Integrand size = 89, antiderivative size = 27 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=x-\log (4-x)+\log (x) \left (-e^{7+x}+\log (x)+\log (1+x)\right ) \]
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Time = 0.69 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78, number of steps used = 32, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {1608, 6860, 84, 630, 31, 646, 2404, 2353, 2352, 2354, 2438, 2338, 2326} \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=x+\log ^2(x)-e^{x+7} \log (x)+\log (x+1) \log (x)-2 \log (4) \log (8-2 x)-\log (4-x)+2 \log (4) \log (x-4) \]
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Rule 31
Rule 84
Rule 630
Rule 646
Rule 1608
Rule 2326
Rule 2338
Rule 2352
Rule 2353
Rule 2354
Rule 2404
Rule 2438
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{x \left (-4-3 x+x^2\right )} \, dx \\ & = \int \left (\frac {x^2}{(-4+x) (1+x)}-\frac {5}{-4-3 x+x^2}-\frac {4 x}{-4-3 x+x^2}+\frac {3 x \log (x)}{(-4+x) (1+x)}-\frac {10 \log (x)}{-4-3 x+x^2}-\frac {8 \log (x)}{x \left (-4-3 x+x^2\right )}-\frac {e^{7+x} (1+x \log (x))}{x}+\frac {\log (1+x)}{x}\right ) \, dx \\ & = 3 \int \frac {x \log (x)}{(-4+x) (1+x)} \, dx-4 \int \frac {x}{-4-3 x+x^2} \, dx-5 \int \frac {1}{-4-3 x+x^2} \, dx-8 \int \frac {\log (x)}{x \left (-4-3 x+x^2\right )} \, dx-10 \int \frac {\log (x)}{-4-3 x+x^2} \, dx+\int \frac {x^2}{(-4+x) (1+x)} \, dx-\int \frac {e^{7+x} (1+x \log (x))}{x} \, dx+\int \frac {\log (1+x)}{x} \, dx \\ & = -e^{7+x} \log (x)-\operatorname {PolyLog}(2,-x)-\frac {4}{5} \int \frac {1}{1+x} \, dx+3 \int \left (\frac {4 \log (x)}{5 (-4+x)}+\frac {\log (x)}{5 (1+x)}\right ) \, dx-\frac {16}{5} \int \frac {1}{-4+x} \, dx-8 \int \left (\frac {\log (x)}{20 (-4+x)}-\frac {\log (x)}{4 x}+\frac {\log (x)}{5 (1+x)}\right ) \, dx-10 \int \left (-\frac {2 \log (x)}{5 (8-2 x)}-\frac {2 \log (x)}{5 (2+2 x)}\right ) \, dx-\int \frac {1}{-4+x} \, dx+\int \frac {1}{1+x} \, dx+\int \left (1+\frac {16}{5 (-4+x)}-\frac {1}{5 (1+x)}\right ) \, dx \\ & = x-\log (4-x)-e^{7+x} \log (x)-\operatorname {PolyLog}(2,-x)-\frac {2}{5} \int \frac {\log (x)}{-4+x} \, dx+\frac {3}{5} \int \frac {\log (x)}{1+x} \, dx-\frac {8}{5} \int \frac {\log (x)}{1+x} \, dx+2 \int \frac {\log (x)}{x} \, dx+\frac {12}{5} \int \frac {\log (x)}{-4+x} \, dx+4 \int \frac {\log (x)}{8-2 x} \, dx+4 \int \frac {\log (x)}{2+2 x} \, dx \\ & = x-2 \log (4) \log (8-2 x)-\log (4-x)+2 \log (4) \log (-4+x)-e^{7+x} \log (x)+\log ^2(x)+\log (x) \log (1+x)-\operatorname {PolyLog}(2,-x)-\frac {2}{5} \int \frac {\log \left (\frac {x}{4}\right )}{-4+x} \, dx-\frac {3}{5} \int \frac {\log (1+x)}{x} \, dx+\frac {8}{5} \int \frac {\log (1+x)}{x} \, dx-2 \int \frac {\log (1+x)}{x} \, dx+\frac {12}{5} \int \frac {\log \left (\frac {x}{4}\right )}{-4+x} \, dx+4 \int \frac {\log \left (\frac {x}{4}\right )}{8-2 x} \, dx \\ & = x-2 \log (4) \log (8-2 x)-\log (4-x)+2 \log (4) \log (-4+x)-e^{7+x} \log (x)+\log ^2(x)+\log (x) \log (1+x) \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=x-\log (4-x)-e^{7+x} \log (x)+\log ^2(x)+\log (x) \log (1+x) \]
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Time = 1.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
default | \(x -\ln \left (x -4\right )-\ln \left (x \right ) {\mathrm e}^{x +7}+\ln \left (x \right )^{2}+\ln \left (x \right ) \ln \left (1+x \right )\) | \(28\) |
risch | \(x -\ln \left (x -4\right )-\ln \left (x \right ) {\mathrm e}^{x +7}+\ln \left (x \right )^{2}+\ln \left (x \right ) \ln \left (1+x \right )\) | \(28\) |
parts | \(x -\ln \left (x -4\right )-\ln \left (x \right ) {\mathrm e}^{x +7}+\ln \left (x \right )^{2}+\ln \left (x \right ) \ln \left (1+x \right )\) | \(28\) |
parallelrisch | \(\ln \left (x \right )^{2}+\ln \left (x \right ) \ln \left (1+x \right )-\ln \left (x \right ) {\mathrm e}^{x +7}-\ln \left (x -4\right )+x -\frac {1}{2}\) | \(29\) |
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=-e^{\left (x + 7\right )} \log \left (x\right ) + \log \left (x + 1\right ) \log \left (x\right ) + \log \left (x\right )^{2} + x - \log \left (x - 4\right ) \]
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Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=x - e^{x + 7} \log {\left (x \right )} + \log {\left (x \right )}^{2} + \log {\left (x \right )} \log {\left (x + 1 \right )} - \log {\left (x - 4 \right )} \]
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Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=-e^{\left (x + 7\right )} \log \left (x\right ) + \log \left (x + 1\right ) \log \left (x\right ) + \log \left (x\right )^{2} + x - \log \left (x - 4\right ) \]
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Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=-e^{\left (x + 7\right )} \log \left (x\right ) + \log \left (x + 1\right ) \log \left (x\right ) + \log \left (x\right )^{2} + x - \log \left (x - 4\right ) \]
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Time = 17.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=x-\ln \left (x-4\right )+{\ln \left (x\right )}^2-{\mathrm {e}}^{x+7}\,\ln \left (x\right )+\ln \left (x+1\right )\,\ln \left (x\right ) \]
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