Integrand size = 117, antiderivative size = 25 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=-4+e^x-\frac {x}{2}-\frac {\log (x)}{(3+x) \log (2+x)} \]
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\[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=\int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x \log (x) (3+x+(2+x) \log (2+x))+\left (6+5 x+x^2\right ) \log (2+x) \left (-2+\left (-1+2 e^x\right ) x (3+x) \log (2+x)\right )}{2 x (2+x) (3+x)^2 \log ^2(2+x)} \, dx \\ & = \frac {1}{2} \int \frac {2 x \log (x) (3+x+(2+x) \log (2+x))+\left (6+5 x+x^2\right ) \log (2+x) \left (-2+\left (-1+2 e^x\right ) x (3+x) \log (2+x)\right )}{x (2+x) (3+x)^2 \log ^2(2+x)} \, dx \\ & = \frac {1}{2} \int \left (2 e^x+\frac {6 x \log (x)+2 x^2 \log (x)-12 \log (2+x)-10 x \log (2+x)-2 x^2 \log (2+x)+4 x \log (x) \log (2+x)+2 x^2 \log (x) \log (2+x)-18 x \log ^2(2+x)-21 x^2 \log ^2(2+x)-8 x^3 \log ^2(2+x)-x^4 \log ^2(2+x)}{x (2+x) (3+x)^2 \log ^2(2+x)}\right ) \, dx \\ & = \frac {1}{2} \int \frac {6 x \log (x)+2 x^2 \log (x)-12 \log (2+x)-10 x \log (2+x)-2 x^2 \log (2+x)+4 x \log (x) \log (2+x)+2 x^2 \log (x) \log (2+x)-18 x \log ^2(2+x)-21 x^2 \log ^2(2+x)-8 x^3 \log ^2(2+x)-x^4 \log ^2(2+x)}{x (2+x) (3+x)^2 \log ^2(2+x)} \, dx+\int e^x \, dx \\ & = e^x+\frac {1}{2} \int \frac {2 x \log (x) (3+x+(2+x) \log (2+x))-\left (6+5 x+x^2\right ) \log (2+x) (2+x (3+x) \log (2+x))}{x (2+x) (3+x)^2 \log ^2(2+x)} \, dx \\ & = e^x+\frac {1}{2} \int \left (-1+\frac {2 \log (x)}{(2+x) (3+x) \log ^2(2+x)}+\frac {2 (-3-x+x \log (x))}{x (3+x)^2 \log (2+x)}\right ) \, dx \\ & = e^x-\frac {x}{2}+\int \frac {\log (x)}{(2+x) (3+x) \log ^2(2+x)} \, dx+\int \frac {-3-x+x \log (x)}{x (3+x)^2 \log (2+x)} \, dx \\ & = e^x-\frac {x}{2}+\int \left (\frac {\log (x)}{(2+x) \log ^2(2+x)}-\frac {\log (x)}{(3+x) \log ^2(2+x)}\right ) \, dx+\int \left (\frac {3+x-x \log (x)}{3 (3+x)^2 \log (2+x)}+\frac {3+x-x \log (x)}{9 (3+x) \log (2+x)}+\frac {-3-x+x \log (x)}{9 x \log (2+x)}\right ) \, dx \\ & = e^x-\frac {x}{2}+\frac {1}{9} \int \frac {3+x-x \log (x)}{(3+x) \log (2+x)} \, dx+\frac {1}{9} \int \frac {-3-x+x \log (x)}{x \log (2+x)} \, dx+\frac {1}{3} \int \frac {3+x-x \log (x)}{(3+x)^2 \log (2+x)} \, dx+\int \frac {\log (x)}{(2+x) \log ^2(2+x)} \, dx-\int \frac {\log (x)}{(3+x) \log ^2(2+x)} \, dx \\ & = e^x-\frac {x}{2}+\frac {1}{9} \int \left (-\frac {1}{\log (2+x)}-\frac {3}{x \log (2+x)}+\frac {\log (x)}{\log (2+x)}\right ) \, dx+\frac {1}{9} \int \left (\frac {3}{(3+x) \log (2+x)}+\frac {x}{(3+x) \log (2+x)}-\frac {x \log (x)}{(3+x) \log (2+x)}\right ) \, dx+\frac {1}{3} \int \left (\frac {3}{(3+x)^2 \log (2+x)}+\frac {x}{(3+x)^2 \log (2+x)}-\frac {x \log (x)}{(3+x)^2 \log (2+x)}\right ) \, dx-\int \frac {\log (x)}{(3+x) \log ^2(2+x)} \, dx+\text {Subst}\left (\int \frac {\log (-2+x)}{x \log ^2(x)} \, dx,x,2+x\right ) \\ & = e^x-\frac {x}{2}-\frac {1}{9} \int \frac {1}{\log (2+x)} \, dx+\frac {1}{9} \int \frac {x}{(3+x) \log (2+x)} \, dx+\frac {1}{9} \int \frac {\log (x)}{\log (2+x)} \, dx-\frac {1}{9} \int \frac {x \log (x)}{(3+x) \log (2+x)} \, dx-\frac {1}{3} \int \frac {1}{x \log (2+x)} \, dx+\frac {1}{3} \int \frac {x}{(3+x)^2 \log (2+x)} \, dx+\frac {1}{3} \int \frac {1}{(3+x) \log (2+x)} \, dx-\frac {1}{3} \int \frac {x \log (x)}{(3+x)^2 \log (2+x)} \, dx-\int \frac {\log (x)}{(3+x) \log ^2(2+x)} \, dx+\int \frac {1}{(3+x)^2 \log (2+x)} \, dx+\text {Subst}\left (\int \frac {\log (-2+x)}{x \log ^2(x)} \, dx,x,2+x\right ) \\ & = e^x-\frac {x}{2}+\frac {1}{9} \int \left (\frac {1}{\log (2+x)}-\frac {3}{(3+x) \log (2+x)}\right ) \, dx-\frac {1}{9} \int \left (\frac {\log (x)}{\log (2+x)}-\frac {3 \log (x)}{(3+x) \log (2+x)}\right ) \, dx+\frac {1}{9} \int \frac {\log (x)}{\log (2+x)} \, dx-\frac {1}{9} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,2+x\right )+\frac {1}{3} \int \left (-\frac {3}{(3+x)^2 \log (2+x)}+\frac {1}{(3+x) \log (2+x)}\right ) \, dx-\frac {1}{3} \int \left (-\frac {3 \log (x)}{(3+x)^2 \log (2+x)}+\frac {\log (x)}{(3+x) \log (2+x)}\right ) \, dx-\frac {1}{3} \int \frac {1}{x \log (2+x)} \, dx+\frac {1}{3} \int \frac {1}{(3+x) \log (2+x)} \, dx-\int \frac {\log (x)}{(3+x) \log ^2(2+x)} \, dx+\int \frac {1}{(3+x)^2 \log (2+x)} \, dx+\text {Subst}\left (\int \frac {\log (-2+x)}{x \log ^2(x)} \, dx,x,2+x\right ) \\ & = e^x-\frac {x}{2}-\frac {\operatorname {LogIntegral}(2+x)}{9}+\frac {1}{9} \int \frac {1}{\log (2+x)} \, dx-\frac {1}{3} \int \frac {1}{x \log (2+x)} \, dx+\frac {1}{3} \int \frac {1}{(3+x) \log (2+x)} \, dx-\int \frac {\log (x)}{(3+x) \log ^2(2+x)} \, dx+\int \frac {\log (x)}{(3+x)^2 \log (2+x)} \, dx+\text {Subst}\left (\int \frac {\log (-2+x)}{x \log ^2(x)} \, dx,x,2+x\right ) \\ & = e^x-\frac {x}{2}-\frac {\operatorname {LogIntegral}(2+x)}{9}+\frac {1}{9} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,2+x\right )-\frac {1}{3} \int \frac {1}{x \log (2+x)} \, dx+\frac {1}{3} \int \frac {1}{(3+x) \log (2+x)} \, dx-\int \frac {\log (x)}{(3+x) \log ^2(2+x)} \, dx+\int \frac {\log (x)}{(3+x)^2 \log (2+x)} \, dx+\text {Subst}\left (\int \frac {\log (-2+x)}{x \log ^2(x)} \, dx,x,2+x\right ) \\ & = e^x-\frac {x}{2}-\frac {1}{3} \int \frac {1}{x \log (2+x)} \, dx+\frac {1}{3} \int \frac {1}{(3+x) \log (2+x)} \, dx-\int \frac {\log (x)}{(3+x) \log ^2(2+x)} \, dx+\int \frac {\log (x)}{(3+x)^2 \log (2+x)} \, dx+\text {Subst}\left (\int \frac {\log (-2+x)}{x \log ^2(x)} \, dx,x,2+x\right ) \\ \end{align*}
Time = 0.79 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=\frac {1}{2} \left (2 e^x-x-\frac {2 \log (x)}{(3+x) \log (2+x)}\right ) \]
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Time = 11.38 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(-\frac {x}{2}+{\mathrm e}^{x}-\frac {\ln \left (x \right )}{\ln \left (2+x \right ) \left (3+x \right )}\) | \(22\) |
parallelrisch | \(-\frac {7 \ln \left (2+x \right ) x^{2}-14 x \,{\mathrm e}^{x} \ln \left (2+x \right )+5 x \ln \left (2+x \right )-42 \,{\mathrm e}^{x} \ln \left (2+x \right )+14 \ln \left (x \right )-48 \ln \left (2+x \right )}{14 \left (3+x \right ) \ln \left (2+x \right )}\) | \(58\) |
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Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=-\frac {{\left (x^{2} - 2 \, {\left (x + 3\right )} e^{x} + 3 \, x\right )} \log \left (x + 2\right ) + 2 \, \log \left (x\right )}{2 \, {\left (x + 3\right )} \log \left (x + 2\right )} \]
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Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=- \frac {x}{2} + e^{x} - \frac {\log {\left (x \right )}}{\left (x + 3\right ) \log {\left (x + 2 \right )}} \]
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Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=-\frac {{\left (x^{2} - 2 \, {\left (x + 3\right )} e^{x} + 3 \, x\right )} \log \left (x + 2\right ) + 2 \, \log \left (x\right )}{2 \, {\left (x + 3\right )} \log \left (x + 2\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=-\frac {x^{2} \log \left (x + 2\right ) - 2 \, x e^{x} \log \left (x + 2\right ) + 3 \, x \log \left (x + 2\right ) - 6 \, e^{x} \log \left (x + 2\right ) + 2 \, \log \left (x\right )}{2 \, {\left (x \log \left (x + 2\right ) + 3 \, \log \left (x + 2\right )\right )}} \]
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Time = 14.37 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.12 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx={\mathrm {e}}^x-\frac {x}{2}-\frac {x+2}{x^2+3\,x}-\frac {\frac {\ln \left (x\right )}{x+3}-\frac {\ln \left (x+2\right )\,\left (x+2\right )\,\left (x-x\,\ln \left (x\right )+3\right )}{x\,{\left (x+3\right )}^2}}{\ln \left (x+2\right )}+\frac {\ln \left (x\right )\,\left (x+2\right )}{x^2+6\,x+9} \]
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