Integrand size = 80, antiderivative size = 27 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=\log \left ((-2+x)^2\right ) \log \left (\frac {5-\frac {x}{2}}{-16+(5-x) x}\right ) \]
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Result contains complex when optimal does not.
Time = 0.41 (sec) , antiderivative size = 204, normalized size of antiderivative = 7.56, number of steps used = 30, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6874, 2465, 2441, 2440, 2438, 2604, 12} \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=2 \log \left (-\frac {10-x}{2 \left (x^2-5 x+16\right )}\right ) \log (x-2)+2 \log \left (\frac {-2 x-i \sqrt {39}+5}{1-i \sqrt {39}}\right ) \log (x-2)+2 \log \left (\frac {-2 x+i \sqrt {39}+5}{1+i \sqrt {39}}\right ) \log (x-2)-2 \log \left (\frac {10-x}{8}\right ) \log (x-2)-\log \left (\frac {-2 x-i \sqrt {39}+5}{1-i \sqrt {39}}\right ) \log \left ((x-2)^2\right )-\log \left (\frac {-2 x+i \sqrt {39}+5}{1+i \sqrt {39}}\right ) \log \left ((x-2)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((x-2)^2\right ) \]
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Rule 12
Rule 2438
Rule 2440
Rule 2441
Rule 2465
Rule 2604
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {\left (34-20 x+x^2\right ) \log \left ((-2+x)^2\right )}{(-10+x) \left (16-5 x+x^2\right )}+\frac {2 \log \left (\frac {-10+x}{2 \left (16-5 x+x^2\right )}\right )}{-2+x}\right ) \, dx \\ & = 2 \int \frac {\log \left (\frac {-10+x}{2 \left (16-5 x+x^2\right )}\right )}{-2+x} \, dx-\int \frac {\left (34-20 x+x^2\right ) \log \left ((-2+x)^2\right )}{(-10+x) \left (16-5 x+x^2\right )} \, dx \\ & = 2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \int \frac {2 \left (16-5 x+x^2\right ) \left (-\frac {(-10+x) (-5+2 x)}{2 \left (16-5 x+x^2\right )^2}+\frac {1}{2 \left (16-5 x+x^2\right )}\right ) \log (-2+x)}{-10+x} \, dx-\int \left (\frac {\log \left ((-2+x)^2\right )}{10-x}+\frac {(-5+2 x) \log \left ((-2+x)^2\right )}{16-5 x+x^2}\right ) \, dx \\ & = 2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-4 \int \frac {\left (16-5 x+x^2\right ) \left (-\frac {(-10+x) (-5+2 x)}{2 \left (16-5 x+x^2\right )^2}+\frac {1}{2 \left (16-5 x+x^2\right )}\right ) \log (-2+x)}{-10+x} \, dx-\int \frac {\log \left ((-2+x)^2\right )}{10-x} \, dx-\int \frac {(-5+2 x) \log \left ((-2+x)^2\right )}{16-5 x+x^2} \, dx \\ & = \log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \int \frac {\log \left (\frac {10-x}{8}\right )}{-2+x} \, dx-4 \int \left (\frac {\log (-2+x)}{2 (-10+x)}+\frac {(5-2 x) \log (-2+x)}{2 \left (16-5 x+x^2\right )}\right ) \, dx-\int \left (\frac {2 \log \left ((-2+x)^2\right )}{-5-i \sqrt {39}+2 x}+\frac {2 \log \left ((-2+x)^2\right )}{-5+i \sqrt {39}+2 x}\right ) \, dx \\ & = \log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \int \frac {\log (-2+x)}{-10+x} \, dx-2 \int \frac {(5-2 x) \log (-2+x)}{16-5 x+x^2} \, dx-2 \int \frac {\log \left ((-2+x)^2\right )}{-5-i \sqrt {39}+2 x} \, dx-2 \int \frac {\log \left ((-2+x)^2\right )}{-5+i \sqrt {39}+2 x} \, dx-2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{8}\right )}{x} \, dx,x,-2+x\right ) \\ & = -2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )+2 \text {Li}_2\left (\frac {1}{8} (-2+x)\right )+2 \int \frac {\log \left (\frac {10-x}{8}\right )}{-2+x} \, dx-2 \int \left (-\frac {2 \log (-2+x)}{-5-i \sqrt {39}+2 x}-\frac {2 \log (-2+x)}{-5+i \sqrt {39}+2 x}\right ) \, dx+2 \int \frac {\log \left (\frac {-5-i \sqrt {39}+2 x}{-1-i \sqrt {39}}\right )}{-2+x} \, dx+2 \int \frac {\log \left (\frac {-5+i \sqrt {39}+2 x}{-1+i \sqrt {39}}\right )}{-2+x} \, dx \\ & = -2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )+2 \text {Li}_2\left (\frac {1}{8} (-2+x)\right )+2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{8}\right )}{x} \, dx,x,-2+x\right )+2 \text {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1-i \sqrt {39}}\right )}{x} \, dx,x,-2+x\right )+2 \text {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1+i \sqrt {39}}\right )}{x} \, dx,x,-2+x\right )+4 \int \frac {\log (-2+x)}{-5-i \sqrt {39}+2 x} \, dx+4 \int \frac {\log (-2+x)}{-5+i \sqrt {39}+2 x} \, dx \\ & = 2 \log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log (-2+x)+2 \log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log (-2+x)-2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \text {Li}_2\left (-\frac {2 (2-x)}{1-i \sqrt {39}}\right )-2 \text {Li}_2\left (-\frac {2 (2-x)}{1+i \sqrt {39}}\right )-2 \int \frac {\log \left (\frac {-5-i \sqrt {39}+2 x}{-1-i \sqrt {39}}\right )}{-2+x} \, dx-2 \int \frac {\log \left (\frac {-5+i \sqrt {39}+2 x}{-1+i \sqrt {39}}\right )}{-2+x} \, dx \\ & = 2 \log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log (-2+x)+2 \log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log (-2+x)-2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \text {Li}_2\left (-\frac {2 (2-x)}{1-i \sqrt {39}}\right )-2 \text {Li}_2\left (-\frac {2 (2-x)}{1+i \sqrt {39}}\right )-2 \text {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1-i \sqrt {39}}\right )}{x} \, dx,x,-2+x\right )-2 \text {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1+i \sqrt {39}}\right )}{x} \, dx,x,-2+x\right ) \\ & = 2 \log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log (-2+x)+2 \log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log (-2+x)-2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right ) \\ \end{align*}
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.18 (sec) , antiderivative size = 220, normalized size of antiderivative = 8.15 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=-2 \log (8) \log (-10+x)+2 \log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log (-2+x)+2 \log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )+2 \operatorname {PolyLog}\left (2,\frac {10-x}{8}\right )+2 \operatorname {PolyLog}\left (2,\frac {1}{8} (-2+x)\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.45 (sec) , antiderivative size = 247, normalized size of antiderivative = 9.15
method | result | size |
default | \(-2 \ln \left (2\right ) \ln \left (-2+x \right )+2 \ln \left (-2+x \right ) \ln \left (\frac {x -10}{x^{2}-5 x +16}\right )-2 \left (\ln \left (-2+x \right )-\ln \left (-\frac {1}{4}+\frac {x}{8}\right )\right ) \ln \left (\frac {5}{4}-\frac {x}{8}\right )+2 \ln \left (-2+x \right ) \ln \left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \ln \left (-2+x \right ) \ln \left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )+2 \operatorname {dilog}\left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \operatorname {dilog}\left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )-\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2}-5 \textit {\_Z} +16\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}-4 x +4\right )-2 \operatorname {dilog}\left (\frac {-2+x}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {-2+x}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )+\ln \left (x -10\right ) \ln \left (x^{2}-4 x +4\right )-2 \ln \left (x -10\right ) \ln \left (-\frac {1}{4}+\frac {x}{8}\right )\) | \(247\) |
parts | \(-2 \ln \left (2\right ) \ln \left (-2+x \right )+2 \ln \left (-2+x \right ) \ln \left (\frac {x -10}{x^{2}-5 x +16}\right )-2 \left (\ln \left (-2+x \right )-\ln \left (-\frac {1}{4}+\frac {x}{8}\right )\right ) \ln \left (\frac {5}{4}-\frac {x}{8}\right )+2 \ln \left (-2+x \right ) \ln \left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \ln \left (-2+x \right ) \ln \left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )+2 \operatorname {dilog}\left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \operatorname {dilog}\left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )-\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2}-5 \textit {\_Z} +16\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}-4 x +4\right )-2 \operatorname {dilog}\left (\frac {-2+x}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {-2+x}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )+\ln \left (x -10\right ) \ln \left (x^{2}-4 x +4\right )-2 \ln \left (x -10\right ) \ln \left (-\frac {1}{4}+\frac {x}{8}\right )\) | \(247\) |
risch | \(\text {Expression too large to display}\) | \(50326\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=\log \left (x^{2} - 4 \, x + 4\right ) \log \left (\frac {x - 10}{2 \, {\left (x^{2} - 5 \, x + 16\right )}}\right ) \]
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Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=\log {\left (\frac {x - 10}{2 x^{2} - 10 x + 32} \right )} \log {\left (x^{2} - 4 x + 4 \right )} \]
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Time = 0.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=-2 \, {\left (\log \left (2\right ) - \log \left (x - 10\right )\right )} \log \left (x - 2\right ) - 2 \, \log \left (x^{2} - 5 \, x + 16\right ) \log \left (x - 2\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (22) = 44\).
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=-{\left (\log \left (x^{2} - 5 \, x + 16\right ) - \log \left (x - 10\right )\right )} \log \left (x^{2} - 4 \, x + 4\right ) - 2 \, \log \left (2 \, x^{2} - 10 \, x + 32\right ) \log \left (x - 2\right ) + 2 \, \log \left (x^{2} - 5 \, x + 16\right ) \log \left (x - 2\right ) \]
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Time = 10.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=\ln \left (x^2-4\,x+4\right )\,\left (\ln \left (x-10\right )-\ln \left (2\,x^2-10\,x+32\right )\right ) \]
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