Integrand size = 89, antiderivative size = 20 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^2}{\log (3) (-4+\log (15-x \log (5)))} \]
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Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(20)=40\).
Time = 0.47 (sec) , antiderivative size = 89, normalized size of antiderivative = 4.45, number of steps used = 23, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.169, Rules used = {6820, 12, 6874, 2458, 2395, 2334, 2336, 2209, 2339, 30, 2343, 2346, 2446, 2436, 2437} \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=-\frac {(15-x \log (5))^2}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}+\frac {30 (15-x \log (5))}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {225}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))} \]
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Rule 12
Rule 30
Rule 2209
Rule 2334
Rule 2336
Rule 2339
Rule 2343
Rule 2346
Rule 2395
Rule 2436
Rule 2437
Rule 2446
Rule 2458
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {x (-120+9 x \log (5)-2 (-15+x \log (5)) \log (15-x \log (5)))}{\log (3) (15-x \log (5)) (4-\log (15-x \log (5)))^2} \, dx \\ & = \frac {\int \frac {x (-120+9 x \log (5)-2 (-15+x \log (5)) \log (15-x \log (5)))}{(15-x \log (5)) (4-\log (15-x \log (5)))^2} \, dx}{\log (3)} \\ & = \frac {\int \left (-\frac {x^2 \log (5)}{(-15+x \log (5)) (-4+\log (15-x \log (5)))^2}+\frac {2 x}{-4+\log (15-x \log (5))}\right ) \, dx}{\log (3)} \\ & = \frac {2 \int \frac {x}{-4+\log (15-x \log (5))} \, dx}{\log (3)}-\frac {\log (5) \int \frac {x^2}{(-15+x \log (5)) (-4+\log (15-x \log (5)))^2} \, dx}{\log (3)} \\ & = -\frac {\text {Subst}\left (\int \frac {\left (\frac {15}{\log (5)}-\frac {x}{\log (5)}\right )^2}{x (-4+\log (x))^2} \, dx,x,15-x \log (5)\right )}{\log (3)}+\frac {2 \int \left (\frac {15}{\log (5) (-4+\log (15-x \log (5)))}-\frac {15-x \log (5)}{\log (5) (-4+\log (15-x \log (5)))}\right ) \, dx}{\log (3)} \\ & = -\frac {\text {Subst}\left (\int \left (-\frac {30}{\log ^2(5) (-4+\log (x))^2}+\frac {225}{x \log ^2(5) (-4+\log (x))^2}+\frac {x}{\log ^2(5) (-4+\log (x))^2}\right ) \, dx,x,15-x \log (5)\right )}{\log (3)}-\frac {2 \int \frac {15-x \log (5)}{-4+\log (15-x \log (5))} \, dx}{\log (3) \log (5)}+\frac {30 \int \frac {1}{-4+\log (15-x \log (5))} \, dx}{\log (3) \log (5)} \\ & = -\frac {\text {Subst}\left (\int \frac {x}{(-4+\log (x))^2} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}+\frac {2 \text {Subst}\left (\int \frac {x}{-4+\log (x)} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}+\frac {30 \text {Subst}\left (\int \frac {1}{(-4+\log (x))^2} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}-\frac {30 \text {Subst}\left (\int \frac {1}{-4+\log (x)} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}-\frac {225 \text {Subst}\left (\int \frac {1}{x (-4+\log (x))^2} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)} \\ & = \frac {30 (15-x \log (5))}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {(15-x \log (5))^2}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}+\frac {2 \text {Subst}\left (\int \frac {e^{2 x}}{-4+x} \, dx,x,\log (15-x \log (5))\right )}{\log (3) \log ^2(5)}-\frac {2 \text {Subst}\left (\int \frac {x}{-4+\log (x)} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}-\frac {30 \text {Subst}\left (\int \frac {e^x}{-4+x} \, dx,x,\log (15-x \log (5))\right )}{\log (3) \log ^2(5)}+\frac {30 \text {Subst}\left (\int \frac {1}{-4+\log (x)} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}-\frac {225 \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,-4+\log (15-x \log (5))\right )}{\log (3) \log ^2(5)} \\ & = \frac {2 e^8 \text {Ei}(-2 (4-\log (15-x \log (5))))}{\log (3) \log ^2(5)}-\frac {30 e^4 \text {Ei}(-4+\log (15-x \log (5)))}{\log (3) \log ^2(5)}-\frac {225}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}+\frac {30 (15-x \log (5))}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {(15-x \log (5))^2}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {2 \text {Subst}\left (\int \frac {e^{2 x}}{-4+x} \, dx,x,\log (15-x \log (5))\right )}{\log (3) \log ^2(5)}+\frac {30 \text {Subst}\left (\int \frac {e^x}{-4+x} \, dx,x,\log (15-x \log (5))\right )}{\log (3) \log ^2(5)} \\ & = -\frac {225}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}+\frac {30 (15-x \log (5))}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {(15-x \log (5))^2}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^2}{\log (3) (-4+\log (15-x \log (5)))} \]
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Time = 0.86 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05
method | result | size |
norman | \(\frac {x^{2}}{\left (\ln \left (-x \ln \left (5\right )+15\right )-4\right ) \ln \left (3\right )}\) | \(21\) |
risch | \(\frac {x^{2}}{\left (\ln \left (-x \ln \left (5\right )+15\right )-4\right ) \ln \left (3\right )}\) | \(21\) |
parallelrisch | \(\frac {x^{2}}{\left (\ln \left (-x \ln \left (5\right )+15\right )-4\right ) \ln \left (3\right )}\) | \(21\) |
derivativedivides | \(\frac {\frac {225}{\ln \left (-x \ln \left (5\right )+15\right )-4}-\frac {30 \left (-x \ln \left (5\right )+15\right )}{\ln \left (-x \ln \left (5\right )+15\right )-4}+\frac {\left (-x \ln \left (5\right )+15\right )^{2}}{\ln \left (-x \ln \left (5\right )+15\right )-4}}{\ln \left (5\right )^{2} \ln \left (3\right )}\) | \(68\) |
default | \(\frac {\frac {225}{\ln \left (-x \ln \left (5\right )+15\right )-4}-\frac {30 \left (-x \ln \left (5\right )+15\right )}{\ln \left (-x \ln \left (5\right )+15\right )-4}+\frac {\left (-x \ln \left (5\right )+15\right )^{2}}{\ln \left (-x \ln \left (5\right )+15\right )-4}}{\ln \left (5\right )^{2} \ln \left (3\right )}\) | \(68\) |
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^{2}}{\log \left (3\right ) \log \left (-x \log \left (5\right ) + 15\right ) - 4 \, \log \left (3\right )} \]
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Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^{2}}{\log {\left (3 \right )} \log {\left (- x \log {\left (5 \right )} + 15 \right )} - 4 \log {\left (3 \right )}} \]
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Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^{2}}{\log \left (3\right ) \log \left (-x \log \left (5\right ) + 15\right ) - 4 \, \log \left (3\right )} \]
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^{2}}{\log \left (3\right ) \log \left (-x \log \left (5\right ) + 15\right ) - 4 \, \log \left (3\right )} \]
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Time = 0.44 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^2}{\ln \left (3\right )\,\left (\ln \left (15-x\,\ln \left (5\right )\right )-4\right )} \]
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