Optimal. Leaf size=30 \[ x \left (x+\frac {(-4+x) (5+x)}{5 \left (e^x-\frac {25}{x^2}+e^x x\right )}\right ) \]
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Rubi [F] time = 2.73, 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 {6250 x+1500 x^2-100 x^3-125 x^4+10 e^{2 x} x^5+10 e^{2 x} x^7+e^x x \left (-500 x^3+21 x^5+20 e^x x^5+x^6-x^7\right )+e^x \left (-500 x^3-20 x^4+22 x^5+2 x^6-x^7\right )}{3125-250 e^x x^2+5 e^{2 x} x^4+5 e^{2 x} x^6+e^x x \left (-250 x^2+10 e^x 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 {x \left (6250+1500 x-100 \left (1+5 e^x\right ) x^2-5 \left (25+104 e^x\right ) x^3+2 e^x \left (11+5 e^x\right ) x^4+e^x \left (23+20 e^x\right ) x^5+10 e^{2 x} x^6-e^x x^7\right )}{5 \left (25-e^x x^2 (1+x)\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {x \left (6250+1500 x-100 \left (1+5 e^x\right ) x^2-5 \left (25+104 e^x\right ) x^3+2 e^x \left (11+5 e^x\right ) x^4+e^x \left (23+20 e^x\right ) x^5+10 e^{2 x} x^6-e^x x^7\right )}{\left (25-e^x x^2 (1+x)\right )^2} \, dx\\ &=\frac {1}{5} \int \left (10 x-\frac {x^2 \left (20-22 x-23 x^2+x^4\right )}{(1+x) \left (-25+e^x x^2+e^x x^3\right )}-\frac {25 x^2 \left (-40-78 x-14 x^2+5 x^3+x^4\right )}{(1+x) \left (-25+e^x x^2+e^x x^3\right )^2}\right ) \, dx\\ &=x^2-\frac {1}{5} \int \frac {x^2 \left (20-22 x-23 x^2+x^4\right )}{(1+x) \left (-25+e^x x^2+e^x x^3\right )} \, dx-5 \int \frac {x^2 \left (-40-78 x-14 x^2+5 x^3+x^4\right )}{(1+x) \left (-25+e^x x^2+e^x x^3\right )^2} \, dx\\ &=x^2-\frac {1}{5} \int \left (-\frac {20}{-25+e^x x^2+e^x x^3}+\frac {20 x}{-25+e^x x^2+e^x x^3}-\frac {22 x^3}{-25+e^x x^2+e^x x^3}-\frac {x^4}{-25+e^x x^2+e^x x^3}+\frac {x^5}{-25+e^x x^2+e^x x^3}+\frac {20}{(1+x) \left (-25+e^x x^2+e^x x^3\right )}\right ) \, dx-5 \int \left (-\frac {20}{\left (-25+e^x x^2+e^x x^3\right )^2}+\frac {20 x}{\left (-25+e^x x^2+e^x x^3\right )^2}-\frac {60 x^2}{\left (-25+e^x x^2+e^x x^3\right )^2}-\frac {18 x^3}{\left (-25+e^x x^2+e^x x^3\right )^2}+\frac {4 x^4}{\left (-25+e^x x^2+e^x x^3\right )^2}+\frac {x^5}{\left (-25+e^x x^2+e^x x^3\right )^2}+\frac {20}{(1+x) \left (-25+e^x x^2+e^x x^3\right )^2}\right ) \, dx\\ &=x^2+\frac {1}{5} \int \frac {x^4}{-25+e^x x^2+e^x x^3} \, dx-\frac {1}{5} \int \frac {x^5}{-25+e^x x^2+e^x x^3} \, dx+4 \int \frac {1}{-25+e^x x^2+e^x x^3} \, dx-4 \int \frac {x}{-25+e^x x^2+e^x x^3} \, dx-4 \int \frac {1}{(1+x) \left (-25+e^x x^2+e^x x^3\right )} \, dx+\frac {22}{5} \int \frac {x^3}{-25+e^x x^2+e^x x^3} \, dx-5 \int \frac {x^5}{\left (-25+e^x x^2+e^x x^3\right )^2} \, dx-20 \int \frac {x^4}{\left (-25+e^x x^2+e^x x^3\right )^2} \, dx+90 \int \frac {x^3}{\left (-25+e^x x^2+e^x x^3\right )^2} \, dx+100 \int \frac {1}{\left (-25+e^x x^2+e^x x^3\right )^2} \, dx-100 \int \frac {x}{\left (-25+e^x x^2+e^x x^3\right )^2} \, dx-100 \int \frac {1}{(1+x) \left (-25+e^x x^2+e^x x^3\right )^2} \, dx+300 \int \frac {x^2}{\left (-25+e^x x^2+e^x x^3\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.06, size = 31, normalized size = 1.03 \begin {gather*} \frac {1}{5} x^2 \left (5+\frac {x \left (-20+x+x^2\right )}{-25+e^x x^2 (1+x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 48, normalized size = 1.60 \begin {gather*} \frac {x^{5} + x^{4} - 20 \, x^{3} - 125 \, x^{2} + 5 \, {\left (x^{4} + x^{3}\right )} e^{\left (x + \log \relax (x)\right )}}{5 \, {\left ({\left (x^{2} + x\right )} e^{\left (x + \log \relax (x)\right )} - 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.26, size = 49, normalized size = 1.63 \begin {gather*} \frac {5 \, x^{5} e^{x} + x^{5} + 5 \, x^{4} e^{x} + x^{4} - 20 \, x^{3} - 125 \, x^{2}}{5 \, {\left (x^{3} e^{x} + x^{2} e^{x} - 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 32, normalized size = 1.07
method | result | size |
risch | \(x^{2}+\frac {\left (x^{2}+x -20\right ) x^{3}}{5 \,{\mathrm e}^{x} x^{3}+5 \,{\mathrm e}^{x} x^{2}-125}\) | \(32\) |
norman | \(\frac {x^{5} {\mathrm e}^{x}+{\mathrm e}^{x} x^{4}-25 x^{2}-4 x^{3}+\frac {x^{4}}{5}+\frac {x^{5}}{5}}{{\mathrm e}^{x} x^{3}+{\mathrm e}^{x} x^{2}-25}\) | \(51\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 44, normalized size = 1.47 \begin {gather*} \frac {x^{5} + x^{4} - 20 \, x^{3} - 125 \, x^{2} + 5 \, {\left (x^{5} + x^{4}\right )} e^{x}}{5 \, {\left ({\left (x^{3} + x^{2}\right )} e^{x} - 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {6250\,x+10\,x^5\,{\mathrm {e}}^{2\,x+2\,\ln \relax (x)}-{\mathrm {e}}^x\,\left (x^7-2\,x^6-22\,x^5+20\,x^4+500\,x^3\right )+{\mathrm {e}}^{x+\ln \relax (x)}\,\left (20\,x^5\,{\mathrm {e}}^x-500\,x^3+21\,x^5+x^6-x^7\right )+10\,x^5\,{\mathrm {e}}^{2\,x}+1500\,x^2-100\,x^3-125\,x^4}{5\,x^4\,{\mathrm {e}}^{2\,x+2\,\ln \relax (x)}-250\,x^2\,{\mathrm {e}}^x+{\mathrm {e}}^{x+\ln \relax (x)}\,\left (10\,x^4\,{\mathrm {e}}^x-250\,x^2\right )+5\,x^4\,{\mathrm {e}}^{2\,x}+3125} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.22, size = 29, normalized size = 0.97 \begin {gather*} x^{2} + \frac {x^{5} + x^{4} - 20 x^{3}}{\left (5 x^{3} + 5 x^{2}\right ) e^{x} - 125} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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