Optimal. Leaf size=23 \[ x-\frac {x}{3-e^x-25 e^9 \log ^2(4)} \]
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Rubi [B] time = 0.90, antiderivative size = 190, normalized size of antiderivative = 8.26, number of steps used = 17, number of rules used = 12, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {6741, 6742, 2185, 2184, 2190, 2279, 2391, 2191, 2282, 36, 29, 31} \begin {gather*} -\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}+x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {\log \left (-e^x+3-25 e^9 \log ^2(4)\right )}{3-25 e^9 \log ^2(4)}+\frac {x \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {x}{-e^x+3-25 e^9 \log ^2(4)}+\frac {x}{3-25 e^9 \log ^2(4)} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 2184
Rule 2185
Rule 2190
Rule 2191
Rule 2279
Rule 2282
Rule 2391
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x}+e^x \left (-5-x+50 e^9 \log ^2(4)\right )+6 \left (1+\frac {125}{6} e^9 \log ^2(4) \left (-1+5 e^9 \log ^2(4)\right )\right )}{\left (e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )\right )^2} \, dx\\ &=\int \left (1+\frac {x \left (-3+25 e^9 \log ^2(4)\right )}{\left (e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )\right )^2}+\frac {1-x}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right ) \, dx\\ &=x+\left (-3+25 e^9 \log ^2(4)\right ) \int \frac {x}{\left (e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )\right )^2} \, dx+\int \frac {1-x}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}+\frac {\int \frac {e^x (1-x)}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx}{3-25 e^9 \log ^2(4)}-\int \frac {e^x x}{\left (e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )\right )^2} \, dx+\int \frac {x}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x}{3-e^x-25 e^9 \log ^2(4)}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}+\frac {\int \frac {e^x x}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx}{3-25 e^9 \log ^2(4)}+\frac {\int \log \left (1-\frac {e^x}{3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right ) \, dx}{3-25 e^9 \log ^2(4)}-\int \frac {1}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x}{3-e^x-25 e^9 \log ^2(4)}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}+\frac {x \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {\int \log \left (1-\frac {e^x}{3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right ) \, dx}{3-25 e^9 \log ^2(4)}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right )}{x} \, dx,x,e^x\right )}{3-25 e^9 \log ^2(4)}-\operatorname {Subst}\left (\int \frac {1}{x \left (-3+x+25 e^9 \log ^2(4)\right )} \, dx,x,e^x\right )\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x}{3-e^x-25 e^9 \log ^2(4)}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}+\frac {x \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {\text {Li}_2\left (\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right )}{x} \, dx,x,e^x\right )}{3-25 e^9 \log ^2(4)}-\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{-3+25 e^9 \log ^2(4)}+\frac {\operatorname {Subst}\left (\int \frac {1}{-3+x+25 e^9 \log ^2(4)} \, dx,x,e^x\right )}{-3+25 e^9 \log ^2(4)}\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}+\frac {x}{3-25 e^9 \log ^2(4)}-\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x}{3-e^x-25 e^9 \log ^2(4)}-\frac {\log \left (3-e^x-25 e^9 \log ^2(4)\right )}{3-25 e^9 \log ^2(4)}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}+\frac {x \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.16, size = 20, normalized size = 0.87 \begin {gather*} x+\frac {x}{-3+e^x+25 e^9 \log ^2(4)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.84, size = 32, normalized size = 1.39 \begin {gather*} \frac {100 \, x e^{9} \log \relax (2)^{2} + x e^{x} - 2 \, x}{100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.55, size = 359, normalized size = 15.61 \begin {gather*} \frac {1000000 \, x e^{27} \log \relax (2)^{6} + 1000000 \, e^{27} \log \relax (2)^{6} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) - 1000000 \, e^{27} \log \relax (2)^{6} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) - 80000 \, x e^{18} \log \relax (2)^{4} + 10000 \, x e^{\left (x + 18\right )} \log \relax (2)^{4} - 30000 \, e^{18} \log \relax (2)^{4} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) + 10000 \, e^{\left (x + 18\right )} \log \relax (2)^{4} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) + 30000 \, e^{18} \log \relax (2)^{4} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) - 10000 \, e^{\left (x + 18\right )} \log \relax (2)^{4} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) + 2100 \, x e^{9} \log \relax (2)^{2} - 600 \, x e^{\left (x + 9\right )} \log \relax (2)^{2} + 600 \, e^{9} \log \relax (2)^{2} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) - 600 \, e^{9} \log \relax (2)^{2} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) + 9 \, x e^{x} + 6 \, e^{x} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) - 6 \, e^{x} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) - 18 \, x - 18 \, \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) + 18 \, \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right )}{1000000 \, e^{27} \log \relax (2)^{6} - 90000 \, e^{18} \log \relax (2)^{4} + 10000 \, e^{\left (x + 18\right )} \log \relax (2)^{4} + 2700 \, e^{9} \log \relax (2)^{2} - 600 \, e^{\left (x + 9\right )} \log \relax (2)^{2} + 9 \, e^{x} - 27} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 19, normalized size = 0.83
method | result | size |
risch | \(x +\frac {x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}\) | \(19\) |
norman | \(\frac {\left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-2\right ) x +{\mathrm e}^{x} x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}\) | \(33\) |
default | \(-\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}+\frac {2}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}+\ln \left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3\right )+\frac {\frac {6}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}+\frac {6 x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}+\frac {6 x \,{\mathrm e}^{x}}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}-\frac {6 \ln \left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3\right )}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}+\frac {-\frac {500 \,{\mathrm e}^{9} \ln \relax (2)^{2}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}-\frac {500 \,{\mathrm e}^{9} \ln \relax (2)^{2} x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}-\frac {500 \,{\mathrm e}^{9} \ln \relax (2)^{2} x \,{\mathrm e}^{x}}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}+\frac {500 \,{\mathrm e}^{9} \ln \relax (2)^{2} \ln \left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3\right )}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}+\frac {\frac {10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}+\frac {10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4} x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}+\frac {10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4} x \,{\mathrm e}^{x}}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}-\frac {10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4} \ln \left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3\right )}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}-\frac {x \left (\ln \left (\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}+{\mathrm e}^{x}-3}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}-3}\right )-\ln \left (\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}+{\mathrm e}^{x}-3}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}-3}\right )\right )}{200 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}}-\frac {\dilog \left (\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}+{\mathrm e}^{x}-3}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}-3}\right )}{200 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}}+\frac {\dilog \left (\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}+{\mathrm e}^{x}-3}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}-3}\right )}{200 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}}\) | \(786\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.81, size = 437, normalized size = 19.00 \begin {gather*} 10000 \, {\left (\frac {x}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} - \frac {\log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right )}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} + \frac {1}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + {\left (100 \, e^{9} \log \relax (2)^{2} - 3\right )} e^{x} + 9}\right )} e^{18} \log \relax (2)^{4} - 500 \, {\left (\frac {x}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} - \frac {\log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right )}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} + \frac {1}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + {\left (100 \, e^{9} \log \relax (2)^{2} - 3\right )} e^{x} + 9}\right )} e^{9} \log \relax (2)^{2} - \frac {200 \, e^{18} \log \relax (2)^{2}}{100 \, e^{18} \log \relax (2)^{2} - 3 \, e^{9} + e^{\left (x + 9\right )}} - \frac {x e^{x}}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + {\left (100 \, e^{9} \log \relax (2)^{2} - 3\right )} e^{x} + 9} + \frac {100 \, e^{9} \log \relax (2)^{2} - 3}{100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3} + \frac {6 \, x}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} - \frac {6 \, \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right )}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} + \frac {\log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right )}{100 \, e^{9} \log \relax (2)^{2} - 3} + \frac {6}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + {\left (100 \, e^{9} \log \relax (2)^{2} - 3\right )} e^{x} + 9} + \frac {5}{100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3} + \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.23, size = 18, normalized size = 0.78 \begin {gather*} x+\frac {x}{{\mathrm {e}}^x+100\,{\mathrm {e}}^9\,{\ln \relax (2)}^2-3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.12, size = 17, normalized size = 0.74 \begin {gather*} x + \frac {x}{e^{x} - 3 + 100 e^{9} \log {\relax (2 )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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