Optimal. Leaf size=25 \[ \frac {5 x \left (x^2-2 x (5+x)\right )}{16 \left (10+4 e^x+x\right )} \]
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Rubi [F] time = 0.93, 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 {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x \left (-(10+x)^2+2 e^x \left (-20+7 x+x^2\right )\right )}{8 \left (10+4 e^x+x\right )^2} \, dx\\ &=\frac {5}{8} \int \frac {x \left (-(10+x)^2+2 e^x \left (-20+7 x+x^2\right )\right )}{\left (10+4 e^x+x\right )^2} \, dx\\ &=\frac {5}{8} \int \left (\frac {x \left (-20+7 x+x^2\right )}{2 \left (10+4 e^x+x\right )}-\frac {x^2 \left (90+19 x+x^2\right )}{2 \left (10+4 e^x+x\right )^2}\right ) \, dx\\ &=\frac {5}{16} \int \frac {x \left (-20+7 x+x^2\right )}{10+4 e^x+x} \, dx-\frac {5}{16} \int \frac {x^2 \left (90+19 x+x^2\right )}{\left (10+4 e^x+x\right )^2} \, dx\\ &=-\left (\frac {5}{16} \int \left (\frac {90 x^2}{\left (10+4 e^x+x\right )^2}+\frac {19 x^3}{\left (10+4 e^x+x\right )^2}+\frac {x^4}{\left (10+4 e^x+x\right )^2}\right ) \, dx\right )+\frac {5}{16} \int \left (-\frac {20 x}{10+4 e^x+x}+\frac {7 x^2}{10+4 e^x+x}+\frac {x^3}{10+4 e^x+x}\right ) \, dx\\ &=-\left (\frac {5}{16} \int \frac {x^4}{\left (10+4 e^x+x\right )^2} \, dx\right )+\frac {5}{16} \int \frac {x^3}{10+4 e^x+x} \, dx+\frac {35}{16} \int \frac {x^2}{10+4 e^x+x} \, dx-\frac {95}{16} \int \frac {x^3}{\left (10+4 e^x+x\right )^2} \, dx-\frac {25}{4} \int \frac {x}{10+4 e^x+x} \, dx-\frac {225}{8} \int \frac {x^2}{\left (10+4 e^x+x\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.20, size = 20, normalized size = 0.80 \begin {gather*} -\frac {5 x^2 (10+x)}{16 \left (10+4 e^x+x\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 20, normalized size = 0.80 \begin {gather*} -\frac {5 \, {\left (x^{3} + 10 \, x^{2}\right )}}{16 \, {\left (x + 4 \, e^{x} + 10\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 20, normalized size = 0.80 \begin {gather*} -\frac {5 \, {\left (x^{3} + 10 \, x^{2}\right )}}{16 \, {\left (x + 4 \, e^{x} + 10\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 18, normalized size = 0.72
method | result | size |
risch | \(-\frac {5 \left (x +10\right ) x^{2}}{16 \left (4 \,{\mathrm e}^{x}+10+x \right )}\) | \(18\) |
norman | \(\frac {-\frac {25}{8} x^{2}-\frac {5}{16} x^{3}}{4 \,{\mathrm e}^{x}+10+x}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 20, normalized size = 0.80 \begin {gather*} -\frac {5 \, {\left (x^{3} + 10 \, x^{2}\right )}}{16 \, {\left (x + 4 \, e^{x} + 10\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.12, size = 24, normalized size = 0.96 \begin {gather*} -\frac {\frac {5\,x^3}{2}+25\,x^2}{8\,x+32\,{\mathrm {e}}^x+80} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.13, size = 20, normalized size = 0.80 \begin {gather*} \frac {- 5 x^{3} - 50 x^{2}}{16 x + 64 e^{x} + 160} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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