Integrand size = 54, antiderivative size = 21 \[ \int \frac {-25 x^2-10 x^3-x^4+\left (-100 x-10 x^2+22 x^3+4 x^4\right ) \log (-2+x)}{(-150+75 x) \log ^2(-2+x)} \, dx=\frac {\left (x+\frac {x^2}{5}\right )^2}{3 \log (-2+x)} \]
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Leaf count is larger than twice the leaf count of optimal. \(78\) vs. \(2(21)=42\).
Time = 0.73 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.71, number of steps used = 77, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6820, 12, 6874, 2465, 2436, 2334, 2335, 2437, 2339, 30, 2447, 2446, 2346, 2209, 2464} \[ \int \frac {-25 x^2-10 x^3-x^4+\left (-100 x-10 x^2+22 x^3+4 x^4\right ) \log (-2+x)}{(-150+75 x) \log ^2(-2+x)} \, dx=-\frac {(2-x) x^3}{75 \log (x-2)}-\frac {4 (2-x) x^2}{25 \log (x-2)}-\frac {49 (2-x) x}{75 \log (x-2)}-\frac {98 (2-x)}{75 \log (x-2)}+\frac {196}{75 \log (x-2)} \]
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Rule 12
Rule 30
Rule 2209
Rule 2334
Rule 2335
Rule 2339
Rule 2346
Rule 2436
Rule 2437
Rule 2446
Rule 2447
Rule 2464
Rule 2465
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {x (5+x) \left (x (5+x)-2 \left (-10+x+2 x^2\right ) \log (-2+x)\right )}{75 (2-x) \log ^2(-2+x)} \, dx \\ & = \frac {1}{75} \int \frac {x (5+x) \left (x (5+x)-2 \left (-10+x+2 x^2\right ) \log (-2+x)\right )}{(2-x) \log ^2(-2+x)} \, dx \\ & = \frac {1}{75} \int \left (-\frac {x^2 (5+x)^2}{(-2+x) \log ^2(-2+x)}+\frac {2 x (5+x) (5+2 x)}{\log (-2+x)}\right ) \, dx \\ & = -\left (\frac {1}{75} \int \frac {x^2 (5+x)^2}{(-2+x) \log ^2(-2+x)} \, dx\right )+\frac {2}{75} \int \frac {x (5+x) (5+2 x)}{\log (-2+x)} \, dx \\ & = -\left (\frac {1}{75} \int \left (\frac {98}{\log ^2(-2+x)}+\frac {196}{(-2+x) \log ^2(-2+x)}+\frac {49 x}{\log ^2(-2+x)}+\frac {12 x^2}{\log ^2(-2+x)}+\frac {x^3}{\log ^2(-2+x)}\right ) \, dx\right )+\frac {2}{75} \int \left (\frac {126}{\log (-2+x)}+\frac {109 (-2+x)}{\log (-2+x)}+\frac {27 (-2+x)^2}{\log (-2+x)}+\frac {2 (-2+x)^3}{\log (-2+x)}\right ) \, dx \\ & = -\left (\frac {1}{75} \int \frac {x^3}{\log ^2(-2+x)} \, dx\right )+\frac {4}{75} \int \frac {(-2+x)^3}{\log (-2+x)} \, dx-\frac {4}{25} \int \frac {x^2}{\log ^2(-2+x)} \, dx-\frac {49}{75} \int \frac {x}{\log ^2(-2+x)} \, dx+\frac {18}{25} \int \frac {(-2+x)^2}{\log (-2+x)} \, dx-\frac {98}{75} \int \frac {1}{\log ^2(-2+x)} \, dx-\frac {196}{75} \int \frac {1}{(-2+x) \log ^2(-2+x)} \, dx+\frac {218}{75} \int \frac {-2+x}{\log (-2+x)} \, dx+\frac {84}{25} \int \frac {1}{\log (-2+x)} \, dx \\ & = -\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}-\frac {4}{75} \int \frac {x^3}{\log (-2+x)} \, dx+\frac {4}{75} \text {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,-2+x\right )+\frac {2}{25} \int \frac {x^2}{\log (-2+x)} \, dx-\frac {12}{25} \int \frac {x^2}{\log (-2+x)} \, dx+\frac {16}{25} \int \frac {x}{\log (-2+x)} \, dx+\frac {18}{25} \text {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-2+x\right )+\frac {98}{75} \int \frac {1}{\log (-2+x)} \, dx-\frac {98}{75} \int \frac {x}{\log (-2+x)} \, dx-\frac {98}{75} \text {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-2+x\right )-\frac {196}{75} \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-2+x\right )+\frac {218}{75} \text {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-2+x\right )+\frac {84}{25} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right ) \\ & = -\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}+\frac {84 \operatorname {LogIntegral}(-2+x)}{25}-\frac {4}{75} \int \left (\frac {8}{\log (-2+x)}+\frac {12 (-2+x)}{\log (-2+x)}+\frac {6 (-2+x)^2}{\log (-2+x)}+\frac {(-2+x)^3}{\log (-2+x)}\right ) \, dx+\frac {4}{75} \text {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (-2+x)\right )+\frac {2}{25} \int \left (\frac {4}{\log (-2+x)}+\frac {4 (-2+x)}{\log (-2+x)}+\frac {(-2+x)^2}{\log (-2+x)}\right ) \, dx-\frac {12}{25} \int \left (\frac {4}{\log (-2+x)}+\frac {4 (-2+x)}{\log (-2+x)}+\frac {(-2+x)^2}{\log (-2+x)}\right ) \, dx+\frac {16}{25} \int \left (\frac {2}{\log (-2+x)}+\frac {-2+x}{\log (-2+x)}\right ) \, dx+\frac {18}{25} \text {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {98}{75} \int \left (\frac {2}{\log (-2+x)}+\frac {-2+x}{\log (-2+x)}\right ) \, dx-\frac {196}{75} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-2+x)\right )+\frac {218}{75} \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-2+x)\right ) \\ & = \frac {218}{75} \operatorname {ExpIntegralEi}(2 \log (-2+x))+\frac {18}{25} \operatorname {ExpIntegralEi}(3 \log (-2+x))+\frac {4}{75} \operatorname {ExpIntegralEi}(4 \log (-2+x))+\frac {196}{75 \log (-2+x)}-\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}+\frac {84 \operatorname {LogIntegral}(-2+x)}{25}-\frac {4}{75} \int \frac {(-2+x)^3}{\log (-2+x)} \, dx+\frac {2}{25} \int \frac {(-2+x)^2}{\log (-2+x)} \, dx+\frac {8}{25} \int \frac {1}{\log (-2+x)} \, dx+\frac {8}{25} \int \frac {-2+x}{\log (-2+x)} \, dx-\frac {8}{25} \int \frac {(-2+x)^2}{\log (-2+x)} \, dx-\frac {32}{75} \int \frac {1}{\log (-2+x)} \, dx-\frac {12}{25} \int \frac {(-2+x)^2}{\log (-2+x)} \, dx+\frac {32}{25} \int \frac {1}{\log (-2+x)} \, dx-\frac {98}{75} \int \frac {-2+x}{\log (-2+x)} \, dx-\frac {48}{25} \int \frac {1}{\log (-2+x)} \, dx-\frac {48}{25} \int \frac {-2+x}{\log (-2+x)} \, dx-\frac {196}{75} \int \frac {1}{\log (-2+x)} \, dx \\ & = \frac {218}{75} \operatorname {ExpIntegralEi}(2 \log (-2+x))+\frac {18}{25} \operatorname {ExpIntegralEi}(3 \log (-2+x))+\frac {4}{75} \operatorname {ExpIntegralEi}(4 \log (-2+x))+\frac {196}{75 \log (-2+x)}-\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}+\frac {84 \operatorname {LogIntegral}(-2+x)}{25}-\frac {4}{75} \text {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,-2+x\right )+\frac {2}{25} \text {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-2+x\right )+\frac {8}{25} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )+\frac {8}{25} \text {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-2+x\right )-\frac {8}{25} \text {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-2+x\right )-\frac {32}{75} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )-\frac {12}{25} \text {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-2+x\right )+\frac {32}{25} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )-\frac {98}{75} \text {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-2+x\right )-\frac {48}{25} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )-\frac {48}{25} \text {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-2+x\right )-\frac {196}{75} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right ) \\ & = \frac {218}{75} \operatorname {ExpIntegralEi}(2 \log (-2+x))+\frac {18}{25} \operatorname {ExpIntegralEi}(3 \log (-2+x))+\frac {4}{75} \operatorname {ExpIntegralEi}(4 \log (-2+x))+\frac {196}{75 \log (-2+x)}-\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}-\frac {4}{75} \text {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (-2+x)\right )+\frac {2}{25} \text {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-2+x)\right )+\frac {8}{25} \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {8}{25} \text {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {12}{25} \text {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {98}{75} \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {48}{25} \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-2+x)\right ) \\ & = \frac {196}{75 \log (-2+x)}-\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {-25 x^2-10 x^3-x^4+\left (-100 x-10 x^2+22 x^3+4 x^4\right ) \log (-2+x)}{(-150+75 x) \log ^2(-2+x)} \, dx=\frac {x^2 (5+x)^2}{75 \log (-2+x)} \]
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Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {x^{2} \left (x^{2}+10 x +25\right )}{75 \ln \left (-2+x \right )}\) | \(20\) |
parallelrisch | \(\frac {x^{4}+10 x^{3}+25 x^{2}}{75 \ln \left (-2+x \right )}\) | \(23\) |
norman | \(\frac {\frac {1}{3} x^{2}+\frac {2}{15} x^{3}+\frac {1}{75} x^{4}}{\ln \left (-2+x \right )}\) | \(24\) |
derivativedivides | \(\frac {\left (-2+x \right )^{4}}{75 \ln \left (-2+x \right )}+\frac {6 \left (-2+x \right )^{3}}{25 \ln \left (-2+x \right )}+\frac {109 \left (-2+x \right )^{2}}{75 \ln \left (-2+x \right )}+\frac {-\frac {168}{25}+\frac {84 x}{25}}{\ln \left (-2+x \right )}+\frac {196}{75 \ln \left (-2+x \right )}\) | \(60\) |
default | \(\frac {\left (-2+x \right )^{4}}{75 \ln \left (-2+x \right )}+\frac {6 \left (-2+x \right )^{3}}{25 \ln \left (-2+x \right )}+\frac {109 \left (-2+x \right )^{2}}{75 \ln \left (-2+x \right )}+\frac {-\frac {168}{25}+\frac {84 x}{25}}{\ln \left (-2+x \right )}+\frac {196}{75 \ln \left (-2+x \right )}\) | \(60\) |
parts | \(\frac {\left (-2+x \right )^{4}}{75 \ln \left (-2+x \right )}+\frac {6 \left (-2+x \right )^{3}}{25 \ln \left (-2+x \right )}+\frac {109 \left (-2+x \right )^{2}}{75 \ln \left (-2+x \right )}+\frac {-\frac {168}{25}+\frac {84 x}{25}}{\ln \left (-2+x \right )}+\frac {196}{75 \ln \left (-2+x \right )}\) | \(60\) |
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-25 x^2-10 x^3-x^4+\left (-100 x-10 x^2+22 x^3+4 x^4\right ) \log (-2+x)}{(-150+75 x) \log ^2(-2+x)} \, dx=\frac {x^{4} + 10 \, x^{3} + 25 \, x^{2}}{75 \, \log \left (x - 2\right )} \]
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Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-25 x^2-10 x^3-x^4+\left (-100 x-10 x^2+22 x^3+4 x^4\right ) \log (-2+x)}{(-150+75 x) \log ^2(-2+x)} \, dx=\frac {x^{4} + 10 x^{3} + 25 x^{2}}{75 \log {\left (x - 2 \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-25 x^2-10 x^3-x^4+\left (-100 x-10 x^2+22 x^3+4 x^4\right ) \log (-2+x)}{(-150+75 x) \log ^2(-2+x)} \, dx=\frac {x^{4} + 10 \, x^{3} + 25 \, x^{2}}{75 \, \log \left (x - 2\right )} \]
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-25 x^2-10 x^3-x^4+\left (-100 x-10 x^2+22 x^3+4 x^4\right ) \log (-2+x)}{(-150+75 x) \log ^2(-2+x)} \, dx=\frac {x^{4} + 10 \, x^{3} + 25 \, x^{2}}{75 \, \log \left (x - 2\right )} \]
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Time = 12.99 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {-25 x^2-10 x^3-x^4+\left (-100 x-10 x^2+22 x^3+4 x^4\right ) \log (-2+x)}{(-150+75 x) \log ^2(-2+x)} \, dx=\frac {x^2\,{\left (x+5\right )}^2}{75\,\ln \left (x-2\right )} \]
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