25 x^{2}\right )}\) | \(35\) |
default | \(-\frac {2 \,{\mathrm e}^{3} {\mathrm e}^{8 \,{\mathrm e}^{5}} \left (\munderset {\textit {\_R} =\RootOf \left (-{\mathrm e}^{9}+15 \textit {\_Z} \,{\mathrm e}^{6}-75 \textit {\_Z}^{2} {\mathrm e}^{3}+125 \textit {\_Z}^{3}\right )}{\sum }\frac {\textit {\_R} \ln \left (x -\textit {\_R} \right )}{{\mathrm e}^{6}-10 \textit {\_R} \,{\mathrm e}^{3}+25 \textit {\_R}^{2}}\right )}{15}\) | \(61\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.38, size = 32, normalized size = 1.45 \begin {gather*} \frac {{\left (10 \, x - e^{3}\right )} e^{\left (8 \, e^{5} + 3\right )}}{25 \, {\left (25 \, x^{2} - 10 \, x e^{3} + e^{6}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.12, size = 37, normalized size = 1.68 \begin {gather*} -\frac {\frac {{\mathrm {e}}^{8\,{\mathrm {e}}^5+6}}{25}-\frac {2\,x\,{\mathrm {e}}^{8\,{\mathrm {e}}^5+3}}{5}}{25\,x^2-10\,{\mathrm {e}}^3\,x+{\mathrm {e}}^6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.15, size = 36, normalized size = 1.64 \begin {gather*} - \frac {2 \left (- 10 x + e^{3}\right ) e^{3} e^{8 e^{5}}}{1250 x^{2} - 500 x e^{3} + 50 e^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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