Optimal. Leaf size=34 \[ 10-\frac {-x+\frac {3 (1-x)}{\left (\frac {18}{5}-x\right ) \left (e^x+x\right )}}{e^5} \]
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Rubi [F] time = 1.13, 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 {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^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 {270+e^{2 x} (18-5 x)^2-150 x+399 x^2-180 x^3+25 x^4+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^5 (18-5 x)^2 \left (e^x+x\right )^2} \, dx\\ &=\frac {\int \frac {270+e^{2 x} (18-5 x)^2-150 x+399 x^2-180 x^3+25 x^4+e^x \left (465+303 x-285 x^2+50 x^3\right )}{(18-5 x)^2 \left (e^x+x\right )^2} \, dx}{e^5}\\ &=\frac {\int \left (1-\frac {15 (-1+x)^2}{\left (e^x+x\right )^2 (-18+5 x)}+\frac {15 \left (31-23 x+5 x^2\right )}{\left (e^x+x\right ) (-18+5 x)^2}\right ) \, dx}{e^5}\\ &=\frac {x}{e^5}-\frac {15 \int \frac {(-1+x)^2}{\left (e^x+x\right )^2 (-18+5 x)} \, dx}{e^5}+\frac {15 \int \frac {31-23 x+5 x^2}{\left (e^x+x\right ) (-18+5 x)^2} \, dx}{e^5}\\ &=\frac {x}{e^5}-\frac {15 \int \left (\frac {8}{25 \left (e^x+x\right )^2}+\frac {x}{5 \left (e^x+x\right )^2}+\frac {169}{25 \left (e^x+x\right )^2 (-18+5 x)}\right ) \, dx}{e^5}+\frac {15 \int \left (\frac {1}{5 \left (e^x+x\right )}+\frac {13}{\left (e^x+x\right ) (-18+5 x)^2}+\frac {13}{5 \left (e^x+x\right ) (-18+5 x)}\right ) \, dx}{e^5}\\ &=\frac {x}{e^5}-\frac {3 \int \frac {x}{\left (e^x+x\right )^2} \, dx}{e^5}+\frac {3 \int \frac {1}{e^x+x} \, dx}{e^5}-\frac {24 \int \frac {1}{\left (e^x+x\right )^2} \, dx}{5 e^5}+\frac {39 \int \frac {1}{\left (e^x+x\right ) (-18+5 x)} \, dx}{e^5}-\frac {507 \int \frac {1}{\left (e^x+x\right )^2 (-18+5 x)} \, dx}{5 e^5}+\frac {195 \int \frac {1}{\left (e^x+x\right ) (-18+5 x)^2} \, dx}{e^5}\\ &=\frac {x}{e^5}-\frac {3}{e^5 \left (e^x+x\right )}-\frac {3 \int \frac {1}{\left (e^x+x\right )^2} \, dx}{e^5}-\frac {24 \int \frac {1}{\left (e^x+x\right )^2} \, dx}{5 e^5}+\frac {39 \int \frac {1}{\left (e^x+x\right ) (-18+5 x)} \, dx}{e^5}-\frac {507 \int \frac {1}{\left (e^x+x\right )^2 (-18+5 x)} \, dx}{5 e^5}+\frac {195 \int \frac {1}{\left (e^x+x\right ) (-18+5 x)^2} \, dx}{e^5}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.25, size = 25, normalized size = 0.74 \begin {gather*} \frac {x-\frac {15 (-1+x)}{\left (e^x+x\right ) (-18+5 x)}}{e^5} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.73, size = 59, normalized size = 1.74 \begin {gather*} \frac {{\left (5 \, x^{3} - 18 \, x^{2} - 15 \, x + 15\right )} e^{5} + {\left (5 \, x^{2} - 18 \, x\right )} e^{\left (x + 5\right )}}{{\left (5 \, x^{2} - 18 \, x\right )} e^{10} + {\left (5 \, x - 18\right )} e^{\left (x + 10\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.19, size = 56, normalized size = 1.65 \begin {gather*} \frac {5 \, x^{3} + 5 \, x^{2} e^{x} - 18 \, x^{2} - 18 \, x e^{x} - 15 \, x + 15}{5 \, x^{2} e^{5} - 18 \, x e^{5} + 5 \, x e^{\left (x + 5\right )} - 18 \, e^{\left (x + 5\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 26, normalized size = 0.76
method | result | size |
risch | \(x \,{\mathrm e}^{-5}-\frac {15 \left (x -1\right ) {\mathrm e}^{-5}}{\left (5 x -18\right ) \left ({\mathrm e}^{x}+x \right )}\) | \(26\) |
norman | \(\frac {-\frac {399 x \,{\mathrm e}^{-5}}{5}-\frac {324 \,{\mathrm e}^{-5} {\mathrm e}^{x}}{5}+15 \,{\mathrm e}^{-5}+5 x^{3} {\mathrm e}^{-5}+5 x^{2} {\mathrm e}^{-5} {\mathrm e}^{x}}{5 \,{\mathrm e}^{x} x +5 x^{2}-18 \,{\mathrm e}^{x}-18 x}\) | \(64\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.43, size = 56, normalized size = 1.65 \begin {gather*} \frac {5 \, x^{3} - 18 \, x^{2} + {\left (5 \, x^{2} - 18 \, x\right )} e^{x} - 15 \, x + 15}{5 \, x^{2} e^{5} - 18 \, x e^{5} + {\left (5 \, x e^{5} - 18 \, e^{5}\right )} e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.27, size = 42, normalized size = 1.24 \begin {gather*} x\,{\mathrm {e}}^{-5}-\frac {15\,{\mathrm {e}}^{-5}\,\left (5\,x^3-28\,x^2+41\,x-18\right )}{{\left (5\,x-18\right )}^2\,\left (x+{\mathrm {e}}^x\right )\,\left (x-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.19, size = 39, normalized size = 1.15 \begin {gather*} \frac {x}{e^{5}} + \frac {15 - 15 x}{5 x^{2} e^{5} - 18 x e^{5} + \left (5 x e^{5} - 18 e^{5}\right ) e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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