Integrand size = 226, antiderivative size = 30 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {25}{\left (-2+\frac {1}{3} \left (1+\log \left (\left (1-e^x-\log (20-x)\right )^2\right )\right )\right )^2} \]
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Time = 0.43 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6820, 12, 6818} \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\left (5-\log \left (\left (e^x+\log (20-x)-1\right )^2\right )\right )^2} \]
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Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {900 \left (1+e^x (-20+x)\right )}{(20-x) \left (1-e^x-\log (20-x)\right ) \left (5-\log \left (\left (-1+e^x+\log (20-x)\right )^2\right )\right )^3} \, dx \\ & = 900 \int \frac {1+e^x (-20+x)}{(20-x) \left (1-e^x-\log (20-x)\right ) \left (5-\log \left (\left (-1+e^x+\log (20-x)\right )^2\right )\right )^3} \, dx \\ & = \frac {225}{\left (5-\log \left (\left (-1+e^x+\log (20-x)\right )^2\right )\right )^2} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\left (-5+\log \left (\left (-1+e^x+\log (20-x)\right )^2\right )\right )^2} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(74\) vs. \(2(25)=50\).
Time = 7.52 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.50
method | result | size |
parallelrisch | \(\frac {225}{\ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right )^{2}-10 \ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right )+25}\) | \(75\) |
risch | \(-\frac {900}{{\left (\pi {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )\right )}^{2} \operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )\right ) {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )^{2}\right )}^{2}+\pi {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )^{2}\right )}^{3}+4 i \ln \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )-10 i\right )}^{2}}\) | \(111\) |
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Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (19) = 38\).
Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.40 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\log \left (2 \, {\left (e^{x} - 1\right )} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )^{2} - 10 \, \log \left (2 \, {\left (e^{x} - 1\right )} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right ) + 25} \]
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Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (20) = 40\).
Time = 0.56 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.33 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\log {\left (\left (2 e^{x} - 2\right ) \log {\left (20 - x \right )} + e^{2 x} - 2 e^{x} + \log {\left (20 - x \right )}^{2} + 1 \right )}^{2} - 10 \log {\left (\left (2 e^{x} - 2\right ) \log {\left (20 - x \right )} + e^{2 x} - 2 e^{x} + \log {\left (20 - x \right )}^{2} + 1 \right )} + 25} \]
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Time = 1.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{4 \, \log \left (e^{x} + \log \left (-x + 20\right ) - 1\right )^{2} - 20 \, \log \left (e^{x} + \log \left (-x + 20\right ) - 1\right ) + 25} \]
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Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (19) = 38\).
Time = 1.52 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.80 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\log \left (2 \, e^{x} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} - 2 \, \log \left (-x + 20\right ) + 1\right )^{2} - 10 \, \log \left (2 \, e^{x} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} - 2 \, \log \left (-x + 20\right ) + 1\right ) + 25} \]
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Time = 0.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{{\left (\ln \left ({\ln \left (20-x\right )}^2+\left (2\,{\mathrm {e}}^x-2\right )\,\ln \left (20-x\right )+{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )-5\right )}^2} \]
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