Integrand size = 68, antiderivative size = 23 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=-5+x (2+x) \log \left (\frac {e^{25} \left (5-\log ^2(x)\right )}{x}\right ) \]
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\[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=\int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {(2+x) \left (-5-2 \log (x)+\log ^2(x)\right )}{-5+\log ^2(x)}+2 (1+x) \left (25+\log \left (-\frac {-5+\log ^2(x)}{x}\right )\right )\right ) \, dx \\ & = 2 \int (1+x) \left (25+\log \left (-\frac {-5+\log ^2(x)}{x}\right )\right ) \, dx-\int \frac {(2+x) \left (-5-2 \log (x)+\log ^2(x)\right )}{-5+\log ^2(x)} \, dx \\ & = 2 \int \left (25 (1+x)+(1+x) \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right )\right ) \, dx-\int \left (2+x-\frac {2 (2+x) \log (x)}{-5+\log ^2(x)}\right ) \, dx \\ & = -2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \frac {(2+x) \log (x)}{-5+\log ^2(x)} \, dx+2 \int (1+x) \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx \\ & = -2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \left (\frac {2 \log (x)}{-5+\log ^2(x)}+\frac {x \log (x)}{-5+\log ^2(x)}\right ) \, dx+2 \int \left (\log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right )+x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right )\right ) \, dx \\ & = -2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+4 \int \frac {\log (x)}{-5+\log ^2(x)} \, dx \\ & = -2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+4 \int \left (-\frac {1}{2 \left (\sqrt {5}-\log (x)\right )}+\frac {1}{2 \left (\sqrt {5}+\log (x)\right )}\right ) \, dx \\ & = -2 x-\frac {x^2}{2}+25 (1+x)^2-2 \int \frac {1}{\sqrt {5}-\log (x)} \, dx+2 \int \frac {1}{\sqrt {5}+\log (x)} \, dx+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx \\ & = -2 x-\frac {x^2}{2}+25 (1+x)^2+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx-2 \text {Subst}\left (\int \frac {e^x}{\sqrt {5}-x} \, dx,x,\log (x)\right )+2 \text {Subst}\left (\int \frac {e^x}{\sqrt {5}+x} \, dx,x,\log (x)\right ) \\ & = -2 x-\frac {x^2}{2}+25 (1+x)^2+2 e^{\sqrt {5}} \text {Ei}\left (-\sqrt {5}+\log (x)\right )+2 e^{-\sqrt {5}} \text {Ei}\left (\sqrt {5}+\log (x)\right )+2 \int \frac {x \log (x)}{-5+\log ^2(x)} \, dx+2 \int \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx+2 \int x \log \left (\frac {5}{x}-\frac {\log ^2(x)}{x}\right ) \, dx \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=x (2+x) \left (25+\log \left (-\frac {-5+\log ^2(x)}{x}\right )\right ) \]
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Time = 3.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61
method | result | size |
parallelrisch | \(x^{2} \ln \left (-\frac {{\mathrm e}^{25} \left (\ln \left (x \right )^{2}-5\right )}{x}\right )+2 \ln \left (-\frac {{\mathrm e}^{25} \left (\ln \left (x \right )^{2}-5\right )}{x}\right ) x\) | \(37\) |
norman | \(x^{2} \ln \left (\frac {-{\mathrm e}^{25} \ln \left (x \right )^{2}+5 \,{\mathrm e}^{25}}{x}\right )+2 x \ln \left (\frac {-{\mathrm e}^{25} \ln \left (x \right )^{2}+5 \,{\mathrm e}^{25}}{x}\right )\) | \(45\) |
risch | \(\left (x^{2}+2 x \right ) \ln \left (\ln \left (x \right )^{2}-5\right )-x^{2} \ln \left (x \right )-2 x \ln \left (x \right )+i \pi \,x^{2}-2 i \pi x {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2}-i \pi x \,\operatorname {csgn}\left (i \left (\ln \left (x \right )^{2}-5\right )\right ) \operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+\frac {i \pi \,x^{2} {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+i \pi x {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 i x \pi +\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )^{2}-5\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2}}{2}+i \pi x {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{3}+i \pi x \,\operatorname {csgn}\left (i \left (\ln \left (x \right )^{2}-5\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2}-i \pi \,x^{2} {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )^{2}-5\right )\right ) \operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi \,x^{2} {\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )^{2}-5\right )}{x}\right )}^{3}}{2}+25 x^{2}+50 x\) | \(323\) |
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Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx={\left (x^{2} + 2 \, x\right )} \log \left (-\frac {e^{25} \log \left (x\right )^{2} - 5 \, e^{25}}{x}\right ) \]
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Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=\left (x^{2} + 2 x\right ) \log {\left (\frac {- e^{25} \log {\left (x \right )}^{2} + 5 e^{25}}{x} \right )} \]
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Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=25 \, x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (-\log \left (x\right )^{2} + 5\right ) - {\left (x^{2} + 2 \, x\right )} \log \left (x\right ) + 50 \, x \]
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Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=25 \, x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (-\log \left (x\right )^{2} + 5\right ) - {\left (x^{2} + 2 \, x\right )} \log \left (x\right ) + 50 \, x \]
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Time = 12.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {10+5 x+(4+2 x) \log (x)+(-2-x) \log ^2(x)+\left (-10-10 x+(2+2 x) \log ^2(x)\right ) \log \left (\frac {5 e^{25}-e^{25} \log ^2(x)}{x}\right )}{-5+\log ^2(x)} \, dx=x\,\ln \left (\frac {5\,{\mathrm {e}}^{25}-{\mathrm {e}}^{25}\,{\ln \left (x\right )}^2}{x}\right )\,\left (x+2\right ) \]
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