Integrand size = 164, antiderivative size = 25 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x \log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \]
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\[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right ) \left (\frac {4 x \left (-5+e^{3+2 x} (2+x)^2\right ) \log (x)}{(2+x) \left (-5 x+e^{3+2 x} (2+x)\right )}+(-2+\log (x)) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )\right )}{\log ^3(x)} \, dx \\ & = \int \left (\frac {20 x \left (-1+2 x+x^2\right ) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)}+\frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right ) \left (4 x \log (x)-2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )+\log (x) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )\right )}{\log ^3(x)}\right ) \, dx \\ & = 20 \int \frac {x \left (-1+2 x+x^2\right ) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+\int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right ) \left (4 x \log (x)-2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )+\log (x) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )\right )}{\log ^3(x)} \, dx \\ & = 20 \int \left (-\frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)}+\frac {x^2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)}+\frac {2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)}\right ) \, dx+\int \left (\frac {4 x \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)}+\frac {(-2+\log (x)) \log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^3(x)}\right ) \, dx \\ & = 4 \int \frac {x \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \, dx-20 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+20 \int \frac {x^2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+40 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+\int \frac {(-2+\log (x)) \log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^3(x)} \, dx \\ & = 4 \int \frac {x \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \, dx-20 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (-5 x+e^{3+2 x} (2+x)\right ) \log ^2(x)} \, dx+20 \int \frac {x^2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (-5 x+e^{3+2 x} (2+x)\right ) \log ^2(x)} \, dx+40 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+\int \left (-\frac {2 \log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^3(x)}+\frac {\log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)}\right ) \, dx \\ & = -\left (2 \int \frac {\log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^3(x)} \, dx\right )+4 \int \frac {x \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \, dx-20 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (-5 x+e^{3+2 x} (2+x)\right ) \log ^2(x)} \, dx+20 \int \frac {x^2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (-5 x+e^{3+2 x} (2+x)\right ) \log ^2(x)} \, dx+40 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+\int \frac {\log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \, dx \\ \end{align*}
\[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx \]
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Time = 23.67 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20
method | result | size |
parallelrisch | \(\frac {\ln \left (\frac {\left (2+x \right ) {\mathrm e}^{3+2 x}-5 x}{2+x}\right )^{2} x}{\ln \left (x \right )^{2}}\) | \(30\) |
risch | \(\text {Expression too large to display}\) | \(1011\) |
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x \log \left (\frac {{\left (x + 2\right )} e^{\left (2 \, x + 3\right )} - 5 \, x}{x + 2}\right )^{2}}{\log \left (x\right )^{2}} \]
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Exception generated. \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\text {Exception raised: TypeError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (24) = 48\).
Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x \log \left ({\left (x e^{3} + 2 \, e^{3}\right )} e^{\left (2 \, x\right )} - 5 \, x\right )^{2} - 2 \, x \log \left ({\left (x e^{3} + 2 \, e^{3}\right )} e^{\left (2 \, x\right )} - 5 \, x\right ) \log \left (x + 2\right ) + x \log \left (x + 2\right )^{2}}{\log \left (x\right )^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (24) = 48\).
Time = 0.86 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.68 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x \log \left (x e^{\left (2 \, x + 3\right )} - 5 \, x + 2 \, e^{\left (2 \, x + 3\right )}\right )^{2} - 2 \, x \log \left (x e^{\left (2 \, x + 3\right )} - 5 \, x + 2 \, e^{\left (2 \, x + 3\right )}\right ) \log \left (x + 2\right ) + x \log \left (x + 2\right )^{2}}{\log \left (x\right )^{2}} \]
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Time = 12.93 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx=\frac {x\,{\ln \left (-\frac {5\,x-{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3\,\left (x+2\right )}{x+2}\right )}^2}{{\ln \left (x\right )}^2} \]
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