Integrand size = 63, antiderivative size = 25 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=8 \left (e^2 x^2+\frac {9 x^2 \log (5+x)}{5 \log (x)}\right ) \]
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\[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {8}{5} x \left (10 e^2-\frac {9 \log (5+x)}{\log ^2(x)}+\frac {9 (x+2 (5+x) \log (5+x))}{(5+x) \log (x)}\right ) \, dx \\ & = \frac {8}{5} \int x \left (10 e^2-\frac {9 \log (5+x)}{\log ^2(x)}+\frac {9 (x+2 (5+x) \log (5+x))}{(5+x) \log (x)}\right ) \, dx \\ & = \frac {8}{5} \int \left (\frac {x \left (9 x+50 e^2 \log (x)+10 e^2 x \log (x)\right )}{(5+x) \log (x)}+\frac {9 x (-1+2 \log (x)) \log (5+x)}{\log ^2(x)}\right ) \, dx \\ & = \frac {8}{5} \int \frac {x \left (9 x+50 e^2 \log (x)+10 e^2 x \log (x)\right )}{(5+x) \log (x)} \, dx+\frac {72}{5} \int \frac {x (-1+2 \log (x)) \log (5+x)}{\log ^2(x)} \, dx \\ & = \frac {8}{5} \int x \left (10 e^2+\frac {9 x}{(5+x) \log (x)}\right ) \, dx+\frac {72}{5} \int \left (-\frac {x \log (5+x)}{\log ^2(x)}+\frac {2 x \log (5+x)}{\log (x)}\right ) \, dx \\ & = \frac {8}{5} \int \left (10 e^2 x+\frac {9 x^2}{(5+x) \log (x)}\right ) \, dx-\frac {72}{5} \int \frac {x \log (5+x)}{\log ^2(x)} \, dx+\frac {144}{5} \int \frac {x \log (5+x)}{\log (x)} \, dx \\ & = 8 e^2 x^2+\frac {72}{5} \int \frac {x^2}{(5+x) \log (x)} \, dx-\frac {72}{5} \int \frac {x \log (5+x)}{\log ^2(x)} \, dx+\frac {144}{5} \int \frac {x \log (5+x)}{\log (x)} \, dx \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\frac {8}{5} \left (5 e^2 x^2+\frac {9 x^2 \log (5+x)}{\log (x)}\right ) \]
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Time = 4.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(8 x^{2} {\mathrm e}^{2}+\frac {72 x^{2} \ln \left (5+x \right )}{5 \ln \left (x \right )}\) | \(22\) |
parallelrisch | \(\frac {40 x^{2} {\mathrm e}^{2} \ln \left (x \right )+72 \ln \left (5+x \right ) x^{2}-1000 \,{\mathrm e}^{2} \ln \left (x \right )}{5 \ln \left (x \right )}\) | \(32\) |
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Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\frac {8 \, {\left (5 \, x^{2} e^{2} \log \left (x\right ) + 9 \, x^{2} \log \left (x + 5\right )\right )}}{5 \, \log \left (x\right )} \]
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Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=8 x^{2} e^{2} + \frac {72 x^{2} \log {\left (x + 5 \right )}}{5 \log {\left (x \right )}} \]
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Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\frac {8 \, {\left (5 \, x^{2} e^{2} \log \left (x\right ) + 9 \, x^{2} \log \left (x + 5\right )\right )}}{5 \, \log \left (x\right )} \]
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Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\frac {8 \, {\left (5 \, x^{2} e^{2} \log \left (x\right ) + 9 \, x^{2} \log \left (x + 5\right )\right )}}{5 \, \log \left (x\right )} \]
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Time = 14.74 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=8\,x^2\,{\mathrm {e}}^2+\frac {72\,x^2\,\ln \left (x+5\right )}{5\,\ln \left (x\right )} \]
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